## Pre-Calculus For Dummies, 2nd Edition (2012)

### Part I. Set It Up, Solve It, Graph It

### Chapter 4. Digging Out and Using Roots to Graph Polynomial Functions

*In This Chapter*

Exploring the factoring of quadratic equations

Solving quadratic equations that you can’t factor

Deciphering and counting a polynomial’s roots

Employing solutions to find factors

Plotting polynomials on the coordinate plane

Ever since those bygone days of algebra, variables have been standing in for unknowns in equations. You’re probably very comfortable with using them by now, so you’re ready to move on and find out how to deal with equations that use multiple terms and figure out how to graph them.

When variables and constants start multiplying, the result is called a *monomial, *which means “one term.” Examples of monomials include 3, *x*^{2}, and 4*ab*^{3}*c*^{2}. When you start adding and subtracting distinct monomials, you get *polynomials,* because you create one or more terms. Usually, *monomial* refers to a polynomial with one term only, *binomial* refers to two terms, *trinomial* refers to three, and the word *polynomial* is reserved for four or more. Think of a polynomial as the umbrella under which are binomial and trinomial. Each part of a polynomial that’s added or subtracted is a term; so, for example, the polynomial 2*x* + 3 has two terms: 2*x* and 3.

Part of the official definition of a polynomial is that it can never have a variable in the denominator of a fraction, it can’t have negative exponents, and it can’t have fractional exponents.

In this chapter you go searching for the *solution(s)* of the given polynomial equation — the value(s) that make it true. When the given equation is equal to zero, these solutions are called *roots* or *zeros*. Textbooks and teachers use these words interchangeably because they represent the same idea — where the graph crosses the *x-*axis (a point called the *x-*intercept). We show you how to find the roots of polynomial functions and how to represent the functions with graphs.

**Understanding Degrees and Roots**

The *degree* of a polynomial is closely related to its exponents, and it determines how you work with the polynomial to find the roots. To find the degree of a polynomial, you simply find the degree of each term by adding the exponents of variables (remembering that when no exponent is given, 1 is implied). The greatest of these sums is the degree of the whole polynomial. For example, consider the expression 3*x*^{4}*y*^{6} – 2*x*^{4}*y *– 5*xy* + 2:

The degree of the first term is 4 + 6, or 10.

The degree of the second term is 4 + 1, or 5.

The degree of the third term is 1 + 1, or 2.

The degree of the last term is 0, because it has no variables.

Therefore, this polynomial has a degree of 10.

A *quadratic expression *is a polynomial in which the highest degree is two. One example of a quadratic polynomial is 3*x*^{2} – 10*x *+ 5. The *x*^{2} term in the polynomial is called the *quadratic term* because it’s the one that makes the whole expression quadratic. The number in front of *x*^{2} is called the *leading coefficient *(in the example above it’s the 3). The *x* term is called the *linear term* (–10*x*)*, *and the number by itself is called the *constant* (5)*.*

Without taking calculus, getting a perfectly accurate graph of a polynomial function by plotting points is nearly impossible. However, in pre-calc you can find the roots of a polynomial (if it has any), and you can use those roots as a guide to get an idea of what the graph of that polynomial looks like. You simply plug in an *x *value between the two roots, which are *x-*intercepts, to see if the function is positive or negative in between those roots. For example, you may be asked to graph the equation *y* = 3*x*^{2} – 10*x* + 5. You now know that this equation is a second-degree polynomial, so it has two roots and, therefore, can cross the *x-*axis up to two times (more on why later).

We begin this chapter by looking at solving quadratics because the techniques required to solve them are specific: factoring, completing the square, and using the quadratic formula are excellent methods to solve quadratics; however, they don’t work for polynomials of higher degrees. We then move on to higher-degree polynomials (like *x*^{3} or *x*^{5}, for example) because the steps required to solve them are oftentimes longer and more complicated.

You can solve *any* polynomial equation (including quadratics) using the steps described near the end of this chapter. However, solving quadratics using the techniques specifically reserved for them saves you time and effort.

**Factoring a Polynomial Expression**

Recall that when two or more terms are multiplied to get a product, each term is called a *factor*. You first ran into factors when multiplication was introduced (remember factor trees, prime factorization, and so on?). For example, one set of factors of 24 is 6 and 4 because 6 · 4 = 24.

In mathematics, *factorization* or *factoring* is the breaking apart of a polynomial into a product of other smaller polynomials. If you choose, you can then multiply these factors together to get the original polynomial (which is a great way to check yourself on your factoring skills).

One way of solving a polynomial equation is to factor the polynomial into the product of two or more binomials (two-term expressions). After the polynomial is fully factored, you can use the zero-product property to solve the equation. We discuss this idea later in “Finding the Roots of a Factored Equation.”

You have multiple options to choose from when factoring:

For a polynomial, no matter how many terms it has, always check for a greatest common factor (GCF) first. The *greatest common factor* is the biggest expression that goes into all of the terms. Using the GCF is like doing the distributive property backward (see Chapter 1).

If the expression is a *trinomial* — it has three terms — you can use the FOIL method (for multiplying binomials) backward.

If the expression is a binomial, look for difference of squares, difference of cubes, or sum of cubes.

The following sections show each of these methods in detail.

If a polynomial doesn’t factor, it’s called *prime* because its only factors are 1 and itself. When you have tried all the factoring tricks in your bag (GCF, backward FOIL, difference of squares, and so on), and the quadratic equation doesn’t factor, then you can either complete the square or use the quadratic formula to solve the equation. The choice is yours. You could even potentially choose to *always *use either completing the square or quadratic formula (and skip the factoring) to solve an equation. Factoring can sometimes be quicker, which is why we recommend that you try it first.

Standard form for a quadratic expression (simply a quadratic equation without the equal sign) is the *x*^{2} term followed by the *x* term followed by the constant — in other words, *ax*^{2} + *bx* + *c*. If you’re given a quadratic expression that isn’t in standard form, rewrite it in standard form by putting the degrees in descending order. This step makes factoring easier (and is sometimes even necessary to factor).

**Always the first step: Looking for a GCF**

No matter how many terms a polynomial has, you always want to check for a greatest common factor (GCF) first. If the polynomial has a GCF, factoring the polynomial is much easier because the number of factors of each term is lower (because you factored one or more of them out!).* *If the GCF includes a variable, your job becomes even easier.

If you forget to factor out the GCF, you may also forget to find a solution, and that could mix you up in more ways than one! Without that solution, you could miss a root, and then you could end up with an incorrect graph for your polynomial. And then all your work would be for nothing!

To factor the polynomial 6*x*^{4} – 12*x*^{3} + 4*x*^{2}, for example, follow these steps:

**1.** **Break down every term into prime factors.**

This step expands the original expression to (3 · 2 · *x* · *x* · *x* · *x*) – (2 · 2 · 3 · *x* · *x *· *x*) + (2 · 2 · *x *· *x*).

**2.** **Look for factors that appear in every single term to determine the GCF.**

In this example, you can see one 2 and two *x*’s in every term: (3 · __2 · x · x __·

*x*·

*x*) – (2 ·

__2__· 3 ·

__x____·__·

*x**x*) + (2 ·

__2 ·__)

*x*·*x**.*The GCF here is 2

*x*

^{2}.

**3.** **Factor the GCF out from every term in front of parentheses and group the remnants inside the parentheses.**

You now have 2 · *x* · *x*(3 · *x *· *x* – 2 · 3 · *x* + 2).

**4.** **Multiply each term to simplify.**

The simplified form of the expression you find in Step 3 is 2*x*^{2}(3*x*^{2} – 6*x* + 2).

**5.** **Distribute to make sure the GCF is correct.**

If you multiply the 2*x*^{2} inside the parentheses, you get 6*x*^{4} – 12*x*^{3} + 4*x*^{2}. You can now say with confidence that 2*x*^{2} is the GCF.

**Unwrapping the FOIL method for trinomials**

After you check a polynomial for a GCF (regardless of whether it had one or not), try to factor again. You may find that it is easier to factor after the GCF has been factored out. The polynomial in the last section had two factors: 2*x*^{2} and 3*x*^{2} – 6*x* + 2. The first factor, 2*x*^{2}, is unfactorable because it’s a monomial. However, the second factor may be able to factor again because it’s a trinomial, and if it does, you’ll have two more factors that are both binomials.

Most teachers show the guess-and-check method of factoring, where you write down two sets of parentheses — ( )( ) — and literally plug in guesses for the factors to see if anything works. Maybe your first guess for this example would be (3*x *– 2)(*x* – 1), but if you FOILed it out, you would get 3*x*^{2} – 5*x* + 2, and you’d have to guess again. Fortunately, here the leading coefficient (3) and the constant (2) are both prime numbers, so you don’t have many number combinations to try. The signs in parentheses have to be negative (to produce a positive constant and a negative middle term), so the only other factorization to test is (3*x* – 1) (*x* – 2), which gives you 3*x*^{2} – 7*x* + 2. Hence this particular quadratic is *prime.*

For polynomials with nonprime leading coefficient and constant term, this guess-and-check method can be tedious and frustrating. If you’re in pre-calculus and your teacher is using the guess-and-check method of factoring, which just isn’t working for you, you’ve come to the right section. The following procedure, called the *FOIL method* of factoring (sometimes called the *British Method*), always works for factoring trinomials and is a very helpful tool if you can’t wrap your brain around guess-and-check. When the FOIL method fails, you know for certain the given quadratic is prime.

The FOIL method of factoring calls for you to follow the steps required to FOIL binomials, only backward. Remember that when you FOIL, you multiply the *first, outside, inside,* and *last* terms together. Then you combine any like terms, which usually come from the multiplication of the outside and inside terms.

For example, to factor *x*^{2} + 3*x* – 10, follow these steps:

**1.** **Check for the GCF first.**

The expression *x*^{2} + 3*x* – 10 doesn’t have a GCF when you break it down and look at it, according to the steps in the last section. The breakdown looks like this: (*x* · *x*) + (3 · *x*) – (2 · 5).* *No factors are common to all terms, so the expression has no GCF. You get to move on to the next step.

**2.** **Multiply the quadratic term and the constant term.**

Be careful of the signs when you do this step. In this example, the

quadratic term is 1*x*^{2} and the constant is –10, hence 1*x*^{2} · (–10) = –10*x*^{2}.

**3.** **Write down all the factors of the result in pairs.**

The factors of –10*x*^{2} are –1*x* and 10*x*, 1*x* and –10*x*, –2*x* and 5*x*, and 2*x *and –5*x*.

**4.** **From this list, find the pair that adds to produce the coefficient of the linear term.**

You want the pair whose sum is +3*x.* For this problem, the answer is –2*x* and 5*x* because –2*x* · 5*x* = –10*x*^{2} and –2*x* + 5*x* = 3*x.*

**5.** **Break up the linear term into two terms, using the numbers from Step 4 as the coefficients.**

Written out, you now have *x*^{2} – 2*x* + 5*x* – 10.

Life is easier in the long run if you always arrange the linear term with the smallest coefficient first. That’s why we put the –2*x* in front of the +5*x.*

**6.** **Group the four terms into two sets of two.**

Always put a plus sign between the two sets: (*x*^{2} – 2*x*) + (5*x* – 10).

**7.** **Find the GCF for each set and factor it out.**

What do the first two terms have in common? An *x.* If you factor out the *x,* you have *x*(*x* – 2). Now, look at the second two terms. They share a 5. If you factor out the 5, you have 5(*x* – 2). The polynomial is now written as *x*(*x* – 2) + 5(*x* – 2).

**8.** **Find the GCF of the two new terms.**

As you can see, (*x* – 2) appears in both terms, so it’s a GCF. Factor out the GCF from both terms (it’s always the expression inside the parentheses) to the front and leave the remaining terms inside the parentheses. Thus *x*(*x*– 2) + 5(*x *– 2) becomes (*x* – 2)(*x* + 5). The (*x* + 5) is the leftover from taking away the GCF.

Sometimes the sign has to change in Step 6 in order to correctly factor out the GCF. But if you don’t start off with a plus sign between the two sets, you may lose a negative sign you need to factor all the way. For example, in factoring *x*^{2} – 13*x* + 36, you end up in Step 5 with the following polynomial: *x*^{2} – 9*x* – 4*x* + 36. When you group the terms, you get (*x*^{2} – 9*x*) + (–4*x* + 36). Factor out the *x* in the first set and the 4 in the second set to get *x*(*x* – 9) + 4(–*x* + 9). Notice that the second set is the exact opposite of the first one? In order for you to move to the next step, the sets have to match exactly. To fix this, change the +4 in the middle to –4 and get *x*(*x* – 9) – 4(*x* – 9). Now that they match, you can factor again.

Even when an expression has a leading coefficient besides 1, the FOIL method still works. The monkey wrench comes only if in Step 2 you can’t find any factors that add to give you the linear coefficient. In this case, the answer is prime. For example, in 2*x*^{2} + 13*x* + 4, when you multiply the quadratic term of 2*x*^{2} and the constant of 4, you get 8*x*^{2}. However, no factors of 8*x*^{2} also add to be 13*x*, so 2*x*^{2} + 13*x* + 4 is prime.

**Recognizing and factoring special types of polynomials**

The whole point of factoring is to discover the original polynomial factors that give you an end product. You spend a long time in algebra FOILing polynomials, and factoring just undoes that process. It’s a little like *Jeopardy!*— you know the answer and are looking for the question.

Special cases can occur when FOILing binomials (and also pop back up in factoring); you should recognize them quickly so that you can save time on factoring:

**Perfect squares: **When you FOIL a binomial times itself, the product is called a *perfect square.* For example, (*a* + *b*)^{2} gives you the perfect-square trinomial *a*^{2} + 2*ab* + *b*^{2}.

**Difference of squares: **When you FOIL a binomial and its conjugate, the product is called a *difference of squares.* The product of (*a* – *b*)(*a* + *b*) is *a*^{2} – *b*^{2}. Factoring a difference of squares also requires its own set of steps, which we explain for you in this section.

Two other special types of factoring didn’t come up when you were learning how to FOIL, because they aren’t the product of two binomials:

**Sum of cubes:** One factor is a binomial and the other is a trinomial. (*a*^{3} + *b*^{3}) can be factored to (*a *+ *b*)(*a*^{2} – *ab* + *b*^{2}).

**Difference of cubes:** These expressions factor almost like a sum of cubes, except that some signs are different in the factors: (*a*^{3} – *b*^{3}) = (*a* – *b*)(*a*^{2} + *ab* + *b*^{3}).

No matter what type of problem you face, you should always check for the GCF first; however, none of the following examples has a GCF, so we skip over that step in the directions. In another section, you find out how to factor more than once when you can.

**Seeing double with perfect squares**

Because a perfect-square trinomial is still a trinomial, you follow the steps in the backward FOIL method of factoring (see the previous section). However,* *you must account for one extra step at the very end where you express the answer as a binomial squared.

For example, to factor the polynomial 4*x*^{2} – 12*x *+ 9, follow these steps:

**1.** **Multiply the quadratic term and the constant term.**

The product of the quadratic term 4*x*^{2} and the constant 9 is 36*x*^{2}, so that made your job easy.

**2.** **Write down all the factors of the result in pairs.**

Following are the factors of 36*x*^{2} in pairs:

• 1*x *and 36*x*

• –1*x* and –36*x*

• 2*x* and 18*x*

• –2*x* and –18*x*

• 3*x* and 12*x*

• –3*x* and –12*x*

• 4*x* and 9*x*

• –4*x* and –9*x*

• 6*x* and 6*x*

• –6*x* and –6*x*

If you think ahead to the next step, you can skip writing out the positive factors, because they produce only *x* terms with a positive coefficient.

**3.** **From this list, find the pair that adds to produce the coefficient of the linear term.**

You want to get a sum of –12*x* in this case. The only way to do that is to use –6*x* and –6*x.*

**4.** **Break up the linear term into two terms, using the terms from Step 3.**

You now get 4*x*^{2} – 6*x *– 6*x* + 9.

**5.** **Group the four terms into two sets of two.**

Remember to include the plus sign between the two groups, resulting in (4*x*^{2} – 6*x*) + (–6*x* + 9).

**6.** **Find the GCF for each set and factor it out.**

The GCF of the first two terms is *x,* and the GCF of the next two terms is –8; when you factor them out, you get 2*x*(2*x* – 3) – 3(2*x* – 3).

**7.** **Find the GCF of the two new terms.**

This time the GCF is (2*x* – 3); when you factor it out, you get (2*x* – 3)(2*x* – 3). Aha! That’s a binomial times itself, which means you have one extra step.

**8.** **Express the resulting product as a binomial squared.**

This step is easy: (2*x* – 3)^{2}.

**Working with differences of squares**

You can recognize a *difference of squares* because it’s always a binomial where both terms are perfect squares and a subtraction sign appears between them. It *always* appears as *a*^{2} – *b*^{2}, or (something)^{2} – (something else)^{2}. When you do have a difference of squares on your hands — after checking it for a GCF in both terms — you follow a simple procedure: *a*^{2} – *b*^{2} = (*a* – *b*)(*a* + *b*).

For example, you can factor 25*y*^{4} – 9 with these steps:

**1.** **Rewrite each term as (something)**^{2}**.**

This example becomes (5*y*^{2})^{2} – (3)^{2}, which clearly shows the difference of squares (“difference of” meaning subtraction).

**2.** **Factor the difference of squares ( a)**

^{2}**– (**

*b*)

^{2}**to (**

*a – b*)(*a + b*).Each difference of squares (*a*)^{2} – (*b*)^{2} always factors to (*a* – *b*)(*a* + *b*). This example factors to (5*y*^{2}* *– 3)(5*y*^{2} + 3).

**Breaking down a cubic difference or sum**

After you’ve checked to see if there’s a GCF in the given polynomial and discovered it’s a binomial that isn’t a difference of squares, consider that it may be a sum or difference of cubes.

A *difference of cubes* sounds an awful lot like a difference of squares (see the last section), but it factors quite differently. A difference of cubes always starts off as a binomial with a subtraction sign in between, but it’s written as (something)^{3} – (something else)^{3}. To factor any difference of cubes, you use the formula *a*^{3} – *b*^{3} = (*a* – *b*)(*a*^{2} + *ab* + *b*^{2}).

A *sum of cubes* is always a binomial with a plus sign in between — the only one where that happens: (something)^{3} + (something else)^{3}. When you recognize a sum of cubes *a*^{3} + *b*^{3}, it factors as (*a* + *b*)(*a*^{2} – *ab* + *b*^{2}).

For example, to factor 8*x*^{3} + 27, you first look for the GCF. You find none, so now you use the following steps:

**1.** **Check to see if the expression is a difference of squares.**

You want to consider the possibility because the expression has two terms, but the plus sign between the two terms quickly tells you that it isn’t a difference of squares.

**2.** **Determine if you must use a sum or difference of cubes.**

The plus sign tells you that it may be a sum of cubes, but that clue isn’t foolproof. Time for some trial and error: Try to rewrite the expression as the sum of cubes; if you try (2*x*)^{3} + (3)^{3}, you’ve found a winner.

**3.** **Break down the sum or difference of cubes by using the factoring shortcut.**

Replace *a* with 2*x* and *b* with 3. The formula becomes [(2*x*) + (3)] [(2*x*)^{2} – (2*x*)(3) + (3)^{2}].

**4.** **Simplify the factoring formula.**

This example simplifies to (2*x* + 3)(4*x*^{2} – 6*x* + 9).

**5.** **Check the factored polynomial to see if it will factor again.**

You’re not done factoring until you’re done. Always look at the “leftovers” to see if they’ll factor again. Sometimes the binomial term may factor again as the difference of squares. However, the trinomial factor *never* factors again.

In the previous example, the binomial term 2*x* + 3 is a first-degree binomial (the exponent on the variable is 1) without a GCF, so it won’t factor again. Therefore, (2*x* + 3)(4*x*^{2} – 6*x *+ 9) is your final answer.

**Grouping to factor four or more terms**

When a polynomial has four or more terms, the easiest way to factor it is to use *grouping. *In this method, you look at only two terms at a time to see if any of the previous techniques becomes apparent (you may see a GCF in two terms, or you may recognize a trinomial as a perfect square). In fact, in previous sections when we show you how to break up the linear term in a trinomial into two separate terms (such as *x*^{2} – 13*x* + 36 = *x*^{2} – 9*x* – 4*x* + 36) and then factor out the GCF twice, we’re showing you a grouping tactic. The ways you can factor by using grouping far outnumber that one example, however, so here we show you how to group when the given polynomial *starts off* with four (or more) terms.

Sometimes you can group a polynomial into sets with two terms each to find a GCF in each set. You should try this method first when faced with a polynomial with four or more terms. This type of grouping is the most common method in a pre-calculus text.

For example, you can factor *x*^{3} + *x*^{2} – *x* – 1 by using grouping. Just follow these steps:

**1.** **Break up the polynomial into sets of two.**

You can go with (*x*^{3} + *x*^{2}) + (–*x* – 1). Put the plus sign between the sets, just like when you factor trinomials.

**2.** **Find the GCF of each set and factor it out.**

The square *x*^{2} is the GCF of the first set, and –1 is the GCF of the second set. Factoring out both of them, you get *x*^{2}(*x* + 1) – 1(*x* + 1).

**3.** **Factor again as many times as you can.**

The two terms you’ve created have a GCF of (*x* + 1). When factored out, you get (*x* + 1)(*x*^{2} – 1).

However, *x*^{2} – 1 is a difference of squares and factors again. In the end, you get the following factors after grouping: (*x* + 1)(*x *+ 1)(*x* – 1), or (*x* + 1)^{2}(*x* – 1).

If the previous method doesn’t work, you may have to group the polynomial some other way. Of course, after all your effort, the polynomial may end up being prime, which is okay.

For example, look at the polynomial *x*^{2} – 4*xy *+ 4*y*^{2} – 16. You can group it into sets of two, and it becomes *x*(*x* – 4*y*) + 4(*y*^{2} – 4). This expression, however, doesn’t factor again. Bells and whistles should go off inside your head at this point, telling you to look again at the original. You must try grouping it in some other way. In this case, if you look at the first three terms, you’ll discover a perfect-square trinomial, which factors to (*x* – 2*y*)^{2} – 16. Now you have a difference of squares, which factors again to [(*x* – 2*y*) – 4][(*x* – 2*y*) + 4].

**Finding the Roots of a Factored Equation**

Sometimes after you’ve factored, the two factors can be factorable again, in which case you should continue factoring. In other cases, they may unfactorable, in which case you can solve them only by using the quadratic formula. For example, 6*x*^{4} – 12*x*^{3} + 4*x*^{2} = 0 factors to 2*x*^{2}(3*x*^{2} – 6*x* + 2) = 0. The first term, 2*x*^{2} = 0, is solvable using algebra, but the second factor, 3*x*^{2} – 6*x* + 2 = 0, is unfactorable and requires the quadratic formula (see the following section).

After you factor a polynomial into its different pieces, you can set each piece equal to zero to solve for the roots with the zero-product property. The *zero-product property* says that if several factors are multiplying to give you zero, at least one of them has to be zero. Your job is to find all the values of *x* that make the polynomial equal to zero. This task is much easier if the polynomial is factored because you can set each factor equal to zero and solve for *x.*

Factoring *x*^{2} + 3*x* – 10 = 0 gives you (*x* + 5)(*x* – 2). Moving forward is easy because each factor is linear (first degree). The term *x* + 5 = 0 gives you one solution, *x* = –5, and *x* – 2 = 0 gives you the other solution, *x* = 2.

These solutions each become an *x*-intercept on the graph of the polynomial (see the section “Graphing Polynomials”).

**Cracking a Quadratic Equation When It Won’t Factor**

When asked to solve a quadratic equation that* *you just can’t seem to factor (or that just doesn’t factor), you have to employ other ways of solving the equation. The inability to factor means that the equation has solutions that you can’t find by using normal techniques. Perhaps they involve square roots of non-perfect squares; they can even be complex numbers involving imaginary numbers (see Chapter 11).

One such method is to use the *quadratic formula,* which is the formula used to solve for the variable in a quadratic equation in standard form. Another is to *complete the square, *which means to manipulate an expression to create a perfect-square trinomial that you can easily factor*.* The following sections present these methods in detail.

**Using the quadratic formula**

When a quadratic equation just won’t factor, remember your old friend from algebra, the quadratic formula, in order to solve. Given a quadratic equation in standard form *ax*^{2} + *bx* + *c* = 0,

Before you apply the formula, rewrite the equation in standard form (if it isn’t already) and figure out the *a, b,* and *c* values.

For example, to solve *x*^{2} – 3*x* + 1 = 0, you first say that *a* = 1, *b* = –3, and *c* = 1. The *a, b,* and *c* terms simply plug into the formula to give you the values for *x*:

Simplify this formula one time to get

Simplify further to get your final answer, which is two *x* values (the *x*-intercepts):

**Completing the square**

Completing the square comes in handy when you’re asked to solve an unfactorable quadratic equation and when you need to graph conic sections (circles, ellipses, parabolas, and hyperbolas), which we explain in Chapter 12. For now, we recommend that you only find the roots of a quadratic using this technique when you’re specifically asked to do so, because factoring a quadratic and using the quadratic formula work just as well (if not better). Those methods are less complicated than completing the square (a pain in the you-know-where!).

Say your instructor calls for you to complete the square. Follow these steps to solve the equation 2*x*^{2} – 4*x* + 5 = 0 by completing the square:

**1.** **Divide every term by the leading coefficient so that a = 1. **If the equation already has a plain

*x*

^{2}term, you can skip to Step 2.

Be prepared to deal with fractions in this step. Dividing each term by 2, the equation now becomes

**2.** **Subtract the constant term from both sides of the equation to get only terms with the variable on the left side of the equation.**

You can subtract from both sides to get

**3.** **Now to complete the square: Divide the linear coefficient by 2 and write it below the problem for later, square this answer, and then add that value to both sides so that both sides remain equal.**

Divide –2 by 2 to get –1. Square this answer to get 1, and add it to both sides:

**4.** **Simplify the equation.**

The equation becomes

**5.** **Factor the newly created quadratic equation.**

The new equation should be a perfect-square trinomial.

**6.** **Get rid of the square exponent by taking the square root of both sides. Remember that the positive and negative roots could both be squared to get the answer!**

This step gives you

**7.** **Simplify any square roots if possible.**

The example equation doesn’t simplify, but the fraction is imaginary (see Chapter 11) and the denominator needs to be rationalized (see Chapter 2). Do the work to get

**8.** **Solve for the variable by isolating it.**

You add 1 to both sides to get

** Note:** You may be asked to express your answer as one fraction; in this case, find the common denominator and add to get

**Solving Unfactorable Polynomials with a Degree Higher than Two**

By now you’re a professional at solving second-degree polynomial equations (quadratics), and you have various tools at your disposal for solving these types of problems. You may have noticed while solving quadratics that a quadratic equation always has two solutions. Note that sometimes both solutions are the same (as in perfect-square trinomials). Even though you get the same solution twice, they still count as two solutions (how many times a solution is a root is called the *multiplicity* of the solution).

When the polynomial degree is higher than two and the polynomial won’t factor using any of the techniques that we discuss earlier in this chapter, finding the roots gets harder and harder. For example, you may be asked to solve a cubic polynomial that is *not* a sum or difference of cubes or* *any polynomial that is fourth degree or greater that can’t be factored by grouping. The higher the degree, the more roots exist, and the harder it is to find them. To find the roots, many different scenarios can guide you in the right direction. You can make very educated guesses about how many roots a polynomial has, as well as how many of them are positive or negative and how many are real or imaginary.

**Counting a polynomial’s total roots**

Usually, the first step you take before solving a polynomial is to find its *degree,* which helps you determine the number of solutions you’ll find later. The degree tells you the maximum number of possible roots.

When you’re being asked to solve a polynomial, finding its degree is even easier because only one variable is in any term. Therefore, the highest exponent is always the highest term when asked to solve. For example, *f*(*x*) = 2*x*^{4} – 9*x*^{3} – 21*x*^{2} + 88*x* + 48 is a fourth-degree polynomial with up to, but no more than, four total possible solutions.

In the next two sections, we help you figure out how many of those roots may be real roots and how many may be imaginary.

**Tallying the real roots: Descartes’s rule of signs**

The terms *solutions/zeros/roots* are synonymous because they all represent where the graph of the polynomial intersects the *x-*axis. The roots that are found when the graph meets with the *x-*axis are called *real* *roots; *you can see them and deal with them as real numbers in the real world. Also, because they cross the *x-*axis, some roots may be *negative roots* (which means they intersect the negative *x-*axis), and some may be *positive roots *(which intersect the positive *x-*axis).

If you know how many total roots you have (see the last section), you can use a pretty cool theorem called *Descartes’s rule of signs* to count how many roots are real numbers (both positive *and* negative) and how many are imaginary (see Chapter 11). You see, the same man who pretty much invented graphing, Descartes, also came up with a way to figure out how many times a polynomial crosses the *x-*axis — in other words, how many real roots it has. All you have to be able to do is count!

Here’s how Descartes’s rule of signs can give you the numbers of possible real roots, both positive and negative:

**Positive real roots:** For the number of positive real roots, look at the polynomial, written in descending order, and count how many times the sign changes from term to term. This value represents the maximum number of positive roots in the polynomial. For example, in the polynomial *f*(*x*) = 2*x*^{4} – 9*x*^{3} – 21*x*^{2} + 88*x* + 48, you see two changes in sign (don’t forget to include the sign of the first term!) — from the first term to the second and from the third term to the fourth. That means this equation can have up to two positive solutions.

Descartes’s rule of signs says the number of positive roots is equal to changes in sign of *f*(*x*), or is less than that by an even number (so you keep subtracting 2 until you get either 1 or 0). Therefore, the previous *f*(*x*) may have 2 or 0 positive roots.

**Negative real roots: **For the number of negative real roots, find *f*(–*x*) and count again. Because negative numbers raised to even powers are positive and negative numbers raised to odd powers are negative, this change affects only terms with odd powers. This step is the same as changing each term with an odd degree to its opposite sign and counting the sign changes again, which gives you the maximum number of negative roots. The example equation becomes *f*(–*x*) = 2*x*^{4} + 9*x*^{3} – 21*x*^{2} – 88*x *+ 48, which changes signs twice. There can be, at most, two negative roots.* *However, similar to the rule for positive roots, the number of negative roots is equal to the changes in sign for *f*(–*x*), or must be less than that by an even number. Therefore, this example can have either 2 or 0 negative roots.

**Accounting for imaginary roots: The fundamental theorem of algebra**

*Imaginary roots* appear in a quadratic equation when the discriminant of the quadratic equation — the part under the square root sign (*b*^{2} – 4*ac*) — is negative. If this value is negative, you can’t actually take the square root, and the answers are not real. In other words, there is no solution; therefore, the graph won’t cross the *x-*axis.

Using the quadratic formula always gives you two solutions, because the ± sign means you’re both adding and subtracting and getting two completely different answers. When the number underneath the square-root sign in the quadratic formula is negative, the answers are called *complex conjugates.* One is *r* + *si* and the other is *r* – *si.* These numbers have both real (the *r*) and imaginary (the *si*) parts.

*The fundamental theorem of algebra* says that every polynomial function has at least one root in the complex number system. This concept is one you may remember from Algebra II. (For reference, flip to Chapter 11 to read the parts on imaginary and complex numbers first.)

The highest degree of a polynomial gives you the highest possible number of *complex *roots for the polynomial. Between this fact and Descartes’s rule of signs, you can figure out how many complex roots a polynomial has. Pair up every possible number of positive real roots with every possible number of negative real roots (see the previous section); the remaining number of roots for each situation represents the number of complex roots.

Continuing with the example *f*(*x*) = 2*x*^{4} – 9*x*^{3} – 21*x*^{2} + 88*x* + 48 from the previous section, the polynomial has a degree of 4, with two or zero positive real roots, and two or zero negative real roots. Pair up the possible situations:

Two positive and two negative real roots, with zero complex roots

Two positive and zero negative real roots, with two complex roots

Zero positive and two negative real roots, with two complex roots

Zero positive and zero negative real roots, with four complex roots

The following chart makes the information easier to picture:

Complex numbers are written in the form *r* + *si* and have both a real and an imaginary part, which is why every polynomial has at least one root in the complex number system (see Chapter 11). Real and imaginary numbers are both included in the complex number system. Real numbers have no imaginary part, and pure imaginary numbers have no real part. For example, if *x* = 7 is one root of the polynomial, this root is considered both real and complex because it can be rewritten as *x* = 7 + 0*i* (the imaginary part is 0).

The fundamental theorem of algebra gives the total number of complex roots (say there are seven); Descartes’s rule of signs tells you how many possible real roots exist and how many of them are positive and negative (say* *there are, at most, two positive roots but only one negative root). Assume you’ve found them all, using the techniques we discuss throughout this section; they’re *x* = 1, *x* = 7, and *x* = –2. These roots are real, but they’re also complex because they can all be rewritten, as in the previous example.

The first two columns in the previous chart find the real roots and classify them as positive or negative. The third column is actually finding, specifically, the non-real numbers: complex numbers with non-zero imaginary parts.

**Guessing and checking the real roots**

After you work through the preceding section, you can determine exactly how many roots (and what type of roots) exist. Now, the rational root theorem is another method you can use to narrow down the search for roots of polynomials. Descartes’s rule of signs only narrows down the real roots into positive and negative. The rational root theorem says that some real roots are rational (they can be expressed as a fraction). It also helps you create a list of the *possible *rational roots of any polynomial.

The problem? Not every root is rational, because some are irrational. A polynomial may even have *only* irrational roots. But this theorem is always the next place to start in your search for roots; it will at least give you a diving-off point. Besides, the problems you’re presented with in pre-calc more than likely have at least one rational root, so the information in this section greatly improves your odds of finding more!

Follow these general steps to ensure that you find every root (we cover these steps in detail later in this section):

**1.** **Use the rational root theorem to list all possible rational roots.**

We describe this theorem next.

**2.** **Pick one root from the list in Step 1 and use long division or synthetic division to find out if it is, in fact, a root.**

• If the root doesn’t work, try another guess.

• If the root works, proceed to Step 3.

**3.** **Using the depressed polynomial (the one you get after doing the synthetic division in Step 2), test the root that worked to see if it works again.**

• If it works, repeat Step 3 *again.*

• If it doesn’t work, return to Step 2 and try a different root from the list in Step 1.

**4.** **List all the roots you find that work; you should have as many roots as the degree of the polynomial.**

Don’t stop until you’ve found them all. And keep in mind that some will be real and some will be imaginary.

**Listing the possibilities with the rational root theorem**

The rational root theorem says that if you take all the factors of the constant term in a polynomial and divide by all the factors of the leading coefficient, you produce a list of all the possible rational roots of the polynomial. However, keep in mind that you’re finding only the *rational* ones, and sometimes the roots of a polynomial are irrational. Some of your roots can also be imaginary, but save those until the end of your search.

For example, consider the equation *f*(*x*) = 2*x*^{4} – 9*x*^{3} – 21*x*^{2} + 88*x* + 48. The constant term is 48, and its factors are as follows:

±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, ±48

The leading coefficient is 2, and its factors are as follows:

±1, ±2

So the list of possible real roots includes the following:

±1/1, ±2/1, ±3/1, ±4/1, ±6/1, ±8/1, ±12/1, ±16/1, ±24/1, ±48/1, ±1/2, ±2/2, ±3/2, ±4/2, ±6/2, ±8/2, ±12/2, ±16/2, ±24/2, ±48/2

Thankfully, these roots all simplify to ±1/2, ±1, ±3/2, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

**Testing roots by dividing polynomials**

Dividing polynomials follows the same algorithm as long division with real numbers. The polynomial you’re dividing by is called the *divisor.* The polynomial being divided is called the *dividend.* The answer is called the *quotient, *and the leftover polynomial is called the *remainder.*

One way, other than synthetic division, that you can test possible roots from the rational root theorem is to use long division of polynomials and hope that when you divide you get a remainder of 0. For example, when you have your list of possible rational roots (as found in the last section), pick one and assume that it’s a root. If *x* = *c* is a root*,* then *x* – c is a factor. So if you pick *x* = 2 as your guess for the root, *x* – 2 should be a factor. We explain in this section how to use long division to test if *x* – 2 is actually a factor and, therefore, *x* = 2 is a root.

Dividing polynomials to get a specific answer isn’t something you do every day, but the idea of a function or expression that’s written as the quotient of two polynomials is important for pre-calculus. If you divide a polynomial by another and get a remainder of 0, the divisor is a factor, which in turn gives a root. The following sections review two methods of checking your real roots: long division and synthetic division.

In math lingo, the division algorithm states the following: If *f*(*x*) and *d*(*x*) are polynomials such that *d*(*x*) isn’t equal to 0, and the degree of *d*(*x*) isn’t larger than the degree of *f*(*x*), there are unique polynomials *q*(*x*) and *r*(*x*)* *such that *f*(*x*) = *d*(*x*) · *q*(*x*) + *r*(*x*)*.* In plain English, the dividend equals the divisor times the quotient plus the remainder. You can always check your results by remembering this information.

__Long division__

You can use long division to find out if your possible rational roots are actual roots or not. We don’t recommend using this approach, but you can do it. Instead, we suggest you use synthetic division, which we cover later. However, in class you may be asked to do long division. We show you how in the following steps and try to figure out a root at the same time.

Remember the mnemonic device* *__D__irty __M__onkeys __S__mell __B__ad when doing long division to check your roots. Make sure all terms in the polynomial are listed in descending order and that every degree is represented. In other words, if *x*^{2} is missing, put in a placeholder of 0*x*^{2} and then do the division. (This step is just to make the division process easier.)

To divide two polynomials, follow these steps:

**1.** __D__ivide.

Divide the leading term of the dividend by the leading term of the divisor. Write this quotient directly above the term you just divided into.

**2.** __M__ultiply.

Multiply the quotient term from Step 1 by the entire divisor. Write this polynomial under the dividend so that like terms are lined up.

**3.** __S__ubtract.

Subtract the whole line you just wrote from the dividend.

You can change all the signs and add if it makes you feel more comfortable. This way, you won’t forget signs.

**4.** __B__ring down the next term.

Do exactly what this says; bring down the next term in the dividend.

**5.** **Repeat Steps 1–4 over and over until the remainder polynomial has a degree that’s less than the dividend’s.**

The following list explains how to divide 2*x*^{4} – 9*x*^{3} – 21*x*^{2} + 88*x *+ 48 by *x* – 2. Each step corresponds with the numbered step in the illustration in Figure 4-1. (Note that in the earlier section on Descartes’s rule of signs, you find that this particular example may have positive roots, so it’s efficient to try a positive number here. If Descartes’s rule of signs had said that no positive roots existed, you wouldn’t test any positives!)

**1.** ** Divide:** What do you have to multiply

*x*in the divisor by to make it become 2

*x*

^{4}in the dividend? The quotient, 2

*x*

^{3}, goes above the 2

*x*

^{4}term.

**2.** ** Multiply:** Multiply this quotient by the divisor and write it under the dividend.

**3.** ** Subtract:** Subtract this line from the dividend: (2

*x*

^{4}– 9

*x*

^{3}) – (2

*x*

^{4}– 4

*x*

^{3}) = –5

*x*

^{3}. If you’ve done the job right, the subtraction of the first terms always produces 0.

**4.** ** Bring down:** Bring down the other terms of the dividend.

**5.** ** Divide:** What do you have to multiply

*x*by to make it –5

*x*

^{3}? Put the answer, –5

*x*

^{2}, above the –21

*x*

^{2}.

**6.** ** Multiply:** Multiply the –5

*x*

^{2}times the

*x*– 2 to get –5

*x*

^{3}+ 10

*x*

^{2}. Write it under the remainder with the degrees lined up.

**7.** ** Subtract:** You now have (–5

*x*

^{3}– 21

*x*

^{2}) – (–5

*x*

^{3}+ 10

*x*

^{2}) = –31

*x*

^{2}.

**8.** ** Bring down:** The +88

*x*takes its place.

**9.** ** Divide:** What to multiply by to make

*x*become –31

*x*

^{2}? The quotient –31

*x*goes above –21

*x*

^{2}.

**10.** ** Multiply:** The value –31

*x*times (

*x*– 2) is –31

*x*

^{2}+ 62

*x;*write it under the remainder.

**11.** ** Subtract:** You now have (–31

*x*

^{2}+ 88

*x*) – (–31

*x*

^{2}+ 62

*x*), which is 26

*x.*

**12.** ** Bring down:** The +48 comes down.

**13.** ** Divide:** The term 26

*x*divided by

*x*is 26. This answer goes on top.

**14.** ** Multiply:** The constant 26 multiplied by (

*x*– 2) is 26

*x*– 52.

**15.** ** Subtract:** You subtract (26

*x*+ 48) – (26

*x*– 52) to get 100.

**16.** **Stop:** The remainder 100 has a degree that’s less than the divisor of *x* – 2.

Wow . . . now you know why they call it *long* division. You went through all that to find out that *x* – 2 isn’t a factor of the polynomial, which means that *x* = 2 isn’t a root.

If you divide by *c* and the remainder is 0, then the linear expression (*x* – *c*) is a factor and that *c* is a root. A remainder other than 0 implies that (*x* – *c*) isn’t a factor and that *c* isn’t a root.

**Figure 4-1:** The process of long division of polynomials.

__Synthetic division__

Want some good news? A shortcut exists for long division of polynomials, and that shortcut is synthetic division. It’s a special case of division when the divisor is a linear factor the form *x* + *c,* where *c* is a constant.

The bad news, however, is that the shortcut only works if the divisor (*x* + *c*) is a first-degree binomial with a leading coefficient of 1 (you can always make it 1 by dividing everything by the leading coefficient first). The *great*news — yep, more news — is that you can always use synthetic division to figure out if a possible root is actually a root.

Here are the general steps for synthetic division:

**1. Make sure the polynomial is written in descending order.**

The term with the highest exponent comes first.

**2. Write down the coefficients and the constant of the polynomial from left to right, filling in a zero if any degrees are missing; place the root you’re testing outside the synthetic division sign.**

The division sign looks like the left and bottom sides of a rectangle. Leave room below the coefficients to write another row of numbers.

**3. Drop down the first coefficient below the division sign.**

**4. Multiply the root you’re testing by the number you just dropped down and write the answer below the next coefficient.**

**5. Add the coefficient and product from Step 4 and put the answer below the line.**

**6. Multiply the root you’re testing by the answer from Step 5 and put the product below the next coefficient.**

**7. Continue multiplying and adding until you use the last number inside the synthetic division sign.**

If you get a remainder, the number you tested isn’t a root.

If the answer is 0, congratulations! You’ve found a root. The numbers below the synthetic division sign are the coefficients of the quotient polynomial. The degree of this polynomial is one less than the original (the dividend), so the exponent on the first *x* term should be one less than what you started with.

In the previous example, you eliminated *x* = 2 by using long division, so you know not to start there. We chose to do synthetic division in Figure 4-2 for *x* = 4 to show you how it works.

**Figure 4-2:** The synthetic division shortcut when testing possible roots.

The 4 on the outside in Figure 4-2 is the root you’re testing. The numbers on the inside are the coefficients of the polynomial. Here’s the synthetic process, step by step:

1. The 2 below the line just drops down from the line above.

2. Multiply 4 with 2 to get 8 and write that under the next term, –9.

3. Add –9 + 8 to get –1.

4. Multiply 4 with –1 to get –4, and write that under the –21.

5. Add –21 + –4 to get –25.

6. Multiply 4 with –25 to get –100, and write that under 88.

7. Add 88 to –100 to get –12.

8. Multiply 4 with –12 to get –48, and write that under 48.

9. Add 48 to –48 to get 0.

All you do is multiply and add, which is why synthetic division is the shortcut. The last number, 0, is your remainder. Because you get a remainder of 0, *x* = 4 is a root.

The other numbers are the coefficients of the quotient, in order from the greatest degree to the least; however, your answer is always one degree lower than the original. So the quotient in the previous example is 2*x*^{3} – *x*^{2} – 25*x* – 12.

Whenever a root works, you should always automatically test it again in the answer quotient to see if it’s a double root, using the same process. A *double root* occurs when a factor has a multiplicity of two. A double root is one example of multiplicity (as we describe earlier in the “Accounting for imaginary roots: The fundamental theorem of algebra” section). We test *x* = 4 again in Figure 4-3.

**Figure 4-3:**Testing an answer root again, just in case it’s a double root.

Whaddya know? You get a remainder of 0 again, so *x* = 4 is a double root. (In math terms, you say that *x* = 4 is a root with *multiplicity of two.*) You have to check it again, though, to see if it has a higher multiplicity. When you synthetically divide *x* = 4 one more time, it doesn’t work. Figure 4-4 illustrates this failure. Because the remainder isn’t 0, *x* = 4 isn’t a root again.

**Figure 4-4:**Testing the root again shows that it’s only a double root as far as the multiplicity goes.

Always work off the newest quotient when using synthetic division. This way, the degree gets lower and lower until you end up with a quadratic expression. At that point, you can solve the quadratic by using any of the techniques we talk about earlier in this chapter: factoring, completing the square, or the quadratic formula. (Some math teachers require you to use the quadratic formula; what else is the formula for?)

Before you tested *x* = 4 for a final time, the polynomial (called a *depressed polynomial*) was down to a quadratic: 2*x*^{2} + 7*x* + 3. If you factor this expression, you get (2*x* + 1)(*x* + 3). This gives you two more roots of –1/2 and –3. To sum it all up, you’ve found *x* = 4 (multiplicity two), *x* = –1/2, and *x* = –3. You found four complex roots — two of them are negative real numbers, and two of them are positive real numbers.

The *remainder theorem* says that the remainder you get when you* *divide a polynomial by a binomial is the same as the result you get from plugging the number into the polynomial. For example, when you used long division to divide by *x* – 2, you were testing to see if *x* = 2 is a root. You could’ve used synthetic division to do this, because you still get a remainder of 100. And if you plug 2 into *f*(*x*) = 2*x*^{4} – 9*x*^{3} – 21*x*^{2} + 88*x* + 48, you also get 100.

For really hard polynomials, doing the synthetic division to determine the roots is much easier than substituting the number. For example, if you try to plug 8 into the previous polynomial, you have to figure out first what 2(8)^{4} – 9(8)^{3} – 21(8)^{2} + 88(8) + 48 is. That process only leads to bigger (and uglier) numbers, whereas in the synthetic division, all you do is multiply and add — no more exponents!

**Put It in Reverse: Using Solutions to Find Factors**

The *factor theorem* states that you can go back and forth between the roots of a polynomial and the factors of a polynomial. In other words, if you know one, you know the other. At times, your teacher or your textbook may ask you to factor a polynomial with a degree higher than two. If you can find its roots, you can find its factors. We show you how in this section.

In symbols, the factor theorem states that if *x* – *c* is a factor of the polynomial *f*(*x*),* *then *f*(*c*) = 0. The variable *c* is a zero or a root or a solution — whatever you want to call it (the terms all mean the same thing).

In the previous sections of this chapter, you employ many different techniques to find the roots of the polynomial *f*(*x*) = 2*x*^{4} – 9*x*^{3} – 21*x*^{2} + 88*x* + 48. You find that they are *x* = –1/2, *x* = –3, and *x* = 4 (multiplicity two). How do you use those roots to find the factors of the polynomial?

The factor theorem states that if *x* = *c* is a root, (*x* – *c*) is a factor. For example, look at the following roots:

If *x* = –1/2, (*x* – (–1/2)) is your factor, which is the same thing as (*x* + 1/2).

If *x* = –3 is a root, (*x* – (–3)) is a factor, which is also (*x* + 3).

If *x* = 4 is a root, (*x* – 4) is a factor with multiplicity two.

You can now factor *f*(*x*) = 2*x*^{4} – 9*x*^{3} – 21*x*^{2} + 88*x* + 48 to get *f*(*x*) = 2(*x* + 1/2) (*x* + 3)(*x* – 4)^{2}.

**Graphing Polynomials**

The hard graphing work is over after you find the zeros of a polynomial function (using the techniques we present earlier in this chapter). Finding the zeros is very important to graphing the polynomial, because they give you a general template for what your graph should look like. Remember that zeros are *x-*intercepts, and knowing where the graph crosses the *x-*axis is half the battle. The other half is knowing what the graph does in between these points. This section shows you how to figure that out.

If you’re lucky enough to own a graphing calculator *and *have a teacher who allows you to use it, you can enter any quadratic equation into the calculator’s graphing utility and graph the equation. The calculator will not only identify the zeros but also tell you the maximum and minimum values of the graph so that you can draw the best possible representation.

**When all the roots are real numbers**

We use many different techniques in this chapter to find the zeros for the example polynomial *f*(*x*) = 2*x*^{4} – 9*x*^{3} – 21*x*^{2} + 88*x* + 48. The time has come to put this work to use to graph the polynomial. Follow these steps to start graphing like a pro:

**1.** **Plot the critical points on the coordinate plane.**

Mark the zeros that you found previously: *x* = –3, *x* = –1/2, and *x* = 4.

Now plot the *y*-intercept of the polynomial. The *y*-intercept is *always* the constant term of the polynomial — in this case, *y* = 48. If no constant term is written, the *y*-intercept is 0.

**2.** **Determine which way the ends of the graph point.**

You can use a handy test called the *leading coefficient test,* which helps you figure out how the polynomial begins and ends. The degree and leading coefficient of a polynomial always explain the end behavior of its graph (see the section “Understanding Degrees and Roots” for more on finding degree):

• If the degree of the polynomial is even and the leading coefficient is positive, both ends of the graph point up.

• If the degree is even and the leading coefficient is negative, both ends of the graph point down.

• If the degree is odd and the leading coefficient is positive, the left side of the graph points down and the right side points up.

• If the degree is odd and the leading coefficient is negative, the left side of the graph points up and the right side points down.

Figure 4-5 displays this concept in correct mathematical terms.

The function *f*(*x*) = 2*x*^{4} – 9*x*^{3} – 21*x*^{2} + 88*x* + 48 is even in degree and has a positive leading coefficient, so both ends of its graph point up (they go to positive infinity).

**3. Figure out what happens between the critical points by picking any value to the left and right of each intercept and plugging it into the function.**

You can either simplify each one or just figure out whether the end result is positive or negative. For now, you don’t really care about the exact look of the graph. (In calculus, you learn how to find additional values that lead to the most accurate graph you can get.)

A graphing calculator gives a very accurate picture of the graph. Calculus allows you to find the relative max and min exactly, using an algebraic process, but you can easily use the calculator to find them. You can use your graphing calculator to check your work and make sure the graph you’ve created looks like the one the calculator gives you.

Using the zeros for the function, set up a table to help you figure out whether the graph is above or below the *x*-axis between the zeros. See Table 4-1 for our table example.

**Figure 4-5:**Illustrating the leading- coefficient test.

The first interval, (–∞, 3), and the last interval, (4, ∞), both confirm the leading coefficient test from Step 2 — this graph points up (to positive infinity) in both directions.

**4.** **Plot the graph.**

Now that you know where the graph crosses the *x-*axis, how the graph begins and ends, and whether the graph is positive (above the *x-*axis) or negative (below the *x-*axis), you can sketch out the graph of the function. Typically, in pre-calc, this information is all you want or need when graphing. Calculus does show you how to get several other critical points that create an even better graph. If you want, you can always pick more points in the intervals and graph them to get a better idea of what the graph looks like. Figure 4-6 shows the completed graph.

Did you notice that the double root (with multiplicity two) causes the graph to “bounce” on the *x-*axis instead of actually crossing it? This is true for any root with even multiplicity. For any polynomial, if the root has an odd multiplicity at root *c,* the graph of the function crosses the *x-*axis at *x* = *c.* If the root has an even multiplicity at root *c,* the graph meets but doesn’t cross the *x-*axis at *x* = *c.*

**Figure 4-6:**Graphing the polynomial* f*(*x*) = 2*x*^{4} – 9*x*^{3} – 21*x*^{2} + 88*x* + 48.

**When roots are imaginary numbers: Combining all techniques**

In pre-calc and in calculus, certain polynomial functions have non-real roots in addition to real roots (and some of the more complicated functions have *all* imaginary roots). When you must find both, start off by finding the real roots, using all the techniques we describe earlier in this chapter (such as synthetic division). Then you’re left with a depressed quadratic polynomial to solve that’s unsolvable using real number answers. No fear! You just have to use the quadratic formula, through which you’ll end up with a negative number under the square root sign. Therefore, you express the answer as a complex number (for more, see Chapter 11).

For instance, the polynomial *g*(*x*) = *x*^{4} + *x*^{3} – 3*x*^{2} + 7*x* – 6 has non-real roots. Follow these steps to find *all* the roots for this (or any) polynomial; each step involves a major section of this chapter:

**1.** **Classify the real roots as positive and negative by using Descartes’s rule of signs.**

Three changes of sign in the *g*(*x*) function reveals you could have three or one positive real root. One change in sign in the *g*(–*x*) function reveals that you have one negative real root.

**2.** **Find how many roots are possibly imaginary by using the fundamental theorem of algebra.**

The theorem reveals that, in this case, up to four complex roots exist. Combining this fact with Descartes’s rule of signs gives you several possibilities:

• One real positive root and one real negative root means that two roots aren’t real.

• Three real positive roots and one real negative root means that all roots are real.

**3.** **List the possible rational roots, using the rational root theorem.**

The possible rational roots include ±1, ±2, ±3, and ±6.

**4.** **Determine the rational roots (if any), using synthetic division.**

Utilizing the rules of synthetic division, you find that *x* = 1 is a root and that *x* = –3 is another root. These roots are the only real ones.

**5.** **Use the quadratic formula to solve the depressed polynomial.**

Having found all the real roots of the polynomial, you’re left with the depressed polynomial *x*^{2} – *x* + 2. Because this expression is quadratic, you can use the quadratic formula to solve for the last two roots. In this case, you get

**6.** **Graph the results.**

The leading coefficient test (see the previous section) reveals that the graph points up in both directions. The intervals include the following:

• (–∞, –3) is positive.

• (–3, 1) is negative.

• (1, ∞) is positive.

Figure 4-7 shows the graph of this function.

**Figure 4-7:**Graphing the polynomial* g*(*x*) = *x*^{4} + *x*^{3} – 3*x*^{2} + 7*x* – 6.