## High School Algebra I Unlocked (2016)

### Chapter 3. Polynomial Expressions

### Lesson 3.3. Multiplying Polynomial Expressions

We’ve conquered the addition and subtraction of polynomials, so now let’s take a look at how we multiply polynomials together.

Let’s consider the most basic type of polynomial multiplication: the multiplication of two monomials.

**EXAMPLE **

**4 x^{2} × 2x =**

Here, you are going to use the rules of exponents to combine these terms. When terms with the same base are multiplied together, their exponents are added together:

4*x*^{2} × 2*x* = 8*x*^{3}

But what happens if we are multiplying together a monomial and a binomial? The next example walks you through this.

**EXAMPLE **

**Simplify −6 x(4x^{2} + 2x).**

To simplify this expression, you need to distribute the −6*x* to each term in within the parentheses.

−6*x*(4*x*^{2} + 2*x*) =

(−6*x*)(4*x*^{2}) + (−6*x*)(2*x*) =

−24*x*^{3} + (−12*x*^{2}) =

−24*x*^{3} − 12*x*^{2}

By distributing the value of −6*x* to each term of the polynomial 4*x*^{2} + 2*x*, you can eliminate the parentheses and simplify the expression to −24*x*^{3} − 12*x*^{2}.

But wait, there’s more! You may also encounter problems where you have to multiply a binomial by a binomial, or a polynomial by a polynomial, as in the next example.

**EXAMPLE **

**Simplify ( x + 3)(x − 4).**

One way to simplify (*x* + 3)(*x* − 4) is by distribution. Here, however, you will need to distribute twice, multiplying both terms in the first polynomial to each term in the second polynomial. Let’s see how this is done, step by step.

Here, use the same

method we used

in __Example 11__: the

distribution method.

First, distribute each term of the first polynomial to both terms in the second polynomial:

(*x* + 3)(*x* − 4) =

(*x*)(*x* − 4) + (3)(*x* − 4)

Now you have two groups of terms, each that multiply a monomial by a binomial. Tackle each group of terms, distributing each monomial term to the terms in the parentheses:

(*x*)(*x* − 4) + (3)(*x* − 4) =

(*x*)(*x*) − (*x*)(4) + (3)(*x*) − (3)(4)

Then multiply the groups of terms together:

(*x*)(*x*) − (*x*)(4) + (3)(*x*) − (3)(4) =

*x*^{2} − 4*x* + 3*x* − 12

Finally, combine like terms:

*x*^{2} − 4*x* + 3*x* − 12 =

*x*^{2} − *x* − 12

The simplified expression (*x* + 3)(*x* − 4) is *x*^{2} − *x* − 12.

**Here is how you may see polynomials on the SAT.**

[*x*^{3} − 2*x* + 3] + [2*x*^{2} + 2*x* − 4] =

A) *x*^{3} − 2*x*^{2} − 4*x* + 7

B) *x*^{3} + 2*x*^{2} − 1

C) 2*x*^{5} + 2*x*^{4} − 8*x*^{3} + 10*x*^{2} + 14*x* − 12

D) 3*x*^{2} − 1

Another method of multiplying binomials, and *only* binomials, is known as **FOIL**, which stands for **F**irst **O**utside **I**nside **L**ast. What does that mean, you ask? Here’s how the FOIL method is used to guide binomial multiplication.

Let’s look at how we could solve the previous question using the FOIL method.

(*x* + 3)(*x* − 4) =

Start by multiplying the *first* terms of each binomial together:

Then multiply the *outside* terms of each binomial together:

Next, multiply the *inside* terms of each binomial together:

Finally, multiply the *last* terms of each binomial together:

Note that when a binomial includes a subtraction sign, the corresponding term is negative. For example, in the previous question, since there is a negative sign in front of the 4 in the second binomial, the second term in the second binomial is −4.

Multiplying *binomials*? Distribute or FOIL.

Multiplying *polynomials*? Distribute.

Let’s take a look at another example.

There are a lot of terms in this question, but don’t worry! Start by combining the polynomials together:

[*x*^{3} − 2*x* + 3] + [2*x*^{2} + 2*x* − 4] = *x*^{3} − 2*x* + 3 + 2*x*^{2} + 2*x* − 4

Next, group like terms:

*x*^{3} − 2*x* + 3 + 2*x*^{2} + 2*x* − 4 = *x*^{3} + 2*x*^{2} − 2*x* + 2*x* + 3 − 4

Finally, combine like terms:

*x*^{3} + 2*x*^{2} − 2*x* + 2*x* + 3 − 4 = *x*^{3} + 2*x*^{2} − 1

Therefore, the correct answer is (B).

**EXAMPLE **

**Simplify the expression ( x − 2)(x − 3).**

Here, you’re multiplying two binomials together, so you can either distribute or FOIL. Let’s see how we can simplify this expression using the distribution method.

First, distribute each term in the first polynomial to the second polynomial:

(*x* − 2)(*x* − 3) =

(*x*)(*x* − 3) + (−2)(*x* − 3)

Now distribute each monomial term through each term in the binomials:

(*x*)(*x* − 3) + (−2)(*x* − 3) =

(*x*)(*x*) + (*x*)(−3) + (−2)(*x*) + (−2)(−3)

Then multiply your groups of terms together:

(*x*)(*x*) + (*x*)(−3) + (−2)(*x*) + (−2)(−3) =

*x*^{2} + (− 3*x*) + (−2*x*) + (6) =

*x*^{2} − 3*x* − 2*x* + 6

Finally, combine like terms:

*x*^{2} − 3*x* − 2*x* + 6 =

*x*^{2} − 5*x* + 6

Done! The simplified version of (*x* − 2)(*x* − 3) is *x*^{2} − 5*x* + 6.

Now let’s see how we could have used FOIL to solve this question. Start by multiplying the *first* terms of each binomial together:

Then multiply the *outside* terms of each binomial together:

Now multiply the *inside* terms of each binomial together:

Finally, multiply the *last* terms of each binomial together:

(*x*)(*x*) + (*x*)(−3) + (−2)(*x*) + (−2)(−3)

Now simplify the expression:

(*x*)(*x*) + (*x*)(−3) + (−2)(*x*) + (−2)(−3) =

*x*^{2} + (−3*x*) + (−2*x*) + 6 =

*x*^{2} − 3*x* − 2*x* + 6

And, finally, combine like terms:

*x*^{2} − 3*x* − 2*x* + 6 =

*x*^{2} − 5*x* + 6

As you can see, the result is the same as the one we found by using the distribution method.

**Here is how you may see polynomial expressions on the ACT.**

The expression −*w*[*x* − (*y* + *z*)] is equivalent to:

A. −*wx* − *wy* − *wz*

B. −*wx* + *wy* − *wz*

C. −*wx* + *wy* + *wz*

D. −*wx* − *y* + *z*

E. *wx* − *y* + *z*

Now let’s try a multiplication question that is a bit more complicated.

**EXAMPLE **

**Simplify the expression ( x^{2} − 4x + 2)(x + 2).**

When multiplying polynomials, you *cannot* use FOIL; rather, you must use the power of distribution to simplify your expression. First, distribute each term in the first polynomial through both terms in the second polynomial:

(*x*^{2} − 4*x* + 2)(*x* + 2) =

(*x*^{2})(*x* + 2) + (−4*x*)(*x* + 2) + 2(*x* + 2)

Next, distribute each monomial term through each term in the binomial:

(*x*^{2})(*x* + 2) + (−4*x*)(*x* + 2) + 2(*x* + 2) =

(*x*^{2})(*x*) + (*x*^{2})(2) + (−4*x*)(*x*) + (−4*x*)(2) + (2)(*x*) + (2)(2)

Now multiply the groups of terms together:

(*x*^{2})(*x*) + (*x*^{2})(2) + (−4*x*)(*x*) + (−4*x*)(2) + (2)(*x*) + (2)(2) =

*x*^{3} + 2*x*^{2} + (−4*x*^{2}) + (−8*x*) + 2*x* + 4 =

*x*^{3} + 2*x*^{2} − 4*x*^{2} − 8*x* + 2*x* + 4

Finally, combine like terms:

*x*^{3} + 2*x*^{2} − 4*x*^{2} − 8*x* + 2*x* + 4 =

*x*^{3} − 2*x*^{2} − 6*x* + 4

Viola! You successfully simplified the expression (*x*^{2} − 4*x* + 2)(*x* + 2) to *x*^{3} − 2*x*^{2} − 6*x* + 4.