High School Algebra II Unlocked (2016)

Lesson 1.6. Polynomials in the Real World

Everything you’ve learned about polynomials can help you in real-world applications.

A company manufactures solar chargers. They have discovered that the demand curve for this particular solar charger is described by the equation q = −1/10 p² + 490, where q is the number of chargers sold in a month at a price of p dollars. It costs the company $10 per charger to manufacture them. Write and graph a polynomial function representing the company’s monthly profit, t, in terms of p. Using graphing technology, find the solar charger price that will maximize the company’s profits.

Profit is equal to total revenue minus costs. The company’s monthly revenue from selling the solar chargers is found by multiplying the number of solar chargers sold by the price per charger. In this case, the number sold is q, or (−1/10 p² + 490), and the price per charger is p. The company’s revenue for the month is (−1/10 p² + 490)(p) dollars.

The company’s costs for manufacturing q solar chargers at $10 per charger is 10q dollars, or 10(−1/10 p² + 490) dollars.

So, the company’s monthly profit from the solar chargers is given by the equation t = (−1/10 p² + 490)(p) − 10(−1/10 p² + 490). Expand and simplify this equation to put it in standard form.

t = −1/10 p³ + 490p + p² − 4900

t = −1/10 p³ + p² + 490p − 4900

In order for the polynomial
to be in standard form,
the terms must be
arranged in order of
decreasing powers of p.

The function is a cubic with a negative leading coefficient, so the left arm points up and the right arm points down. The y-intercept (or t-intercept) is −4900. To find the x-intercepts, we need to factor the polynomial and solve for its zeros. To factor, look back at the function’s original form, t = (−1/10 p² + 490)(p) − 10(−1/10 p² + 490). The binomial is a common factor in the two terms.

If 0 = −1/10 (p + 70)(p − 70)(p − 10), then p = −70, p = 70, or p = 10.

The function t has zeros of −70, 10, and 70, so these are the x-intercepts (p-intercepts) of its graph. The function generally looks like this.

However, negative values of p do not make sense in the context of the problem. (The company can’t charge a negative amount for each charger. That would mean paying people to take them!) Also, p cannot be greater than 70, because no solar chargers get sold at prices of $70 or more. Our graph should only show the curve for 0 ≤ p ≤ 70.

Let’s use the factored form of the equation to find more points on the graph.

t = −1/10 (p + 70)(p − 70)(p − 10)

When p = 20, t = −1/10 (20 + 70)(20 − 70)(20 − 10) = 4500

When p = 30, t = −1/10 (30 + 70)(30 − 70)(30 − 10) = 8000

When p = 40, t = −1/10 (40 + 70)(40 − 70)(40 − 10) = 9900

When p = 50, t = −1/10 (50 + 70)(50 − 70)(50 − 10) = 9600

When p = 60, t = −1/10 (60 + 70)(60 − 70)(60 − 10) = 6500

The graph passes through the points (20, 4500), (30, 8000), (40, 9900), (50, 9600), and (60, 6500), in addition to the intercepts (0, −4900), (10, 0), and (70, 0).

Here is a better graph of the function t.

You can also use a graphing calculator or graphing program to graph the function t = −1/10 p³ + p² + 490p − 4900. Zoom in to locate the maximum on the curve. It appears to be at (44, 10,077.6). This represents the point of the greatest possible profit for the company. To maximize their profits, t, they should set the price, p, of the solar charger at $44.

DRILL

CHAPTER 1 PRACTICE QUESTIONS

Click here to download a PDF of Chapter 1 Practice Questions.

Directions: Complete the following open-ended problems as specified by each question stem. For extra practice after answering each question, try using an alternative method to solve the problem or check your work.

1. Find all of the zeros of the polynomial function f(x) = x⁴ − x² − 20.

2. Find all the zeros of f(x) = x⁴ − 3x³ + 6x² + 2x − 60 given that 1 + 3i is a zero of f.

3. Find the solutions of the quadratic 3x² − 2x + 5 = 0.

4. Write 6i (5 − 2i) in the form a + bi.

5. Expand and simplify (2 + 3i)² + (2 − 3i)².

6. Expand the binomial (2t − s)⁵ using Pascal’s Triangle to determine the coefficients.

7. Given f(x) = x³ − x² − 2x, find a new function, g, which is created by shifting f(x) 4 units to the right.

8. A group of students are painting a poster to promote the math team. They paint a grid on a piece of poster board, when a centipede starts walking on the board. They step away and watch as the centipede trails the fresh paint as it walks, forming a path that can be described by the cubic function f(x) = x³ + 6x² − x − 30. At what points will the centipede cross the horizontal axis and bring with it more paint to continue its odd drawing?

9. A few friends have recently started a technology company and want to have the best possible launch for their new smartphone. The demand for the phone follows a curve as described by the equation d = c² + 125, where c is the price at which each phone is being sold and d is the number of phones that would be sold per month at that price. Due to cost of labor and technology, however, it costs $15 to manufacture each phone. Write and graph a polynomial function representing the company’s monthly profit, p, in terms of c. At what price point would the company start to make money by selling the phones? At what price point after that would the company begin to lose money?

SOLUTIONS TO CHAPTER 1 PRACTICE QUESTIONS

1. x = ±2i and x = ±

To begin, you should recognize that the trinomial x⁴ − x² − 20 is in the form (x²)² − (x²) − 20. You can then set up parentheses and input the first terms:

(x² )(x² )

Next, identify factors of 20 that will add together to get a sum of −1 to arrive at the middle term. The factors of 20 are 1 and 20, 2 and 10, and 4 and 5. The only possible pairing of numbers are 4 and 5, so place those inside the parentheses.

(x² 4)(x² 5)

Since the middle term of the polynomial is negative, the larger number, 5, must be negative and since the final term is negative, the 4 must be positive.

(x² + 4)(x² − 5)

To find the zeros, set the factored expression equal to zero and solve for each x-value.

2. 1 + 3i, 1 − 3i, 3, and −2

When a complex zero is provided, we need to remember that complex zeros occur in conjugate pairs. What that means is since 1 + 3i is a zero, 1 − 3i must also be a zero. This means that both

[x − (1 + 3i)] and [x − (1 − 3i)]

are factors of f(x). Multiplying these two factors produces

[x − (1 + 3i)][x − (1 − 3i)] = [(x − 1) − 3i][(x − 1) + 3i]

= (x − 1)² − 9i ²

= x² − 2x + 1 −9(−1)

= x² − 2x + 10

Using long division, we can divide x² − 2x + 10 into f(x) to obtain the following.

Therefore, we have

f(x) = (x² − 2x + 10)(x² − x − 6)

 = (x² − 2x + 10)(x − 3)(x + 2)

and we conclude that the zeros of f are

1 + 3i, 1 − 3i, 3, and −2.

3. and .

First, try to determine whether or not the provided quadratic is easy to factor. The quadratic does not look easy to factor, so you should utilize the quadratic formula with a = 3, b = −2, and c = 5.

Thus, the given equation has two solutions:

4. 12 + 30i

When combining concepts such as the distribution property and imaginary numbers, pay close attention to the signs at each step of work. Begin by distributing 6i to each term in the parentheses.

Therefore, the result is 12 + 30i.

5. −10

Proceed with caution here. You can’t just square the individual terms inside the parentheses. First, to expand, FOIL out the individual monomials or use the identity (a + b)² = a² + 2ab + b²:

Next, combine the like terms through addition.

(−5 + 12i) + (−5 − 12i) = −5 + 12i − 5 − 12i = −5 − 5 = −10.

6. 32t⁵ − 80t⁴s + 80t³s² − 40t²s³ + 10ts⁴ − s⁵

Recall that Pascal’s Triangle is a handy tool to determine quickly what the coefficients of each term will be when expanded out, with the top row accounting for the term (x + y)⁰, which is equal to 1. The Binomial Theorem also accounts for the shortcut of figuring the exponents for each term: The first term’s exponents decrease from left to right, from 5 to 0 in this case, whereas the second term’s exponents increase from left to right, from 0 to 5. First, determine the coefficients from Pascal’s Triangle. The binomial is raised to a power of 5, so the 6th row of the triangle is needed: 1, 5, 10, 10, 5, and 1. Next, using the coefficients in conjunction with the Binomial Theorem, begin expanding out the binomial.

(2t − s)⁵ = 1(2t)⁵(−s)⁰ + 5(2t)⁴(−s)¹ + 10(2t)³(−s)² + 10(2t)²(−s)³ + 5(2t)¹(−s)⁴ + 1(2t)⁰(−s)⁵

Carefully apply the exponents to each term and pay close attention to the signs.

1(2t)⁵(−s)⁰ + 5(2t)⁴(−s)¹ + 10(2t)³(−s)² + 10(2t)²(−s)³ + 5(2t)¹(−s)⁴ + 1(2t)⁰(−s)⁵

= 1(32t⁵)(1) + 5(16t⁴)(−s) + 10(8t³)(s²) + 10(4t²)(−s³) + 5(2t)(s⁴) + 1(1)(−s⁵)

= 32t⁵ − 80t⁴s + 80t³s² − 40t²s³ + 10ts⁴ − s⁵.

7. x³ − 13x² + 54x − 72

Recall that when shifting polynomial expressions horizontally, the change occurs inside the parentheses affecting the x term, so the appropriate substitution here is to evaluate f(x − 4) and combine like terms.

f(x − 4) = (x − 4)³ − (x − 4)² − 2(x − 4)

First begin by expanding (x − 4)³.

Next (x − 4)² needs to be expanded, but through the work in the previous step, the result of x² − 8x + 16 was already gathered, so begin combining all like terms to simplify and arrive at the function g.

So g(x) = x³ − 13x² + 54x − 72.

Alternatively, you could factor the given cubic, x³ − x² − 2x, to find the zeros of f(x). It factors as x(x − 2)(x + 1), so the graph of f(x) intersects the x-axis at 0, 2, and −1. If the graph is shifted 4 units to the right, then the new graph has x-intercepts 4, 6, and 3. Using these zeros, you can write the equation of this new function: g(x) = (x − 4)(x − 6)(x − 3). If you expand, this becomes g(x) = x³ − 13x² + 54x − 72.

8. x = −5, −3, and 2

This question is asking for the x-intercepts, or the zeros, of f(x), so you must find the roots of the cubic. Use the Rational Root Theorem first to determine possible roots. In this case, the coefficient of the x³ term is 1, so the possible roots are the positive or negative factors of 30, giving options of ±1, ±2, ±3, ±5, ±6, ±10, ±15, and ±30. Try testing the roots in the original function using the Remainder Theorem:

f(x) = (1)³ + 6(1)² − 1 − 30 = 1 + 6 − 1 − 30 = −24, so 1 is not a root.

f(x) = (−1)³ + 6(−1)² − (−1) − 30 = −1 + 6 + 1 − 30 = −24, so −1 is not a root.

f(x) = (2)³ + 6(2)² − (2) − 30 = 8 + 24 − 2 − 30 = 0, so 2 is a root and therefore (x − 2) is a factor. Divide that out to see what remains:

The remaining quadratic is x² + 8x + 15. This can be factored to (x + 3)(x + 5), so the three binomial factors of the function are (x − 2)(x + 3)(x + 5), and the three points at which the centipede will cross the horizontal axis are at x = −5, −3, and 2.

9. At a price point above $15, the company makes money; at a price point above $50, they lose money again.

First, since profit is the same thing as revenue minus cost, both of those need to be combined to form the desired equation. The revenue is given by the product of the number of phones sold and the price per phone; since the number sold is d and the price for each phone is c, the revenue would be . The cost to make d phones would be 15d, or 15. Given this information, the company’s profit from phone sales would be given by p = . Expand this and simplify where possible to put this in standard form:

This cubic equation has a negative leading coefficient, so it will have a left arm facing up and a right arm facing down. Its y-intercept, when c = 0, is −1875.

The company would break even when they make $0 profit, so factor the polynomial and find the zeros of the function. Going back to the original form rather than standard form, one of the factors is pulled out.

The fractional part of the second piece can be pulled out:

p = −(c − 15)(c² − 2500)

This gives the difference of two squares for the other piece:

p = − (c − 15)(c − 50) (c + 50)

This gives three zeros, at c = −50, 15, and 50. A rough sketch of this information can be seen here:

Because it is impossible to sell phones at a negative price in the real world, we can ignore the x-intercept at −50. The point at which the company starts making money is the x-intercept where the profits change from negative to positive. That would be $15, so they would begin to make money if the phones were priced above $15. Also, if they priced their phones at more than $50 per phone, the company would begin to lose money again, as at that point the function is about to dip back below the x-axis.

REFLECT

Congratulations on completing Chapter 1!
Here’s what we just covered.
Rate your confidence in your ability to:

• Perform arithmetic operations with complex numbers

1 2 3 4 5

• Perform arithmetic operations with polynomials

1 2 3 4 5

• Use polynomial identities to solve for certain quantities, find Pythagorean triples, and factor polynomials

1 2 3 4 5

• Graph polynomial functions and derive polynomial equations from given graphs

1 2 3 4 5

• Factor higher-degree polynomials completely, into any real- and/or imaginary-number roots

1 2 3 4 5

• Create and interpret polynomial functions and their graphs to represent and explain real-world relationships

1 2 3 4 5

If you rated any of these topics lower than you’d like, consider reviewing the corresponding lesson before moving on, especially if you found yourself unable to correctly answer one of the related end-of-chapter questions.

Access your online student tools for a handy, printable list of Key Points for this chapter. These can be helpful for retaining what you’ve learned as you continue to explore these topics.