Lesson 5 Divide and Conquer Solidify Understanding

Learning Focus

Divide polynomials.

Write equivalent multiplication statements after dividing.

Know when one polynomial is a factor of another polynomial.

How is dividing polynomials like dividing whole numbers?

How is factoring related to division?

How can the remainder of a polynomial division problem be used?

Open Up the Math: Launch, Explore, Discuss

We’ve seen how numbers and polynomials relate in addition, subtraction, and multiplication. Now we’re ready to consider division.

Division, you say? Like, long division? Yup, that’s what we’re talking about. Hold the judgment! It’s actually pretty cool.

As usual, let’s start by looking at how the operation works with numbers. Since division is the inverse operation of multiplication, the same models should be useful. The area model that we used with multiplication is also used with division. When we were using area models to factor a quadratic expression, we were actually dividing.

Let’s brush up on that a bit.

1.

The area model for $x^{2} + 7 x + 10$ is shown.

Use the area model to write $x^{2} + 7 x + 10$ in factored form.

2.

We also used number patterns to factor without drawing the area model. Use any strategy to factor the following quadratic polynomials:

a.

$x^{2} + 7 x + 12$

b.

$x^{2} + 2 x - 15$

c.

$x^{2} - 11 x + 24$

d.

$x^{2} - 5 x - 36$

Pause and Reflect:

Factoring works great for quadratics and a few special cases of other polynomials. Let’s look at a more general version of division that is a lot like what we do with numbers. Let’s say we want to divide $1,452$ by $12$ . If we write the analogous polynomial division problem, it would be:

$(x^{3} + 4 x^{2} + 5 x + 2) \div (x + 2)$ .

Let’s use the division process for numbers to create a division process for polynomials. (Don’t panic—in many ways it’s easier with polynomials than numbers!)

Step 1: Start with writing the problem as long division. The polynomial needs to have the terms written in descending order. If there are any missing powers, it’s easier if you leave a little space for them.

$\begin{array}{r} 12 1452 x + 2 x^{3} + 4 x^{2} + 5 x + 2 \end{array}$

Step 2: Determine what you could multiply the divisor by to get the first term of the dividend.

$\begin{array}{r} 1 \\ 1 2 1 452 \end{array} \begin{array}{r} x^{2} \\ x + 2 x^{3} + 4 x^{2} + 5 x + 2 \end{array}$

Step 3: Multiply and put the result below the dividend.

$\begin{array}{r} 1 \\ 1 2 1 452 \\ \underset{―}{- 1200} \end{array} \begin{array}{r} x^{2} \\ x + 2 x^{3} + 4 x^{2} + 5 x + 2 \\ - \underset{―}{(x^{3} + 2 x^{2})} \end{array}$

Step 4: Subtract. (It helps to keep the signs straight if you change the sign on each term and add on the polynomial.)

$\begin{array}{r} 1 \\ 12 1452 \\ \underset{―}{- 1200} \\ 252 \end{array} \begin{array}{r} x^{2} \\ x + 2 x^{3} + 4 x^{2} + 5 x + 2 \\ + (- \underset{―}{x^{3} - 2 x^{2})} \\ 2 x^{2} + 5 x + 2 \end{array}$

Step 5: Repeat the process with the number or expression that remains in the dividend.

$\begin{array}{r} 1 2 \\ 1 2 1452 \\ \underset{―}{- 1200} \\ 2 52 \\ - \underset{―}{240} \\ 12 \end{array} \begin{array}{r} x^{2} + 2 x \\ x + 2 x^{3} + 4 x^{2} + 5 x + 2 \\ + (- \underset{―}{x^{3} - 2 x^{2})} \\ 2 x^{2} + 5 x + 2 \\ - \underset{―}{(2 x^{2} + 4 x)} \\ x + 2 \end{array}$

Step 6: Keep going until the number or expression that remains is smaller than the divisor.

$\begin{array}{r} 12 1 \\ 1 2 1452 \\ \underset{―}{- 1200} \\ 252 \\ - \underset{―}{240} \\ 1 2 \\ \underset{―}{- 12} \\ 0 \end{array} \begin{array}{r} x^{2} + 2 x + 1 \\ x + 2 x^{3} + 4 x^{2} + 5 x + 2 \\ + (- \underset{―}{x^{3} - 2 x^{2})} \\ 2 x^{2} + 5 x + 2 \\ - \underset{―}{(2 x^{2} + 4 x)} \\ x + 2 \\ - \underset{―}{(x + 2)} \\ 0 \end{array}$

In this case, $1452$ divided by $12$ leaves no remainder, so we would say that $12$ is a factor of $1452$ . Similarly, since $(x^{3} + 4 x^{2} + 5 x + 2)$ divided by $(x + 2)$ leaves no remainder, we would say that $(x + 2)$ is a factor of $(x^{3} + 4 x^{2} + 5 x + 2)$ .

Polynomial division doesn’t always match up perfectly to an analogous whole number problem, but the process is always the same. Let’s try it.

3.

Use long division to determine if $(x - 1)$ is a factor of $(x^{3} - 3 x^{2} - 13 x + 15)$ . Don’t worry: the steps for the division process are provided:

Write the problem as long division.
What do you have to multiply $x$ by to get $x^{3}$ ? Write your answer above the bar.
Multiply your answer from step b by $(x - 1)$ and write your answer below the dividend.
Subtract. Be careful to subtract each term. (You might want to change the signs and add.)
Repeat steps a–d until the expression that remains is less than $(x - 1)$ .

4.

Try it again. Use long division to determine if $(2 x + 3)$ is a factor of $2 x^{3} + 7 x^{2} + 2 x + 9$ . No hints this time. You can do it!

When dividing numbers, there are several ways to deal with the remainder. Sometimes, we just write it as the remainder, like this:

$\begin{array}{r} 8 r .1 \\ 3 25 \end{array}$

because $3 (8) + 1 = 25$

You may remember also writing the remainder as a fraction like this:

$\begin{array}{r} 8 \frac{1}{3} \\ 3 25 \end{array}$

because $3 (8 \frac{1}{3}) = 25$

We do the same things with polynomials.

Maybe you found that $(2 x^{3} + 7 x^{2} + 2 x + 9) \div (2 x + 3) = (x^{2} + 2 x - 2), r . 15$ . You can use it to write two multiplication statements:

$(2 x + 3) (x^{2} + 2 x - 2) + 15 = (2 x^{3} + 7 x^{2} + 2 x + 9)$

$and$

$(2 x + 3) (x^{2} + 2 x - 2 + \frac{15}{2 x + 3}) = (2 x^{3} + 7 x^{2} + 2 x + 9)$

5.

Divide each of the following polynomials. Write the two multiplication statements that go with your answers if there is a remainder. Write only one multiplication statement if the divisor is a factor. Use graphing technology to check your work and make the necessary corrections.

a.

$(x^{3} + 6 x^{2} + 13 x + 12) \div (x + 3)$

Multiplication statements:

b.

$(x^{3} - 4 x^{2} + 2 x + 5) \div (x - 2)$

Multiplication statements:

c.

$(6 x^{3} - 11 x^{2} - 4 x + 5) \div (2 x - 1)$

Multiplication statements:

d.

$(x^{4} - 23 x^{3} + 49 x + 4) \div (x^{2} + x + 2)$

Multiplication statements:

6.

There’s one more interesting thing to notice with division of polynomials. Look back at problem 5b: $(x^{3} - 4 x^{2} + 2 x + 5) \div (x - 2)$ . Let $f (x) = x^{3} - 4 x^{2} + 2 x + 5$ (the dividend). Find $f (2)$ and compare it to the remainder you found when you divided $f (x)$ by $(x - 2)$ . What do you see?

7.

Try it with this one: Let $f (x) = x^{3} + 5 x^{2} - 3 x + 4$ . Find $f (- 3)$ and compare it to the remainder of $\frac{f (x)}{x + 3}$ . What do you see?

8.

Based on these two examples, write a conjecture about $f (x)$ and dividing by $x - n$ .

9.

Look back at problem 3: $(x^{3} - 3 x^{2} - 13 x + 15) \div (x - 1)$ . Let $f (x) = (x^{3} - 3 x^{2} - 13 x + 15)$ (the dividend). Find $f (1)$ and compare it to the remainder you found when you divided $(x^{3} - 3 x^{2} - 13 x + 15)$ by $(x - 1)$ . What do you see? How does your observation compare to your answers for problem 8?

Ready for More?

Write your own long division problem with a remainder of $5$ and then trade with a partner and work the problem they have written. Use your work in creating the problem to check their answer.

Takeaways

The Remainder Theorem, given $f (x)$ is a polynomial function and $x - n$ is a linear factor:

Vocabulary

Lesson Summary

In this lesson, we learned that polynomials can be divided using long division like whole numbers. We learned to use technology to check our work and to avoid errors in subtraction by adding the opposite of the terms to be subtracted. We found that, like numbers, a polynomial is a factor of another polynomial if it divides with no remainder. We learned two ways to write equivalent multiplication statements when there was a remainder after dividing.

Retrieval

1.

Solve for $x$ . $x^{2} - 7 x - 18 = 0$

2.

Evaluate the logarithmic expression:

$\log_{12} 1,728 =$