# Lesson 12

Polynomial Division (Part 1)

• Let’s learn a way to divide polynomials.

### 12.1: Notice and Wonder: A Different Use for Diagrams

What do you notice? What do you wonder?

A. $$(x-3)(x+5)=x^2+2x-15$$

$$x$$ 5
$$x$$ $$x^2$$ $$5x$$
-3 $$\text-3x$$ -15

B. $$(x-1)(x^2+3x-4)=x^3+2x^2-7x+4$$

$$x^2$$ $$3x$$ -4
$$x$$ $$x^3$$ $$3x^2$$ $$\text-4x$$
-1 $$\text-x^2$$ $$\text-3x$$ +4

C. $$(x-2)(?)=(x^3 - x^2 - 4x + 4)$$

$$x$$ $$x^3$$
-2

### 12.2: Factoring with Diagrams

Priya wants to sketch a graph of the polynomial $$f$$ defined by $$f(x)=x^3 + 5x^2 + 2x - 8$$. She knows $$f(1)=0$$, so she suspects that $$(x-1)$$ could be a factor of $$x^3 + 5x^2 + 2x - 8$$ and writes $$\displaystyle (x^3 + 5x^2 + 2x - 8) = (x - 1)(?x^2 + ?x + ?)$$ and draws a diagram.

$$x$$ $$x^3$$
-1
1. Finish Priya’s diagram.
2. Write $$f(x)$$ as the product of $$(x-1)$$ and another factor.
3. Write $$f(x)$$ as the product of three linear factors.
4. Make a sketch of $$y=f(x)$$.

### 12.3: More Factoring with Diagrams

Here are some polynomial functions with known factors. Rewrite each polynomial as a product of linear factors. Note: you may not need to use all the columns in each diagram. For some problems, you may need to make another diagram.

1. $$A(x)=x^3 - 7x^2 - 16x + 112$$, $$(x-7)$$

$$x^2$$
$$x$$ $$x^3$$ 0
-7 $$\text-7x^2$$
2. $$B(x)=2x^3 - x^2 - 27x + 36$$, $$\left(x-\frac32\right)$$

$$2x^2$$
$$x$$ $$2x^3$$ $$2x^2$$
$$\text-\frac32$$ $$\text-3x^2$$
3. $$C(x)=x^3 - 3x^2 - 13x + 15$$, $$(x+3)$$

$$x$$
3
4. $$D(x)=x^4 - 13x^2 + 36$$, $$(x-2)$$, $$(x+2)$$

(Hint: $$x^4 - 13x^2 + 36 = x^4 +0x^3 - 13x^2 +0x + 36$$)

5. $$F(x)=4x^4 - 15x^3 - 48x^2 + 109x + 30$$, $$(x-5)$$, $$(x-2)$$, $$(x+3)$$

A diagram can also be used to divide polynomials even when a factor is not linear. Suppose we know $$(x^2-2x+5)$$ is a factor of $$x^4+x^3-5x^2+23x-20$$. We could write $$(x^4+x^3-5x^2+23x-20)=(x^2-2x+5)(?x^2+?x+?)$$. Make a diagram and find the missing factor.

### Summary

What are some things that could be true about the polynomial function defined by $$p(x) = x^3 -5x^2 - 2x + 24$$ if we know $$p(\text-2)=0$$? If we think about the graph of the polynomial, the point $$(\text-2,0)$$ must be on the graph as a horizontal intercept. If we think about the expression written in factored form, $$(x+2)$$ could be one of the factors, since $$x+2=0$$ when $$x=\text-2$$. How can we figure out whether $$(x+2)$$ actually is a factor?

Well, if we assume $$(x+2)$$ is a factor, there is some other polynomial $$q(x)=ax^2+bx+c$$ where $$a$$, $$b$$, and $$c$$ are real numbers and $$p(x)=(x+2)q(x)$$. (Can you see why $$q(x)$$ has to have a degree of 2?) In the past, we have done things like expand $$(x+2)(ax^2+bx+c)$$ to find $$p(x)$$. Since we already know the expression for $$p(x)$$, we can instead work out the values of $$a$$, $$b$$, and $$c$$ by thinking through the calculation.

One way to organize our thinking is to use a diagram. We first fill in $$(x+2)$$ and the leading term of $$p(x)$$, $$x^3$$. From this start, we see the leading term of $$q(x)$$ must be $$x^2$$, meaning $$a=1$$, since $$x \boldcdot x^2 = x^3$$.

$$x^2$$
$$x$$ $$x^3$$
+2

We then fill in the rest of the diagram using similar thinking and paying close attention to the signs of each term. For example, we put in a $$2x^2$$ in the bottom left cell because that’s the product of $$2$$ and $$x^2$$. But that means we need to have a $$\text-7x^2$$ in the middle cell of the middle row, since that’s the only other place we will get an $$x^2$$ term, and we need to get $$\text-5x^2$$ once all the terms are collected. Continuing in this way, we get the completed table:

$$x^2$$ $$\text-7x$$ +12
$$x$$ $$x^3$$ $$\text-7x^2$$ $$+12x$$
+2 $$+2x^2$$ $$\text-14x$$ +24

Collecting all the terms in the interior of the diagram, we see that $$x^3-5x^2-2x+24=(x+2)(x^2-7x+12)$$, so $$q(x) = x^2-7x+12$$. Notice that the 24 in the bottom right was exactly what we needed, and it’s how we know that $$(x+2)$$ is a factor of $$p(x)$$. In a future lesson, we will see why this happened. With a bit more factoring, we can say that $$p(x) = (x+2)(x-3)(x-4)$$.

### Glossary Entries

• end behavior

How the outputs of a function change as we look at input values further and further from 0.

The power to which a factor occurs in the factored form of a polynomial. For example, in the polynomial $$(x-1)^2(x+3)$$, the factor $$x-1$$ has multiplicity 2 and the factor $$x+3$$ has multiplicity 1.