# Factoring \$ax^2+bx+c\$

In the previous lesson, you learned how to factor polynomials of the form \$x^2+bx+c\$, and how to use that to solve equations that look like \$x^2+bx+c=0\$. In this lesson, you will learn how to do the same thing with quadratic polynomials where the \$x^2\$ term could have a coefficient different from 1.

## Factoring out a constant

The equation \$y=a(2x^2-3x-2)\$ is graphed on the grid to the left, with a slider for \$a\$.

 When \$a\$ is set to 1, the graph has two \$x\$-intercepts, one where \$x\$ is positive and one where \$x\$ is negative. Give the positive one.
 Give the positive \$x\$-intercept of the graph when \$a\$ is set to \$-2\$.
 Does setting \$a\$ to any nonzero value affect the \$x\$-intercepts of the graph?
 Does setting \$a=0\$ affect the \$x\$-intercepts of the graph?
 What is a positive solution to the equation \$3(2x^2-3x-2)=0\$?
 What is a positive solution to the equation \$20(2x^2-3x-2)=0\$?

As you saw in factorPict, multiplying the polynomial \$2x^2-3x-2\$ by a nonzero constant doesn’t change when that polynomial is zero. In fact, you can do this for any polynomial, because \$ab=0\$ exactly when \$a=0\$ or \$b=0\$. This can be useful in solving equations.

 We want to solve the equation \$2x^2-6x+4=0\$. By noticing that all three coefficients are multiples of \$2\$, rewrite the equation in the form shown to the right.
 Based on your previous answer, what is a simpler quadratic equation that has the same solutions?
 Factor the left-hand side of that equation as the product of two linear polynomials.
 What are the solutions to the simpler equation?

Check that these solutions solve the original equation \$2x^2-6x+4=0\$.

 Now, we want to solve the equation \$3x^2+12x-15=0\$. What is the largest integer factor of that equation’s left-hand side?
 Rewrite the equation as a multiple of that number.
 What is a simpler equation that has the same solutions?
 Factor this simpler quadratic and find those solutions.

Check that these solutions solve the original equation \$3x^2+12x-15\$.

Write each quadratic polynomial in the table below as the product of a number and a simpler quadratic polynomial. Use the largest possible number.

Write the polynomial \$-3x^2+2x+7\$ as the product of of \$-1\$ and another quadratic polynomial.

## Factoring \$ax^2+bx+c\$

In order to solve quadratic equations, we want to be able to write quadratic polynomials as the product of linear polynomials. Just as we did when the coefficient of \$x^2\$ was \$1\$, we’ll start by looking more closely at the expansion of such products.

Expand each product \$(px+q)(rx+s)\$ below, using scratch paper. Note the values of \$pr\$ and \$qs\$.

\$(px+q)(rx+s)\$Expanded form\$pr\$\$qs\$

Look at the second and third rows of the table. Notice that in the second row we’re multiplying \$(3x+2)\$ by \$(2x+5)\$, while in the third row we’re multiplying \$(-3x-2)\$, which is the opposite of \$(3x+2)\$, by \$(-2x-5)\$, which is the opposite of \$(2x+5)\$.

 Is the product in the third row the same as the product in the second row, or different?

As you saw in expand, when you expand \$(px+q)(rx+s)\$:

• The constant coefficient (the number not multiplied by anything) is the product of \$q\$ and \$s\$.
• The quadratic coefficient (the number multiplied by \$x^2\$) is the product of \$p\$ and \$r\$.
• The product of the opposites of \$(px+q)\$ and \$(rx+s)\$ — that is, \$(-px-q)(-rx-s)\$ — is equal to the product of \$(px+q)\$ and \$(rx+s)\$.

Say we want to write \$3x^2-5x+2\$ as a product of linear polynomials.

 The linear coefficients of those polynomials must multiply to make \$3\$. What are all the pairs of integers whose product is \$3\$?

Let’s start by trying to factor \$3x^2-5x+2\$ as \$(3x+q)(x+s)\$ with \$qs=2\$.

 What are all the pairs of integers whose product is \$2\$?

Write \$(3x+q)(x+s)\$ with \$qs=2\$ in all possible ways, and then expand those products to see if they equal \$3x^2-5x+2\$. (The pairs of integers you just found could be \$q\$ and \$s\$ in either order.)

ProductIs it \$3x^2-5x+2\$?
 Based on the above table, what is the way to factor \$3x^2-5x+2\$ as a product of the form \$(3x+q)(x+s)\$?
 What is a way of factoring \$3x^2-5x+2\$ as a product of the form \$(-3x+q)(-x+s)\$? (Remember that if you take the opposites of both factors in a product, the overall product is unchanged.)
 Are there any other ways of factoring \$3x^2-5x+2\$ as a product of the form \$(-3x+q)(-x+s)\$? (If there is another way, think about taking the opposites of both its factors, and finding the result in the table above.)

If you have a way of factoring a quadratic polynomial into two linear polynomials with negative linear coefficients, the factors will be the opposites of a way of factoring that polynomial into linear polynomials with positive linear coefficients. As a result:

In order to find out if a quadratic polynomial with a positive quadratic coefficient can be factored into linear polynomials, you only need to check whether it can be factored into linear polynomials with positive linear coefficients.

In the next few questions, we’ll solve the equation

\$\$ 8x^2+18x-5=0 \$\$

by factoring its left-hand side.

 We would like to factor the polynomial \$8x^2+18x-5\$. What are all the pairs of positive integers whose product is \$8\$?

From slowFactor2, we know that if we can factor \$8x^2+18x-5\$, we can factor it as either \$(8x+q)(x+s)\$ or \$(4x+q)(2x+s)\$, where \$qs=-5\$.

 What are all the ways to write \$-5\$ as the product of two integers?

We’ll start by trying to factor the polynomial as \$(8x+q)(x+s)\$. Write \$(8x+q)(x+s)\$ with \$qs=-5\$ in all possible ways, and then expand those products.

 Are any of the quadratic polynomials you just found equal to \$8x^2+18x-5\$?

Now, we’ll try factoring it as \$(4x+q)(2x+s)\$. Write \$(4x+q)(2x+s)\$ with \$qs=-5\$ in all possible ways, and then expand those products.

 Are any of the quadratic polynomials in the table just above equal to \$8x^2+18x-5\$?

Using your work from the two tables above, write \$8x^2+18x-5\$ as the product of two linear polynomials.

 What two linear equations does the equation \$8x^2+18x-5=0\$ reduce to?
 What are the solutions to \$8x^2+18x-5\$? (Write them as decimals, or as fractions by using the ‘/’ key.)

## Simplifying quadratic equations

In the last few questions, you learned how to solve certain quadratic equations. In this section, we’ll look at some ways of turning some other quadratic equations into ones you already know how to solve.

Let’s look at the equation \$-3x^2+4x-1=0\$. Unlike the equations you’ve been solving, the coefficient of \$x^2\$ here is negative.

What is the result of multiplying this equation by \$-1\$?

 Now, factor the left-hand side of the equation you just found as the product of two linear polynomials. Use scratch paper.
 What are the two solutions to this equation? (Write fractions by using the ‘/’ key.)

What about quadratic equations that involve fractions? For example, let’s look at how you would solve the equation

\$\$ 2/3x^2 + x + 1/3 = 0 \$\$

As you saw in Questions factorPict and consFactor, multiplying a quadratic equation by a nonzero constant doesn’t change its solutions. What is the result of multiplying this equation by \$3\$?

 Now, factor the left-hand side of the equation you just found as the product of two linear polynomials. (Use scratch paper.)
 What are the solutions to the equation?
To solve a quadratic equation with fractional coefficients, multiply it by some nonzero number that turns all the fractions into integers, and solve the resulting equation. This is called clearing denominators.

Each equation in the table below includes a polynomial with fractional coefficients. Multiply both sides of that equation by a constant, to get an equation where all the coefficients are integers. Use the smallest possible positive constant. You do not have to solve the resulting equation.

Now, say you wanted to solve an equation like

\$\$ 10x^2+5x+11 = 2x^2-13x+16 \$\$

in which two different polynomials are set equal to each other.

 We’ll look at what happens when you subtract \$2x^2-13x+16\$ from both sides of the equation. What is the result of subtracting this polynomial from the right-hand side of the equation? (What is the result of subtracting any quantity from itself?)

What is the result of subtracting \$2x^2-13x+16\$ from the left-hand side of the equation?

 To summarize, what equation do you get when you subtract \$2x^2-13x+16\$ from both sides of the equation \$10x^2+5x+11=2x^2-13x+16\$?

Each row of the table below has an equation involving two quadratic polynomials. By subtracting the same quantity from both sides, rewrite those equations as equations involving only one quadratic polynomial set equal to 0. Use scratch paper.