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

In the last few lessons, you learned how to expand expressions like \$(x+2)(x+3)\$ into \$x^2+5x+6\$. In this lesson, you’ll learn how to reverse this process, starting with a polynomial like \$x^2+5x+6\$ and writing it as a product like \$(x+2)(x+3)\$. You will also learn how this helps you solve equations like \$x^2+5x+6=0\$.

## Graphing \$y=(x-r)(x-s)\$

 The graph of \$y=(x-2)(x-3)\$ is shown to the left. Give the points where that graph crosses the \$x\$-axis.
 Because \$y=0\$ on the \$x\$-axis, the \$x\$-coordinates of the points you just found are solutions to the equation \$0=(x-2)(x-3)\$. Give those \$x\$-coordinates.

Now the equation \$y=(x-r)(x-s)\$ is graphed to the left, with sliders for \$r\$ and \$s\$.

 Set \$r=1\$ and \$s=4\$, so you are looking at the graph of \$y=(x-1)(x-4)\$. Use this graph to give two solutions to \$0=(x-1)(x-4)\$.

Each row of the table below has an equation of the form \$0=(x-r)(x-s)\$. Slide the sliders to find two solutions to that equation.

 Give two solutions to the equation \$0=(x+10)(x-20)\$.

## Solving \$(x+u)(x+v)=0\$

To understand what’s happening algebraically, we want to look at when a product of two numbers can be equal to zero.

 Is \$2(5)=0\$?
 Is \$2(0)=0\$?
 Can you find any two nonzero numbers whose product is zero?
 Can you find any number whose product with zero is not equal to zero?

In fact:

If \$ab = 0\$, then either \$a=0\$ or \$b=0\$. If either \$a=0\$ or \$b=0\$, then \$ab=0\$.

You can solve equations of the form \$(x+u)(x+v)=0\$ by using this fact.

Algebraically solve the equation \$(x+2)(x+4)=0\$, and check your solution.

## Solving \$x^2+bx=0\$ by factoring

The method in algProductQn allows you to solve equations of the form \$(x+u)(x+v)=0\$. This means you can also solve any equation that can be rewritten in this form.

For example, the expression \$x^2+2x\$ can be rewritten as \$x(x+2)\$, as shown on the grids to the left. This means that we can solve the equation \$x^2+2x=0\$ algebraically, by first rewriting it as \$x(x+2)=0\$ and then applying the method of algProductQn.

Solve the equations in the table below, as pictured on the grids to the left. Then check your answers.

Rewriting an expression as a product in this way is called factoring that expression. The values of \$x\$ where a polynomial is zero are the roots of that polynomial.

## Factoring integers

We would like to be able to solve some more quadratic equations by factoring them. In order to do that, we first need to look at how to factor integers (whole numbers or their negatives). That is, we want to look at all the ways of writing one integer as a product of two other integers.

If you want to write the number 4 as the product of two integers, there are four different ways to do it:

\$\$ \cl"tight"{\table 4, =, 1(4); 4, =, 2(2); 4, =, (-1)(-4); 4, =, (-2)(-2); } \$\$

This is illustrated by the grid to the left. The rectangle on that grid has area 4. So whenever the slider is set to a value that makes the rectangle’s width and height both integers (both line up with the grid lines), the grid gives you a way to factor 4.

 Click . What are all the ways to write 6 as the product of two integers?

For each integer given below, what are all the ways to write it as the product of two integers?

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

In order to move from factoring integers to factoring quadratic polynomials, we first want to look more carefully at how expansion works.

Expand each product \$(x+u)(x+v)\$ in the table below, using scratch paper. Then complete the table by computing the numbers \$u+v\$ and \$uv\$. (Remember to type \$x^2\$ as x^2.)

\$(x+u)(x+v)\$Expanded form\$u+v\$\$uv\$

When you expand \$(x+u)(x+v)\$:

This will often let you factor quadratic polynomials. In order to factor \$x^2+bx+c\$, you need to find two numbers whose product is \$c\$ and whose sum is \$b\$. For example, suppose you wanted to factor \$x^2+4x+3\$. You could notice that the numbers \$1\$ and \$3\$ multiply to make \$3\$ (the constant coefficient) and add up to \$4\$ (the linear coefficient). So \$x^2+4x+3\$ factors as \$(x+1)(x+3)\$.

 Say we want to factor the polynomial \$x^2+6x+5\$. What are all the pairs of numbers whose product is 5?
 Of those pairs, which one adds up to 6?
 What is the factored form of \$x^2+6x+5\$?
 Now say we want to factor the polynomial \$x^2+7x+10\$. What are all the pairs of numbers whose product is 10?
 Which of those pairs adds up to 7?
 What is the factored form of \$x^2+7x+10\$?
 What are the roots of \$x^2+7x+10\$? In other words, what are the solutions to the equation \$x^2+7x+10=0\$?

Check those solutions.

Using scratch paper, solve each equation below by factoring the quadratic polynomial on the left-hand side, and then check your answers in the original equation.

EquationFactored formSolutions

All of the quadratic polynomials in the last two questions had positive coefficients. However, you can factor quadratic polynomials with negative coefficients in the same way; just keep in mind that \$(x+u)\$ or \$(x+v)\$ will look like a subtraction if \$u\$ or \$v\$ is negative.

 Say we want to solve the equation \$x^2+2x-3=0\$ by factoring its left-hand side. What are all the pairs of numbers whose product is \$-3\$?
 Which of those pairs adds up to \$2\$?
 What is the factored form of the equation?
 What are the solutions to the equation?

Check those solutions.

 Say we want to solve the equation \$x^2-4=0\$ by factoring its left-hand side. What are all the pairs of numbers whose product is \$-4\$?
 Because there is no linear term, the coefficient of \$x\$ is \$0\$ (another way to write the polynomial is \$x^2+0x-4\$). Which of the pairs you just found adds up to \$0\$?
 What is the factored form of the equation?
 What are the solutions to the equation?

Solve each equation below by factoring the quadratic polynomial on the left-hand side. Use scratch paper, and check your answers in the original equation.

EquationFactored formSolutions
 The equation \$y=x^2-3x+1\$ is graphed on the grid to the left. Based on this graph, does \$x^2-3x+1=0\$ have any solutions?
 What are all the pairs of integers whose product is 1?
 Do either of those pairs of integers add up to \$-3\$?
 Can you find any solutions to this equation by using the method from monicSolveTable?