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Learn how to solve quadratic equations like (x-1)(x+3)=0 and how to use factorization to solve other forms of equations.

## What you will learn in this lesson

So far you have solved linear equations, which include constant terms—plain numbers—and terms with the variable raised to the first power, x, start superscript, 1, end superscript, equals, x.
You may have also solved some quadratic equations, which include the variable raised to the second power, by taking the square root from both sides.
In this lesson, you will learn a new way to solve quadratic equations. Specifically you will learn
• how to solve factored equations like left parenthesis, x, minus, 1, right parenthesis, left parenthesis, x, plus, 3, right parenthesis, equals, 0 and
• how to use factorization methods in order to bring other equations left parenthesislike x, squared, minus, 3, x, minus, 10, equals, 0, right parenthesis to a factored form and solve them.

Suppose we are asked to solve the quadratic equation left parenthesis, x, minus, 1, right parenthesis, left parenthesis, x, plus, 3, right parenthesis, equals, 0.
This is a product of two expressions that is equal to zero. Note that any x value that makes either left parenthesis, x, minus, 1, right parenthesis or left parenthesis, x, plus, 3, right parenthesis zero, will make their product zero.
\begin{aligned} (x-1)&(x+3)=0 \\\\ \swarrow\quad&\quad\searrow \\\\ x-1=0\quad&\quad x+3=0 \\\\ x=1\quad&\quad x=-3 \end{aligned}
Substituting either x, equals, 1 or x, equals, minus, 3 into the equation will result in the true statement 0, equals, 0, so they are both solutions to the equation.
Now solve a few similar equations on your own.
Solve left parenthesis, x, plus, 5, right parenthesis, left parenthesis, x, plus, 7, right parenthesis, equals, 0.

Solve left parenthesis, 2, x, minus, 1, right parenthesis, left parenthesis, 4, x, minus, 3, right parenthesis, equals, 0.

### Reflection question

Can the same solution method be applied to the equation left parenthesis, x, minus, 1, right parenthesis, left parenthesis, x, plus, 3, right parenthesis, equals, 6?

### A note about the zero-product property

How do we know there are no more solutions other than the two we find using our method?
The answer is provided by a simple but very useful property, called the zero-product property:
If the product of two quantities is equal to zero, then at least one of the quantities must be equal to zero.
Substituting any x value except for our solutions results in a product of two non-zero numbers, which means the product is certainly not zero. Therefore, we know that our solutions are the only ones possible.

## Solving by factoring

Suppose we want to solve the equation x, squared, minus, 3, x, minus, 10, equals, 0, then all we have to do is factor x, squared, minus, 3, x, minus, 10 and solve like before!
x, squared, minus, 3, x, minus, 10 can be factored as left parenthesis, x, plus, 2, right parenthesis, left parenthesis, x, minus, 5, right parenthesis.
The complete solution of the equation would go as follows:
\begin{aligned}x^2-3x-10&=0\\\\ (x+2)(x-5)&=0&&\text{Factor.}\end{aligned}
\begin{aligned}&\swarrow&\searrow\\\\ x+2&=0&x-5&=0\\\\ x&=-2&x&=5\end{aligned}
Now it's your turn to solve a few equations on your own. Keep in mind that different equations call for different factorization methods.

### Solve $x^2+5x=0$x, squared, plus, 5, x, equals, 0.

Step 1. Factor x, squared, plus, 5, x as the product of two linear expressions.$\quad$

Step 2. Solve the equation.

### Solve $x^2-11x+28=0$x, squared, minus, 11, x, plus, 28, equals, 0.

Step 1. Factor x, squared, minus, 11, x, plus, 28 as the product of two linear expressions.$\quad$

Step 2. Solve the equation.

### Solve $4x^2+4x+1=0$4, x, squared, plus, 4, x, plus, 1, equals, 0.

Step 1. Factor 4, x, squared, plus, 4, x, plus, 1 as the product of two linear expressions.$\quad$

Step 2. Solve the equation.

### Solve $3x^2+11x-4=0$3, x, squared, plus, 11, x, minus, 4, equals, 0.

Step 1. Factor 3, x, squared, plus, 11, x, minus, 4 as the product of two linear expressions.$\quad$

Step 2. Solve the equation.

## Arranging the equation before factoring

### One of the sides must be zero.

This is how the solution of the equation x, squared, plus, 2, x, equals, 40, minus, x goes:
\begin{aligned}x^2+2x&=40-x\\\\ x^2+2x-40+x&=0&&\text{Subtract 40 and add }x\text{.}\\\\ x^2+3x-40&=0&&\text{Combine like terms.}\\\\ (x+8)(x-5)&=0&&\text{Factor.}\end{aligned}
\begin{aligned}&\swarrow&\searrow\\\\ x+8&=0&x-5&=0\\\\ x&=-8&x&=5\end{aligned}
Before we factored, we manipulated the equation so all the terms were on the same side and the other side was zero. Only then were we able to factor and use our solution method.

### Removing common factors

This is how the solution of the equation 2, x, squared, minus, 12, x, plus, 18, equals, 0 goes:
\begin{aligned}2x^2-12x+18&=0\\\\ x^2-6x+9&=0&&\text{Divide by 2.}\\\\ (x-3)^2&=0&&\text{Factor.}\\\\ &\downarrow\\\\ x-3&=0\\\\ x&=3\end{aligned}
All terms originally had a common factor of 2, so we divided all sides by 2—the zero side remained zero—which made the factorization easier.
Now solve a few similar equations on your own.
Find the solutions of the equation.
2, x, squared, minus, 3, x, minus, 20, equals, x, squared, plus, 34

Find the solutions of the equation.
3, x, squared, plus, 33, x, plus, 30, equals, 0

Find the solutions of the equation.
3, x, squared, minus, 9, x, minus, 20, equals, x, squared, plus, 5, x, plus, 16

## Want to join the conversation?

• In the above equation 3x^2+11x-4 = 0, I understand where we need to find two numbers were a+b need to equal 11 to satisfy the 11x, however, I'm having trouble connecting where -12 came from where it states that we need to find numbers to satisfy (a)(b) = -12. I'm seeing a -4 at the end of the equation. Not sure where -12 came from. Was it from multiplying -4 to the co-effiecient of the 3 in 3x^2?
• In the standard form of quadratic equations, there are three parts to it: ax^2 + bx + c where a is the coefficient of the quadratic term, b is the coefficient of the linear term, and c is the constant. The -4 at the end of the equation is the constant. This hopefully answers your last question. Now, your first question.

So the problem is that you need to find two numbers (a and b) such that the sum of a and b equals 11 and the product equals -12. Correct? Correct indeed. I use a pretty straightforward mental method but I'll introduce my teacher's method of factors first. What you need to do is find all the factors of -12 that are integers. Let's start with 1.
1 * -12
2 * -6
3 * -4
4 * -3
6 * -2
12 * -1
Here we see 6 factor pairs or 12 factors of -12. Let's see which one adds up to 11. It seems like 12 + (-1) = 11. So we'll split 11 to 12 and -1.
My other method is straight out recognising the middle terms. This works well with small numbers. I can clearly see that 12 is close to 11 and all I need is a change of 1. So that leaves out 12 * -1 and -12 * 1. I can see that -12 * 1 makes -11 which is not what I want so I go with 12 * -1.

• Sometimes I don't understand some of the problems. :(
• Just always remember to simplify the expression before u do anything and then make one side equal to zero my subtracting or adding etc. Its easy once u get the hang of it.
• My equation is x^2-2x+1=0 ( find x) How do i factor out the 1? Could i make it 1^2?
• You need 2 factors of 1 that add to -2.
Your choices are to use: 1*1 or (-1)(-1). Which pair adds to -2?
Hope this helps.
• I know this is not related to the above questions, but i could not find where to ask normal day to day questions. If someone could please help me answer this would be great.:
Factorise the expression as far as possible to prime factors:
(x-1)^2-9(y-1)^2 ...
^2(power of 2)
• Expand the squares, then distribute the 9. Combine like terms then factor. Post comment if you still feel stuck.
• I figure out that when A+B=[some number] and AB=[some number] combines, it could be an linear equation. But I don't know how to solve it, because when I'm solving it, a new quadratic equation comes out! For example:
Problem:
Factoring x²+3·x-10=(x+a)(x+b)
Linear equation:
①a+b=3
②a·b=-10
Solve:
a=3-b
Substitute 3-b into equation ②
(3-b)b=-10
3·b-b²=-10
And I don't know how to solve.
• this is how i would continue: guess and check:-) (it is the most annoying way but it works)
• How do you factor it when the leading coefficient is more than 1? For example, something like 3xsquared+ x-2.
• In the last two problems' explanations there seems to be a mistake. Why when they say "divide by 3" in the second step does the equation look like
"x^2+11x+10 = 0"
3(x^2+11x+10) = 0"

The way they wrote it, the right hand side of the equation would equal 0/3, wouldn't it?
• They can divide the entire equation by 3. That's ok. Remember: 0/3 = 0, which is why you see just o.
• In the above equation 3x^2+11x-4 = 0, I understand where we need to find two numbers were a+b need to equal 11 to satisfy the 11x, however, I'm having trouble connecting where -12 came from where it states that we need to find numbers to satisfy (a)(b) = -12. I'm seeing a -4 at the end of the equation. Not sure where -12 came from. Was it from multiplying -4 to the co-effiecient of the 3 in 3x^2?
• Yes, the -12 comes from multiplying -4 with the 3 from 3x^2.
• how do you solve an equation not equal to 0
example 14(squared)+6x=2x