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## Algebra 1 (Eureka Math/EngageNY)

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?
• You multiply the quadratic coefficient with the constant so that when you factor the middle term , it has a common factor with both the coefficient of x^2 and the constant in this case 3 and -4 respectively. if you see the equation 3x^2+11x-4=0 , in order to factor the terms , we need there to be a one of the terms in x which can group with 3x^2 and one of the terms in x which can be grouped with -4. that way if you multiply the coeff. of x^2 and the constant , and then try to breakdown the term which has x to have a sum of the original term , but to have the product equal to that of the coeff of x^2 and the constant , we can easily factor the equation. The following mathematical explaination should help you understand better.
Consider an equation ax^2+bx+c=0:
if we try to split b into two numbers d and e , which add up to b and equal the product of a and c , we are effectively factorizing the equation without changing it.
eg x^2+6x+9=0:
we split 6 into two such numbers which when added give 6 but when multiplied give the product of 1( coeff of x^2 ) and 9( the constant ). These two numbers being 3 and 3 so when we try to group the broken down equation .ie. x^2+3x+3x+9, we will be able to group it without much thinking into x(x+3)+3(x+3) which will then give us (x+3)(x+3)=0 or (x+3)^2=0. Hope this helped you.
• How do you factor it when the leading coefficient is more than 1? For example, something like 3xsquared+ x-2.
• 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)
• Sometimes I don't understand some of the problems. :(
• I know how that can feel, but it is important to feel confidence in one's self. As ChrisTry162 said factoring out common numbers out can help simplify the problem. Having smaller numbers will always help. I believe in you!
• how do you solve an equation not equal to 0
example 14(squared)+6x=2x
• subtract 2x on both sides to make it equal to 0. I also assume you mean 14 x^2, not just 14 squared. If so, you end up with 14x^2 + 4x = 0. If not, subtract 6x to get 144 = -4x and divide by -4.
• 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.
• 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.
• On the problem 4x^2+4x+1=0 I don't understand why 0.5 is not an answer but -0.5 is.
• The reason why its -1/2 is because:
when you factor 4x^2+4x+1=0 it becomes (2x+1)^2=0
if it is equal to 0.5 then that would become (2(0.5)+1)^2=0, which is to 2^2=0 which is not true
but when x = -0.5 it goes like (2(-0.5)+1)^2=0
=(-1+1)^2=0
(0)^2=0
0=0 , This is a true statement
• I need help with this question:
A right triangle with legs of lengths (x+1) and (2x-2) has an area of 80. What is the length of the shorter leg?
(1 vote)
• I have a slightly different approach than David S. has. I think we get the same result.

area [A] = 1/2 bh. A = 80

A = 1/2 (x+1)(2x-2)
Multiply out 1/2(x+1)(2x-2).

A =1/2(2x^2-2x+2x-2)

A = 1/2( 2x^2-2)
A = (x^2 - 1)

You can remove the parenthesis.

80 = x^2-1

Now, here's where it gets a bit different. Add 1 to both sides and solve for x. What do you get? Remember, this is just x. You still have to plug the value for x into the expressions (x+1) and (2x-2) to see which side is the shorter leg. Hope this helps.

Note: (x^2-1) = (x+1)(x-1). This is where David left you.