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, $x^1=x$.
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 $(x-1)(x+3)=0$ and
• how to use factorization methods in order to bring other equations $($like $x^2-3x-10=0)$ to a factored form and solve them.

Suppose we are asked to solve the quadratic equation $(x-1)(x+3)=0$.
You may wonder how we even know that this equation is quadratic, as it doesn't have anything raised to the second power.
The reason this is a quadratic equation is because it has the product of two expressions that both contain the variable. If we expand this product, we will get the quadratic equation $x^2+2x-3=0$.
This is a product of two expressions that is equal to zero. Note that any $x$ value that makes either $(x-1)$ or $(x+3)$ 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 $x=1$ or $x=-3$ into the equation will result in the true statement $0=0$, so they are both solutions to the equation.
Now solve a few similar equations on your own.
Solve $(x+5)(x+7)=0$.

$(x+5)(x+7)=0$
\begin{aligned}&\swarrow&\searrow\\\\ x+5&=0&x+7&=0\\\\ x&=-5&x&=-7\end{aligned}
In conclusion, the solutions are $x=-5$ and $x=-7$.
Solve $(2x-1)(4x-3)=0$.

$(2x-1)(4x-3)=0$
\begin{aligned}&\swarrow&\searrow\\\\ 2x-1&=0&4x-3&=0\\\\ 2x&=1&4x&=3\\\\ x&=\dfrac{1}{2}&x&=\dfrac{3}{4}\end{aligned}
In conclusion, the solutions are $x=\dfrac{1}{2}$ and $x=\dfrac{3}{4}$.

### Reflection question

Can the same solution method be applied to the equation $(x-1)(x+3)=6$?

The solution method you just learned is only applicable in cases where we have a product that is equal to zero.
This is because only when a product is equal to zero do we know that one of the factors must be zero; we don't care what the other factor is.
In the case of a product that is equal to a non-zero number like $6$, there is no special requirement for one of the factors. All we can say is something about the two factors together. For instance:
• if one of the factors is $1$, the other is $6$;
• if one of the factors is $2$, the other is $3$; or
• if one of the factors is $\dfrac{1}{2}$, the other is $12$.
This doesn't allow us to turn the quadratic equation into two separate linear equations.

### 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.
We already know that the product of any number and zero is equal to zero, but this principle tells us that if the product is zero, then it's certain that one of the factors is zero.
Think about the case where both factors are not zero. In this case, the product wouldn't be zero either. For example, there's nothing you can multiply by $2$ to get zero, except zero.
Therefore, in order for a product to be zero, one of the factors must be 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 $x^2-3x-10=0$, then all we have to do is factor $x^2-3x-10$ and solve like before!
$x^2-3x-10$ can be factored as $(x+2)(x-5)$.
Since the leading coefficient—the coefficient of $x^2$—is $1$, we can use the sum-product pattern.
According to the sum-product pattern, if we find two numbers $a$ and $b$ such that $a+b=\tealD{-3}$ and $a\cdot b=\blueD{-10}$, then $x^2\tealD{-3}x\blueD{-10}=(x+a)(x+b)$.
By considering the pairs of integers whose product is $\blueD{-10}$ and then eliminating those whose sum isn't $\tealD{-3}$, we find that the numbers we need are $a=2$ and $b=-5$.
Therefore, the factored expression is $(x+2)(x-5)$.
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 $x^2+5x=0$.

Step 1. Factor $x^2+5x$ as the product of two linear expressions.$\quad$

To factor $x^2+5x$, all we need to do is take a common factor of $x$ from both terms:
$x^2+5x=x(x+5)$
In conclusion, the factored expression is $x(x+5)$.
Step 2. Solve the equation.

We already found that $x^2+5x$ can be factored as $x(x+5)$. Therefore, we can solve the equation as follows.
\begin{aligned}x^2+5x&=0\\\\ x(x+5)&=0&&\text{Factor.}\end{aligned}
\begin{aligned}&\swarrow&\searrow\\\\ x&=0&x+5&=0\\\\ &&x&=-5\end{aligned}
In conclusion, the solutions are $x=0$ and $x=-5$.

### Solve $x^2-11x+28=0$.

Step 1. Factor $x^2-11x+28$ as the product of two linear expressions.$\quad$

Since the leading coefficient—the coefficient of $x^2$—is $1$, we can use the sum-product pattern.
According to the sum-product pattern, if we find two numbers $a$ and $b$ such that $a+b=\tealD{-11}$ and $a\cdot b=\blueD{28}$, then $x^2\tealD{-11}x+\blueD{28}=(x+a)(x+b)$.
By considering the pairs of integers whose product is $\blueD{28}$ and then eliminating those whose sum isn't $\tealD{-11}$, we find that the numbers we need are $a=-4$ and $b=-7$.
Therefore, the factored expression is $(x-4)(x-7)$.
Step 2. Solve the equation.

We already found that $x^2-11x+28$ can be factored as $(x-4)(x-7)$. Therefore, we can solve the equation as follows.
\begin{aligned}x^2-11x+28&=0\\\\ (x-4)(x-7)&=0&&\text{Factor.}\end{aligned}
\begin{aligned}&\swarrow&\searrow\\\\ x-4&=0&x-7&=0\\\\ x&=4&x&=7\end{aligned}
In conclusion, the solutions are $x=4$ and $x=7$.

### Solve $4x^2+4x+1=0$.

Step 1. Factor $4x^2+4x+1$ as the product of two linear expressions.$\quad$

This expression has the form of a perfect square:
$4x^2+4x+1=(\purpleC{2x})^2+2(\purpleC{2x})(\maroonD{1})+(\maroonD{1})^2$
Therefore, it can be factored as $(\purpleC{2x}+\maroonD{1})^2$, which is the same as $(2x+1)(2x+1)$.
Step 2. Solve the equation.

We already found that $4x^2+4x+1$ can be factored as $(2x+1)^2$. Therefore, we can solve the equation as follows.
\begin{aligned}4x^2+4x+1&=0\\\\ (2x+1)^2&=0&&\text{Factor.}\end{aligned}
In this case, we have a product where the two factors are the same: $2x+1$. There is no need to solve two equations. The only way for $(2x+1)^2$ to be zero is if $2x+1$ is zero.
\begin{aligned}2x+1&=0\\\\ 2x&=-1\\\\ x&=-\dfrac{1}{2}\end{aligned}
In conclusion, the equation has a single solution, $x=-\dfrac{1}{2}$.

### Solve $3x^2+11x-4=0$.

Step 1. Factor $3x^2+11x-4$ as the product of two linear expressions.$\quad$

Since the leading coefficient—the coefficient of $x^2$—is other than $1$, we should use the method of grouping.
To do that, we should first find two numbers $a$ and $b$ such that $a+b=\tealD{11}$ and $ab=(\purpleC{3})(\blueD{-4})=-12$.
By considering the pairs of integers whose product is $-12$ and then eliminating those whose sum isn't $\tealD{11}$, we find that the numbers we need are $a=12$ and $b=-1$. Now we rewrite $\tealD{11}x$ as $(12x-x)$ and group.
\begin{aligned}&\phantom{=}\purpleC{3}x^2+\tealD{11}x\blueD{-4}\\\\ &=3x^2+12x-x-4\\\\ &=3x(x+4)-1(x+4)\\\\ &=(3x-1)(x+4)\end{aligned}
In conclusion, $3x^2+11x-4$ can be factored as $(3x-1)(x+4)$.
Step 2. Solve the equation.

We already found that $3x^2+11x-4$ can be factored as $(3x-1)(x+4)$. Therefore, we can solve the equation as follows.
\begin{aligned}3x^2+11x-4&=0\\\\ (3x-1)(x+4)&=0&&\text{Factor.}\end{aligned}
\begin{aligned}&\swarrow&\searrow\\\\ 3x-1&=0&x+4&=0\\\\ 3x&=1&x&=-4\\\\ x&=\dfrac{1}{3}\end{aligned}
In conclusion, the solutions are $x=\dfrac{1}{3}$ and $x=-4$.

# Arranging the equation before factoring

### One of the sides must be zero.

This is how the solution of the equation $x^2+2x=40-x$ 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 $2x^2-12x+18=0$ 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.
$2x^2-3x-20=x^2+34$

\begin{aligned}2x^2-3x-20&=x^2+34\\\\ 2x^2-3x-20-x^2-34&=0&&\text{Subtract }(x^2+34)\text{.}\\\\ x^2-3x-54&=0&&\text{Combine like terms.}\\\\ (x+6)(x-9)&=0&&\text{Factor.}\end{aligned}
\begin{aligned}&\swarrow&\searrow\\\\ x+6&=0&x-9&=0\\\\ x&=-6&x&=9\end{aligned}
In conclusion, the solutions are $x=-6$ and $x=9$.
Find the solutions of the equation.
$3x^2+33x+30=0$
\begin{aligned}3x^2+33x+30&=0\\\\ x^2+11x+10&=0&&\text{Divide by 3.}\\\\ (x+1)(x+10)&=0&&\text{Factor}.\end{aligned}
\begin{aligned}&\swarrow&\searrow\\\\ x+1&=0&x+10&=0\\\\ x&=-1&x&=-10\end{aligned}
In conclusion, the solutions are $x=-1$ and $x=-10$.
$3x^2-9x-20=x^2+5x+16$
\begin{aligned}3x^2-9x-20&=x^2+5x+16\\\\ 3x^2-9x-20-x^2-5x-16&=0&&\text{Subtract }(x^2+5x+16)\text{.}\\\\ 2x^2-14x-36&=0&&\text{Combine like terms.}\\\\ x^2-7x-18&=0&&\text{Divide by 2.}\\\\ (x+2)(x-9)&=0&&\text{Factor.}\end{aligned}
\begin{aligned}&\swarrow&\searrow\\\\ x+2&=0&x-9&=0\\\\ x&=-2&x&=9\end{aligned}
In conclusion, the solutions are $x=-2$ and $x=9$.