# Solving quadratics by factoring

CCSS Math: HSA.REI.B.4, HSA.REI.B.4b, HSA.SSE.B.3, HSA.SSE.B.3a, HSF.IF.C.8, HSF.IF.C.8a

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 should be familiar with before taking this lesson

#### 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.

# Solving factored quadratic equations

Suppose we are asked to solve the quadratic equation $(x-1)(x+3)=0$.

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.

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.

### Reflection question

### 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 $x^2-3x-10=0$, then all we have to do is factor $x^2-3x-10$ and solve like before!

$x^2-3x-10$ can be factored as $(x+2)(x-5)$.

The complete solution of the equation would go as follows:

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 $x^2+5x=0$.

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

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

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

# 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:

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:

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.