Topics
8. Quadratic Equations
topic badge
United States of AmericaVirginia
Algebra 1

8.02 Solve quadratics by factoring

Lesson
Practice

Solve quadratics by factoring

Exploration

What values of the variables make each of the following equations true?

  • z-7=0
  • 13a=0
  • 3(d+4)=0
  • x\cdot y=0

The zero product property states that if a product of two or more factors is equal to 0, then at least one of the factors must be equal to 0. That is, if we know that xy=0 then at least one of x=0 or \\y=0 must be true.

We can use this property to solve quadratic equations by first writing the equation in factored form: a\left(x-x_1\right)\left(x-x_2\right)=0

If we can write a quadratic equation in the factored form, then we know that either x-x_1=0 or \\ x-x_2=0. This means that the solutions to the quadratic equation are x=x_1 and x=x_2. This approach can be useful if the equation has rational solutions.

Given a quadratic function f(x), the following statements are equivalent for any real number, k, such that f(k)=0:

  • k is a zero of the function f(x), located at (k,0)

  • (x-k) is a factor of f(x)

  • k is a solution or root of the equation f(x)=0

  • the point (k,0) is an x-intercept for the graph of y=f(x)

Let's take a look at this for a specific function:

-4
-3
-2
-1
1
2
3
4
x
-7
-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
7
f(x)

Here is the graph of f(x)=(x-3)(x+2)

  • -2 and 3 are zeros of the function f(x), located at (-2,0) and (3,0)

  • (x+2) and (x-3) are factors of f(x)

  • -2 and 3 are solutions or roots of the equation f(x)=0

  • the points (-2,0) and (3,0) are x-intercepts for the graph of y=f(x)

Examples

Example 1

Solve the following equations by factoring:

a

x^2+6x-55=0

Worked Solution
Create a strategy

Since there are no common factors for all three terms, we proceed with finding the value of two integers that multiply to ac = (1)(-55) = -55 and add up to b = 6. After finding these integers, we use them to rewrite the middle term 6x as a sum of two terms, and then factor the trinomial by grouping.

The factors of -55 are \pm1,\, \pm 5,\, \pm11 and \pm 55, and we want a factor pair whose sum is 6.

Apply the idea

The factor pair whose sum is 6 is -5 and 11.

We can use this to rewrite the trinomial and factor by grouping as follows:

\displaystyle x^{2} + 6 x - 55\displaystyle =\displaystyle x^{2} + 11x - 5x - 55Rewrite polynomial with four terms
\displaystyle =\displaystyle x(x+11) -5(x+11)Factor each pair
\displaystyle =\displaystyle (x+11)(x-5)Divide out common factor of (x+11)

There are no more common factors to be divided out, so the fully factored form of the polynomial is (x+11)(x-5).

This leads to the equation \left(x-5\right)\left(x+11\right)=0

We can then solve the equation by setting each factor equal to zero, giving us x-5=0 and {x+11=0}, which gives the solutions x=5 and x=-11.

Reflect and check

The solutions of this equation are the x-intercepts because we solved the equation when it was equal to zero.

-10
-8
-6
-4
-2
2
4
x
-60
-50
-40
-30
-20
-10
10
y
b

3x^2+3x-10=8

Worked Solution
Create a strategy

To solve this equation by factoring, we need to create an equivalent equation first. We want the equation to be equal to zero so we can eventually use the zero product property.

Apply the idea
\displaystyle 3x^2+3x-10\displaystyle =\displaystyle 8Given equation
\displaystyle 3x^2+3x-18\displaystyle =\displaystyle 0Subtraction property of equality

Now, we can solve by factoring.

\displaystyle 3(x^2+x-6)\displaystyle =\displaystyle 0Factor out the GCF
\displaystyle 3(x^2 + 3x - 2x - 6)\displaystyle =\displaystyle 0Rewrite the trinomial as a polynomial with four terms
\displaystyle 3[x(x + 3) - 2(x + 3)]\displaystyle =\displaystyle 0Factor each pair of terms
\displaystyle 3(x + 3)(x-2)\displaystyle =\displaystyle 0Divide out common factor of (x+3)
\displaystyle x+3=0\text{ and }x-2\displaystyle =\displaystyle 0Zero product property
\displaystyle x=-3 \text{ and }x\displaystyle =\displaystyle 2Addition property of equality
Reflect and check

We can check our answers by subsituting them back into the original equation to see if they make the equation true. We will check x=-3 first.

\displaystyle 3x^2+3x-10\displaystyle =\displaystyle 8Original equation
\displaystyle 3(-3)^2+3(-3)-10\displaystyle =\displaystyle 8Substitue x=-3
\displaystyle 27-9-10\displaystyle =\displaystyle 8Evaluate the multiplication
\displaystyle 8\displaystyle =\displaystyle 8Evaluate the subtraction

This is a solution to the equation. Now, we will check x=2.

\displaystyle 3x^2+3x-10\displaystyle =\displaystyle 8Original equation
\displaystyle 3(2)^2+3(2)-10\displaystyle =\displaystyle 8Substitue x=2
\displaystyle 12+6-10\displaystyle =\displaystyle 8Evaluate the multiplication
\displaystyle 8\displaystyle =\displaystyle 8Evaluate the subtraction

This also satisfies the equation, so it is a solution.

Example 2

Luis throws a ball straight into the air. The path of the ball can be modeled by the equation {y=-5x^2+14x+3} where x represents the time the ball is in the air in seconds and y represents the height of the ball in meters. How long will it take the ball to hit the ground?

Worked Solution
Create a strategy

The question is asking us to find the time (the x-value) it takes for the ball to hit the ground (the y-value). The ground represents a height of 0. In other words, the question is asking us to solve the equation -5x^2+14x+3=0.

When we factor, we usually have a positive leading coefficient. To begin, we can factor out -1 which will give us a positive leading coefficient.

Apply the idea
\displaystyle -(5x^2-14x-3)\displaystyle =\displaystyle 0Factor out -1
\displaystyle 5x^2-14x-3\displaystyle =\displaystyle 0Divide both sides by -1

Next, we need to find two numbers that multiply to ac=5\cdot -3= -15 and add to b=-14. The factor pair that satisfies these conditions is -15 and 1.

\displaystyle 5x^2-15x+x-3\displaystyle =\displaystyle 0Rewrite the polynomial with 4 terms
\displaystyle 5x(x-3)+1(x-3)\displaystyle =\displaystyle 0Factor by grouping
\displaystyle (5x+1)(x-3)\displaystyle =\displaystyle 0Factor out the GCF of (x-3)
\displaystyle 5x+1=0\text{ and }x-3\displaystyle =\displaystyle 0Zero product property
\displaystyle 5x=-1\text{ and }x\displaystyle =\displaystyle 3Addition property of equality
\displaystyle x=-\dfrac{1}{5}\text{ and }x\displaystyle =\displaystyle 3Division property of equality

The x-values represent time, so a negative value does not make sense since we cannot go backward in time. This means x=-\dfrac{1}{5} is a nonviable solution, and x=3 is the only viable solution.

The ball hit the ground after 3 seconds.

Reflect and check

As we can see from the graph, x=-\dfrac{1}{5} is an x-intercept. But because it does not make sense in context, it is not a solution to the problem. We can picture Luis standing at the y-axis when he throws the ball since x=0 would represent the present moment.

1
2
3
x
2
4
6
8
10
12
y
Idea summary

We can use the zero product property to solve quadratic equations by first writing the equation in factored form: a\left(x-x_1\right)\left(x-x_2\right)=0

then setting each factor equal to zero and solving for x.

Outcomes

A.EI.3

The student will represent, solve, and interpret the solution to a quadratic equation in one variable.

A.EI.3a

Solve a quadratic equation in one variable over the set of real numbers with rational or irrational solutions, including those that can be used to solve contextual problems.

A.EI.3c

Verify possible solution(s) to a quadratic equation in one variable algebraically, graphically, and with technology to justify the reasonableness of answer(s). Explain the solution method and interpret solutions for problems given in context.

A.F.2

The student will investigate, analyze, and compare characteristics of functions, including quadratic and exponential functions, and model quadratic and exponential relationships.

A.F.2d

Make connections between the algebraic (standard and factored forms) and graphical representation of a quadratic function.

A.F.2g

For any value, x, in the domain of f, determine f(x) of a quadratic or exponential function. Determine x given any value f(x) in the range of f of a quadratic function. Explain the meaning of x and f(x) in context.

What is Mathspace

About Mathspace