Topics
3. Quadratic Equations
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United States of America
Grades 9-12

3.02 Solving by factoring

Lesson
Practice

Introduction

In lesson  11.01 Solving using graphs and tables  , we learned two methods that we can use to solve quadratic equations. In lesson  9.05 Factoring trinomials  , we learned how to factor quadratics in standard form. This lesson will build on those factoring skills and teach us how to use them to solve quadratic equations.

Solving quadratic equations 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.

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 prime 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=-3
\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
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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.

Creating quadratic equations

If we already know the zeros of a function, we can use that information to create an equation.

If x=x_1 is a zero, then (x-x_1) is a factor. And if x=-x_2 is a zero, then (x+x_2) is a factor.

Exploration

Sketch a graph of a quadratic function with the factors (x+2) and (x-3).

  1. Is there a different quadratic function you could graph that would have the same factors?
  2. How many different quadratic functions have those same two factors?
  3. What additional information might we need to describe a unique quadratic function?

The zeros help us find the factors of the quadratic function, but remember that there are many equivalent equations of that function. Using the factors alone will create one equation, but we need to know an additional point on the graph if we are looking for a specific equation.

Examples

Example 3

The graph of a quadratic function has x-intercepts at \left(3,\,0\right) and \left(-1,\,0\right) and passes through the point \left(4,\,10\right). Write an equation in factored form that models this quadratic.

Worked Solution
Create a strategy

To write the equation for this quadratic in factored form we need to first identify the solutions of the equation. We can then substitute these values for x_1 and x_2. Next, we will need to find a by substituting one other point and solving the resulting equation.

The roots of the equation are at \left(-1,\,0\right) and \left(3,\,0\right), so we know the equation has zeros of {x=-1} and x=3.

Apply the idea

Since the zeros are x=-1 and x=3, we can write the factored form as:

y=a(x+1)(x-3)

for some value of a. We can find a by substituting in the coordinates of the additional point, \left(4,\,10\right), into the function.

To find a:

\displaystyle y\displaystyle =\displaystyle a(x+1)(x-3)Factored form
\displaystyle 10\displaystyle =\displaystyle a(4+1)(4-3)Substitute x=4 and y=10
\displaystyle 10\displaystyle =\displaystyle a(5)(1)Evaluate the addition
\displaystyle 10\displaystyle =\displaystyle 5aEvaluate the multiplication
\displaystyle 2\displaystyle =\displaystyle aDivide both sides by 5

The equation of the quadratic function in factored form:

y=2(x+1)(x-3)

Reflect and check

Checking the graph of the equation, we can see that it satisfies the given information.

-4
-3
-2
-1
1
2
3
4
x
-8
-6
-4
-2
2
4
6
8
10
y

Example 4

Find the equation that models the graph shown below.

-4
-2
2
4
6
8
x
1
2
3
4
5
6
7
8
y
Worked Solution
Create a strategy

This quadratic function only has 1 x-intercept, which is also the vertex. When this happens, the function is in the form f(x)=a(x-x_1)^2. Rememeber, we need an additional point, like the y-intercept, to find the exact equation fo this function.

Apply the idea

Since the x-intercept is at (2,0), the function takes the form f(x)=a(x-2)^2.

Next, we can use the y-intercept of \left(0,\dfrac{1}{2}\right) to find the value of the leading coefficient.

\displaystyle f(x)\displaystyle =\displaystyle a(x-2)^2Given equation
\displaystyle \dfrac{1}{2}\displaystyle =\displaystyle a(0-2)^2Substitute x=0 and y=\dfrac{1}{2}
\displaystyle \dfrac{1}{2}\displaystyle =\displaystyle a(-2)^2Evaluate the subtraction
\displaystyle \dfrac{1}{2}\displaystyle =\displaystyle 4aEvaluate the exponent
\displaystyle \dfrac{1}{8}\displaystyle =\displaystyle aDivide both sides by 4

The equation of the graph is f(x)=\dfrac{1}{8}(x-2)^2.

Reflect and check

We could have used the vertex form of a quadratic equation to set up the equation. The vertex is at (2,0), so the equation would take the form f(x)=a(x-2)^2+0. Notice that this is the same thing we got earlier because the vertex is also the x-intercept of the function.

Example 5

The whale jumps out the water at 3 seconds and reenters the water after 6.5 seconds. The whale reaches a maximum height of 49 feet after 4.75 seconds. Determine the equation in standard form that models the whale’s jump.

A humpback whale breaching.
Worked Solution
Create a strategy

If we think of the water level as the x-axis, then the moments where the whale exits and reenters the water would represent the x-intercepts. The maximum height is the vertex of the parabola formed by the whale's jump path.

Apply the idea

Using the x-intercepts of 3 and 6.5, we can create the following equation, where x represents the time in seconds and y represents the height of the jump in feet:

y=a(x-3)(x-6.5)

Next, we can substitute the values of the maximum point, which occurs at (4.75, 49), to find the value of a.

\displaystyle y\displaystyle =\displaystyle a\left(x-3\right)\left(x-6.5\right)Equation for whale's path
\displaystyle 49\displaystyle =\displaystyle a(4.75-3)(4.75-6.5)Substitute y=49 and x=4.75
\displaystyle 49\displaystyle =\displaystyle a(1.75)(-1.75)Evaluate the subtraction
\displaystyle 49\displaystyle =\displaystyle -3.0625aEvaluate the multiplication
\displaystyle -16\displaystyle =\displaystyle aDivision property of equality

Now, we know the factored form of the equation that models the whale's jump:

y=-16(x-3)(x-6.5)

The last step is to get it into standard form. We can do this by multiplying all the factors together.

\displaystyle y\displaystyle =\displaystyle -16\left(x-3\right)\left(x-6.5\right)Equation for whale's path
\displaystyle y\displaystyle =\displaystyle -16(x^2-6.5x-3x+19.5)Distributive property
\displaystyle =\displaystyle -16x^2+104x+48x-312Distributive property
\displaystyle =\displaystyle -16x^2+152x-312Combine like terms

The equation in standard form that models the whale's jump is y=-16x^2+152x-312.

Reflect and check

Notice that the parabola formed by the whale opens downward. If we did not have another point to help us find the value of a, our parabola would have been facing upward. Using the vertex helped us find a negative value for a which is what made the parabola face downward.

Idea summary

We can use the zeros of a function to create an equation. If x=x_1 is a zero, then (x-x_1) is a factor. And if x=-x_2 is a zero, then (x+x_2) is a factor.

To find the specific equation that models a situation, we need to know the zeros and 1 additional point on the parabola.

Outcomes

A.SSE.B.3.A

Factor a quadratic expression to reveal the zeros of the function it defines.

A.CED.A.1

Create equations and inequalities in one variable and use them to solve problems.

A.CED.A.2

Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales.

A.REI.B.4

Solve quadratic equations in one variable.

A.REI.B.4.B

Solve quadratic equations by inspection (e.g. For x^2 = 49), taking square roots, completing the square, the quadratic formula and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them as a ± bi for real numbers a and b.

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