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Quadratic equations can be solved using square roots. Sometimes this method might be inconvenient to use for equations of the form The following lesson introduces another method of solving such equations.

Challenge

Solve a Quadratic Equation

Consider the following quadratic equation for
Can this equation be solved by taking square roots? If not, how can it be solved algebraically — without graphing?

Discussion

Solving an Equation Using the Zero Product Property

An expression written in factored form and set equal to can be solved using the Zero Product Property. When the product of two or more factors is at least one of the factors must equal Consider the following equation.
Since the left-hand side of the equation is in factored form, there are two steps to solve it.
1
Set Each Factor Equal to
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By the Zero Product Property, at least one of the factors must equal Therefore, each of the factors can be set equal to
Note that the word connecting the equations is or. By setting the factors equal to new equations are created. There are as many equations as there were factors — in this case there are two.
2
Solve the Obtained Equations
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Now, inverse operations will be used to solve the new equations.
Solve for
The solutions to the new equations both solve the original equation. Therefore, and are the solutions to

Pop Quiz

Practice Solving Equations Using the Zero Product Property

The following applet shows a quadratic equation whose left-hand side is written in factored form. Using the Zero Product Property find the solutions of the equation. Round to the nearest tenth if necessary.

Interactive applet showing equations in factored form.

Discussion

Solving a Quadratic Equation of the Form

A quadratic equation in the form is obtained when the constant term is equal to Such quadratic equations can be solved using the Zero Product Property. Consider the following equation.
There are three steps to solve it.
1
Factor Out the GCF
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First, the greatest common factor (GCF) of and has to be factored out. In the considered case, and The factors of these two terms will be now listed.
The GCF is Finally, it can be factored out.
Factor out
2
Set Each Factor Equal to
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After factoring out the GCF, the left-hand side of the equation is in factored form. Now, the steps of solving an equation using the Zero Product Property will be followed. By the Zero Product Property each of the factors can be set equal to
3
Solve the Obtained Equations
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The two linear equations obtained in the previous step will be now solved using inverse operations.
The solutions to the original equation are and Please note that makes the expression equal to for any values of and Therefore, it is a solution to any equation of the form

Example

Jumping Dolphins

Dolphins jump out of the water to improve their navigation and to see the surface of the ocean. They also do it for fun.

Dolphins
The following equation models the height in meters of a jumping dolphin seconds after it leaves the water.
After how many seconds is the dolphin back in the water? Write and solve by factoring the appropriate quadratic equation.

Hint

The dolphin is in the water when

Solution

When the dolphin is in the water. An equation for can be written by substituting for in the given equation.
The solutions of the above equation represent the time when the dolphin leaves the water and the time when the dolphin is back in the water. The equation will be now solved. First, it will be rearranged, as it is usually preferred to have the variable terms on the left-hand side.
Next, the greatest common factor (GCF) of the terms on the left-hand side will be factored out. The factors of and are listed below.
The two terms have only one common factor Therefore, the GCF of and is
The left-hand side of the equation is written in factored form and the right-hand side is equal to Therefore, this equation can be solved using the Zero Product Property.
Solve for
The solutions are and It is known that at the dolphin is still in the water. Therefore, it must be back in the water after seconds.

Discussion

Factoring and Solving Quadratic Equations

Method

Factoring a Quadratic Trinomial

When trying to factor a quadratic trinomial of the form it can be difficult to see its factors. Consider the following expression.
Here, and There are six steps to factor this trinomial.
1
Factor Out the GCF of and
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To fully factor a quadratic trinomial, the Greatest Common Factor (GCF) of and has to be factored out first. To identify the GCF of these numbers, their prime factors will be listed.
It can be seen that and share exactly one factor,
Now, can be factored out.
In the remaining steps, the factored coefficient before the parentheses can be ignored. The new considered quadratic trinomial is Therefore, the current values of and are and respectively. If the GCF of the coefficients is this step can be ignored.
2
Find the Factor Pair of Whose Sum Is
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It is known that and so Therefore, the factors must have the same sign. Also, Since the sum of the factors is positive and they must have the same sign, both factors must be positive. All positive factor pairs of can now be listed and their sums checked.

Factors of Sum of Factors
and
and
and

In this case, the correct factor pair is and The following table sums up how to determine the signs of the factors based on the values of and

Factors
Positive Positive Both positive
Positive Negative Both negative
Negative Positive One positive and one negative. The absolute value of the positive factor is greater.
Negative Negative One positive and one negative. The absolute value of the negative factor is greater.

Such analysis makes the list of possible factor pairs shorter.

3
Write as a Sum
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The factor pair obtained in the previous step will be used to rewrite the term — the linear term — of the quadratic trinomial as a sum. Remember that the factors are and
The linear term can be rewritten in the original expression as
4
Factor Out the GCF of the First Two Terms
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The expression has four terms, which can be grouped into the and the Then, the GCF of each group can be factored out.
The first two terms, and can be factored.
The GCF of and is
5
Factor Out the GCF of the Last Two Terms
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The process used in Step will be repeated for the last two terms. In this case, and cannot be factored, so their GCF is
6
Factor Out the Common Factor
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If all the previous steps have been performed correctly, there should now be two terms with a common factor.
The common factor will be factored out.
The factored form of is Remember that the original trinomial was and that the GCF was factored out in Step This GCF has to be included in the final result.

The above method can be used to factor any quadratic trinomial. It is particularly useful for trinomials in the forms or However, in these two cases there are other approaches that might be used as well.

  • A quadratic trinomial is of the form when the constant term is equal to To write this expression in factored form, the GCF of and has to be factored out.
  • A quadratic trinomial is of the form when is equal to Sometimes such expressions can be factored using the Difference of Squares formula.

Please note that not every quadratic trinomial can be factored. When a pair of integers whose product is equal to and whose sum is cannot be found, the trinomial cannot be factored using the described method.

Product of Factors Sum of Factors
The above trinomial cannot be factored at all. Some trinomials can be factored using other methods, like the Difference of Squares formula. However, the coefficients of the obtained factors might no longer be integers. Consider the following example.

Solving a Quadratic Equation by Factoring

Think of a quadratic equation in standard form.
If its left-hand side is factorable, the equation can be solved using the Zero Product Property.

Example

Finding Unknown Dimensions

Zosia is a student studying how many Dolphins swim past a certain area of cove. She marks off a rectangle using rope where she will count how many dolphins pass through in any given day. The study area's rectangle is equal to square meters. However, the dimensions of the rectangle are unknown.

A rectangle
What are the width and the length of the rectangle?

Hint

Based on the given information, write and solve a quadratic equation for

Solution

It is known that the area of the given rectangle is
The area of a rectangle can be calculated by multiplying its dimensions — width and length. In this case, the width and the length of the rectangle are given as algebraic expressions. The product of and represents the area of the rectangle.
Note that both dimensions of the rectangle are given in meters, so there is no need to change the units. The actual dimensions of the rectangle can be determined by solving the above equation. First, the equation will be rewritten in standard form by using the Distributive Property and Subtraction Property of Equality.
The equation is written in the form The values of and can be identified.
It can be seen that and Next, the quadratic trinomial on the left-hand side will be written in factored form following the general method. Note that and have a greatest common factor (GCF) of Next, the product of and will be found so that all pairs of integers whose product is equal to can be listed.
The product and are both negative. Therefore, one of the factors has to be positive and one negative. Additionally, the absolute value of the negative factor has to be greater than the absolute value of the positive factor. The sum of the factors should be
Pair of Factors Sum of Factors
and
and
and
and
and
and
and
and
The pair of factors whose product is equal to and whose sum is is and Now, the linear term can be rewritten as a difference.
Next, the GCF of the first two terms and the last two terms will be factored out. For more information on this process, read about finding the GCF.
Factor out
Note that for the factorization of the polynomial to work, -7 and not the GCF 7 has to be factored out from the last two terms. Finally, the common factor will be factored out.
Since the left-hand side of the equation is written in factored form and the right-hand side is the equation can be solved using the Zero Product Property.
Solve for
Solve for
Two solutions to the quadratic equation have been found. They will now be substituted into the expressions for the width and length of the rectangle.

It can be seen that results in a negative width and length, which is not logical for the dimensions of a rectangle. Therefore, the only valid solution is The width of the rectangle is meters and the length is meters.

Example

Who Is Correct?

Magdalena came to help Zosia in her study. Both were trying to figure out the following quadratic equation.
They want to solve the equation by factoring. Magdalena found that the equation has one solution. Zosia determined that it has no solutions. Who is correct?

Hint

Rewrite the equation in standard form. Factor the left-hand side of the rewritten equation.

Solution

To decide which girl is correct, the given equation will be solved by factoring. First, it will be rewritten in standard form.
In this case and The greatest common factor (GCF) of these numbers is Now, all pairs of integer factors whose product is equal to will be listed. Since is positive and is negative, both factors have to be negative. Their sum will be also calculated in the table, it should be equal to
Pair of Factors Sum of Factors
and
and
and
and
and
It can be noted that the pair of factors whose product is equal to and whose sum is is and The linear term on the left-hand side will be rewritten as a difference and then the trinomial will be factored.
Factor out
Factor out
Note that is a perfect square trinomial, so it could also be factored using the method described here. Since the left-hand side of the equation is written in factored form and the right-hand side is the equation can be solved using the Zero Product Property.
Note that Equation (I) and (II) are the same. Therefore, they have the same solution.
The given quadratic equation has one solution, so Magdalena is correct.

Example

Throwing a Pebble

Zosia, waiting for dolphins to swim by, decides to throw a pebble from an foot cliff into the sea. The following quadratic function describes the height of the pebble above water seconds after the it was thrown.
The height is expressed in feet.
a When does the pebble hit the water?
b Can the pebble reach the height of feet? Write a quadratic equation describing the situation. Solve it by factoring.

Hint

a Substitute for in the given function.
b Substitute for in the given function.

Solution

a The pebble hits the water when its height is equal to By substituting for in the given function the quadratic equation corresponding to this situation is obtained.
The above equation can be solved by factoring. First, it will be rearranged so that its right-hand side is
The equation is written in standard form. In this case and The greatest common factor (GCF) of these numbers will be now identified and factored out to simplify further calculations.
Factor out
The new values of and are and respectively. Now a pair of integer factors whose product is equal to and whose sum is has to be found. Note that and Since the product is negative and is positive the two factors must have different signs. Also, the positive factor must have greater absolute value.
Pair of Factors Sum of Factors
and
and
The pair of factors which meets the desired conditions is and Using these factors, can be rewritten. Then, the expression on the left-hand side of the equation will be fully factored.
Factor
Finally, the Zero Product Property can be applied.
Solve for
Solve for
Two solutions to the equation have been obtained. Remember that represents the number of seconds since the pebble was thrown, so negative values of do not make sense. Therefore, the only valid solution is The pebble hits the water seconds after it was thrown.
b By substituting for an equation corresponding to the given situation will be obtained.
The solution to this equation, if there is one, represents the time that the pebble reaches feet in height. The equation should be solved by factoring. To do this, it will be first rewritten in standard form.
Now, the same steps as in Part A will be followed. Let the GCF be factored and canceled out.
Factor out
The values of and are and respectively. Now all pairs of integer factors whose product is will be listed and their sum calculated. The sum should be Note that since and are both positive, the two factors should be positive as well.
Pair of Factors Sum of Factors
and
and
and
and
As seen above none of the pairs adds up to Therefore, the equation cannot be factored using integers. This might mean that the equation has no solutions. To confirm, the function corresponding to the equation will be graphed.
The quadratic function in standard form will be now graphed. Recall the definitions of the axis of symmetry, the vertex, and the intercept.
The graph of a parabola
As seen above, the function corresponding to the equation has no zeros. Therefore, the equation has no solutions, meaning that the pebble cannot reach a height of feet.

Closure

Usefulness of Factoring

This lesson showed how to use factoring and the Zero Product Property to solve quadratic equations. The factoring method can be also used to solve other types of equations, particularly cubic equations of the following form.
Since the constant term is equal to can be factored out in the equation.
Next, two equations are obtained by the Zero Product Property.
Equation (I) is already solved. Equation (II) is a quadratic equation that can be solved by factoring or by using some other known method. In general, factoring allows the splitting of a more complicated equation into simpler ones.