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Solving Quadratic Equations by Factoring Method | Step by Step


This is the easiest method in solving quadratic equation as long as the binomial or trinomial is easily factorable. Otherwise, we will need other methods such as completing the square, or using the quadratic formula.

The following diagram illustrates the main approach in solving quadratic equation by factoring method.


Factoring Method

general concept of using the factoring method to solve quadratic equations


The diagram above suggests the following key points:

  • One side of the equation is just zero.
  • The opposite side should contain the factors of the given polynomial.
  • After the two conditions stated above are met, then it is now OKAY to set each factor equal to zero then solve for the value of the unknown variable.

Direction: Solve the following quadratic equations using the Factoring Method.

1) (x+7)(x-2)=0 See solution
2) -2x^2+6x=0 See solution
3) x^2+3x-10=0 See solution
4) 3x^3-27x=0 See solution
5) x^2+5x+6=20 See solution
6) x^2-x-42=0 exercisesPractice
7) x^2-6x+4=-1 your turnPractice
8) 2x^2+4x-6=0 solve on your ownPractice

Example 1: Solve the quadratic equation (x+7)(x-2)=0 by Factoring Method.

I consider this type of problem as "freebie" because it is already setup for us to find the solutions. Notice that the left side contains factors of some polynomial, and the right side is just zero!

What we need to do is simply set each factor equal to zero, and solve each equation for x.

(x+7)(x-2)=0; x+7=0 or x-2=0; x=-7 and x=2

The answers are x = −7 and x= 2. You may back-substitute these values of x to the original equation to verify if they are true answers. I will leave it to your as an exercise.


Example 2: Solve the quadratic equation -2x^2+6x=0 by Factoring Method.

The left side of the equation is a binomial. That means I can pull out a monomial factor. If you think about it, between the numerical coefficients -2 and 6, I can factor out -2. More so, between x2 and x, I can factor out x. So to find the overall factor (it's like finding the GCF), I will multiply -2 and x to get -2x.

Note, I can also factor out 2x instead of -2x. The final answer should be the same. Try it out!


-2x^2+6x=0; -2x(x-3)=0; -2x=0 or x-3=0; x=0 and x=3

Example 3: Solve the quadratic equation x^2+3x-10=0 using Factoring Method.

Have you factored a trinomial before where the coefficient of the squared term is +1? If not, it is very simple.

To factor this trinomial into two binomials, I need to find two numbers (by trial and error) that satisfy two given conditions:

  • The product of these two numbers is equal to the constant term (last number) which is -10.
  • The sum of these two numbers is equal to the coefficient of the linear term which is +3.


condition #1: multiply two numbers to get -10; condition #2: Add two numbers to get +3


Since the product of two numbers is negative, I know that these numbers must have opposite signs. More so, having a sum of positive number implies that the number with the larger absolute value must be positive.

If you work it out mentally, or using paper and pencil to run through possible combinations, the two numbers that can satisfy the given conditions are +5 and -2.

To check, their products (+5)(-2) = -10, and their sum (+5) + (-2) = +3. Works out great!


x^2+3x-10=0; (x+5)(x-2)=0; x+5=0 or x-2=0; x=-5 and x=2

The final solutions are x = −5 and x= 2.

Example 4: Solve the quadratic equation 3x^3-27x=0 using Factoring Method.

Between the coefficients 3 and -27, I can pull out 3. And between x3 and x, I can take out x. Therefore the overall expression that I can factor out is their product: (3)(x) = 3x.

Notice that after I factored out 3x, I am left with a "special" binomial called the "Difference of Two Squares" which is very easy to factor.


express 9 as squared term; x^2-9=x^2-3^2=(x+3)(x-3)

It is always the case that the middle signs will be opposites (see yellow).

Here is the complete solution.

3x^3-27x=0; 3x(x^2-9)=0; 3x(x+3)(x-3)=0; 3x=0 or x+3=0 or x-3=0; x=0, x=-3 and x=3

You should back-substitute to verify that x = 0 , x= 3 and x= -3 are the correct solutions.


Example 5: Solve the quadratic equation x^2+5x+6=20 using Factoring Method.

The first thing I realize in this problem is that one side of the equation doesn't contain zero. I can easily create a zero on the right side by subtracting both sides by 20.

After doing so, the left side should have a factorable trinomial that is very similar to problem 3.

To factor this trinomial, think of two numbers when multiplied together gives -14 (constant term) and when added gives +5 (coefficient of x-term). By trial and error, the numbers should be -2 and 7. You may verify this correct combination.


x^2+5x+6=20; x^2+5x-14=0; (x-2)(x+7)=0; x-2=0 or x+7=0; x=2 and x=-7


The final answers are x = 2 and x= −7.

Example 6: Solve the quadratic equation x^2-x-42=0 using Factoring Method.


Hint: Similar to example 3.

choice A x=7 and x=6
choice B x=-7 and x=-6
choice C x=-7 and x=6
choice D x=7 and x=-6


Example 7: Solve the quadratic equation x^2-6x+4=-1 using Factoring Method.


Hint: Similar to example 5.

choice A x=-1 and x=-5
choice B x=1 and x=5
choice C x=1 and x=-5
choice D x=-1 and x=5



Example 8: Solve the quadratic equation 2x^2+4x-6=0 using Factoring Method.


Hint: Factor out 2 first, then work it out just like example 3.


choice A x=-3 and x=-1
choice B x=3 and x=-1
choice C x=-3 and x=1
choice D x=3 and x=1



Practice Problems with Answers
Worksheet 1 Worksheet 2


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