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How to Factor the Difference of Two Perfect Squares

Identify the coefficient, variable, and degree of each term., Look for a greatest common factor., Determine whether the terms are perfect squares., Make sure you are finding the difference., Set up the formula for the difference of squares., Plug...

13 Steps 6 min read Advanced

Step-by-Step Guide

  1. Step 1: Identify the coefficient

    A coefficient is the number in front of a variable, which is multiplied by the variable.The variable is the unknown value, usually denoted by x{\displaystyle x} or y{\displaystyle y}..

    The degree refers to the exponent of the variable.

    For example, a second-degree term has a value to the second power (x2{\displaystyle x^{2}}) and a fourth-degree term has a value to the fourth power (x4{\displaystyle x^{4}}).For example, in the polynomial 36x4−100x2{\displaystyle 36x^{4}-100x^{2}}, the coefficients are 36{\displaystyle 36} and 100{\displaystyle 100}, the variable is x{\displaystyle x}, and the first term (36x4{\displaystyle 36x^{4}}) is a fourth-degree term, and the second term (100x2{\displaystyle 100x^{2}}) is a second-degree term.

  2. Step 2: variable

    A greatest common factor is the highest factor that divides evenly into into two or more terms.If there is a factor common to both terms of the polynomial, factor this out.For example, the two terms in the polynomial 36x4−100x2{\displaystyle 36x^{4}-100x^{2}} have a greatest common factor of 4x2{\displaystyle 4x^{2}}.

    Factoring this out, the problem becomes 4x2(9x2−25){\displaystyle 4x^{2}(9x^{2}-25)} . , If you factored out a greatest common factor, you are only looking at the terms that remain inside the parentheses.

    A perfect square is the result of multiplying an integer by itself.A variable is a perfect square if its exponent is an even number.You can only factor using the difference of squares if each term in the polynomial is a perfect square.

    For example, 9x2{\displaystyle 9x^{2}} is a perfect square, because (3x)(3x)=9x2{\displaystyle (3x)(3x)=9x^{2}}.

    The number 25{\displaystyle 25} is also a perfect square, because (5)(5)=25{\displaystyle (5)(5)=25}.

    Thus, you can factor 9x2−25{\displaystyle 9x^{2}-25} using the difference of squares formula. , You know you are finding the difference if you have a polynomial that subtracts one term from another.

    The difference of squares only applies to these polynomials, and not those in which addition is used.

    For example, you cannot factor 9x2+25{\displaystyle 9x^{2}+25} using the difference of squares formula, because in this polynomial you are finding a sum, not a difference. , The formula is a2−b2=(a−b)(a+b){\displaystyle a^{2}-b^{2}=(a-b)(a+b)}.

    The terms a2{\displaystyle a^{2}} and b2{\displaystyle b^{2}} are the perfect squares in your polynomial, and a{\displaystyle a} and b{\displaystyle b} are the roots of the perfect squares., This is the value for a{\displaystyle a}.

    To find this value, take the square root of the first perfect square in the polynomial.

    Remember that a square root of a number is a factor you multiply by itself to get that number.

    For example, since (3x)(3x)=9x2{\displaystyle (3x)(3x)=9x^{2}}, the square root of 9x2{\displaystyle 9x^{2}} is 3x{\displaystyle 3x}.

    So you should substitute this value for a{\displaystyle a} in the difference of squares formula: 9x2−25=(3x−b)(3x+b){\displaystyle 9x^{2}-25=(3x-b)(3x+b)}. , This is the value for b{\displaystyle b}, which is the square root of the second term in the polynomial.

    For example, since (5)(5)=25{\displaystyle (5)(5)=25}, the square root of 25{\displaystyle 25} is 5{\displaystyle 5}.

    So you should substitute this value for b{\displaystyle b} in the difference of squares formula: 9x2−25=(3x−5)(3x+5){\displaystyle 9x^{2}-25=(3x-5)(3x+5)}. , Use the FOIL method to multiply the two factors.

    If your result is your original polynomial, you know you have factored correctly.

    For example:(3x−5)(3x+5){\displaystyle (3x-5)(3x+5)}=9x2+15x−15x−25{\displaystyle =9x^{2}+15x-15x-25}=9x2−25{\displaystyle =9x^{2}-25}. , Use the difference of two squares formula: 36x4−9{\displaystyle 36x^{4}-9}.

    The terms have no greatest common factor, so there is no need to factor anything out of the polynomial.

    The term 36x4{\displaystyle 36x^{4}} is a perfect square, since (6x2)(6x2)=36x4{\displaystyle (6x^{2})(6x^{2})=36x^{4}}.

    The term 9{\displaystyle 9} is a perfect square, since (3)(3)=9{\displaystyle (3)(3)=9}.

    The difference of squares formula is a2−b2=(a−b)(a+b){\displaystyle a^{2}-b^{2}=(a-b)(a+b)}.

    Thus, 36x4−9=(a−b)(a+b){\displaystyle 36x^{4}-9=(a-b)(a+b)}, where a{\displaystyle a} and b{\displaystyle b} are the square roots of the perfect squares.

    The square root of 36x4{\displaystyle 36x^{4}} is 6x2{\displaystyle 6x^{2}}.

    Plugging in for a{\displaystyle a} you have 36x4−9=(6x2−b)(6x2+b){\displaystyle 36x^{4}-9=(6x^{2}-b)(6x^{2}+b)}.

    The square root of 9{\displaystyle 9} is 3{\displaystyle 3}.

    So plugging in for b{\displaystyle b}, you have 36x4−9=(6x2−3)(6x2+3){\displaystyle 36x^{4}-9=(6x^{2}-3)(6x^{2}+3)}. , Make sure you factor out a greatest common factor, and use the difference of two squares: 48x3−27x{\displaystyle 48x^{3}-27x}.

    Find the greatest common factor of each term.

    This term is 3x{\displaystyle 3x}, so factor this out of the polynomial: 3x(16x2−9){\displaystyle 3x(16x^{2}-9)}.

    The term 16x2{\displaystyle 16x^{2}} is a perfect square, since (4x)(4x)=16x2{\displaystyle (4x)(4x)=16x^{2}}.

    The term 9{\displaystyle 9} is a perfect square, since (3)(3)=9{\displaystyle (3)(3)=9}.

    The difference of squares formula is a2−b2=(a−b)(a+b){\displaystyle a^{2}-b^{2}=(a-b)(a+b)}.

    Thus, 48x3−27x=3x(a−b)(a+b){\displaystyle 48x^{3}-27x=3x(a-b)(a+b)}, where a{\displaystyle a} and b{\displaystyle b} are the square roots of the perfect squares.

    The square root of 16x2{\displaystyle 16x^{2}} is 4x{\displaystyle 4x}.

    Plugging in for a{\displaystyle a} you have 48x3−27x=3x(4x−b)(4x+b){\displaystyle 48x^{3}-27x=3x(4x-b)(4x+b)}.

    The square root of 9{\displaystyle 9} is 3{\displaystyle 3}.

    So plugging in for b{\displaystyle b}, you have 48x3−27x=3x(4x−3)(4x+3){\displaystyle 48x^{3}-27x=3x(4x-3)(4x+3)}. , It has two variables, but it still follows the rules for the difference of squares method: 4x2−81y2{\displaystyle 4x^{2}-81y^{2}}.

    No factor is common to each term in this polynomial, so there is nothing to factor out before you begin factoring the difference of squares.

    The term 4x2{\displaystyle 4x^{2}} is a perfect square, since (2x)(2x)=4x2{\displaystyle (2x)(2x)=4x^{2}}.

    The term 81y2{\displaystyle 81y^{2}} is a perfect square, since (9y)(9y)=81y2{\displaystyle (9y)(9y)=81y^{2}}.

    The difference of squares formula is a2−b2=(a−b)(a+b){\displaystyle a^{2}-b^{2}=(a-b)(a+b)}.

    Thus, 4x2−81y2=(a−b)(a+b){\displaystyle 4x^{2}-81y^{2}=(a-b)(a+b)}, where a{\displaystyle a} and b{\displaystyle b} are the square roots of the perfect squares.

    The square root of 4x2{\displaystyle 4x^{2}} is 2x{\displaystyle 2x}.

    Plugging in for a{\displaystyle a} you have 4x2−81y2=(2x−b)(2x+b){\displaystyle 4x^{2}-81y^{2}=(2x-b)(2x+b)}.

    The square root of 81y2{\displaystyle 81y^{2}} is 9y{\displaystyle 9y}.

    So plugging in for b{\displaystyle b}, you have 4x2−81y2=(2x−9y)(2x+9y){\displaystyle 4x^{2}-81y^{2}=(2x-9y)(2x+9y)}.

  3. Step 3: and degree of each term.

  4. Step 4: Look for a greatest common factor.

  5. Step 5: Determine whether the terms are perfect squares.

  6. Step 6: Make sure you are finding the difference.

  7. Step 7: Set up the formula for the difference of squares.

  8. Step 8: Plug the first term into the formula.

  9. Step 9: Plug the second term into the formula.

  10. Step 10: Check your work.

  11. Step 11: Factor this polynomial.

  12. Step 12: Try factoring this polynomial.

  13. Step 13: Factor the following polynomial.

Detailed Guide

A coefficient is the number in front of a variable, which is multiplied by the variable.The variable is the unknown value, usually denoted by x{\displaystyle x} or y{\displaystyle y}..

The degree refers to the exponent of the variable.

For example, a second-degree term has a value to the second power (x2{\displaystyle x^{2}}) and a fourth-degree term has a value to the fourth power (x4{\displaystyle x^{4}}).For example, in the polynomial 36x4−100x2{\displaystyle 36x^{4}-100x^{2}}, the coefficients are 36{\displaystyle 36} and 100{\displaystyle 100}, the variable is x{\displaystyle x}, and the first term (36x4{\displaystyle 36x^{4}}) is a fourth-degree term, and the second term (100x2{\displaystyle 100x^{2}}) is a second-degree term.

A greatest common factor is the highest factor that divides evenly into into two or more terms.If there is a factor common to both terms of the polynomial, factor this out.For example, the two terms in the polynomial 36x4−100x2{\displaystyle 36x^{4}-100x^{2}} have a greatest common factor of 4x2{\displaystyle 4x^{2}}.

Factoring this out, the problem becomes 4x2(9x2−25){\displaystyle 4x^{2}(9x^{2}-25)} . , If you factored out a greatest common factor, you are only looking at the terms that remain inside the parentheses.

A perfect square is the result of multiplying an integer by itself.A variable is a perfect square if its exponent is an even number.You can only factor using the difference of squares if each term in the polynomial is a perfect square.

For example, 9x2{\displaystyle 9x^{2}} is a perfect square, because (3x)(3x)=9x2{\displaystyle (3x)(3x)=9x^{2}}.

The number 25{\displaystyle 25} is also a perfect square, because (5)(5)=25{\displaystyle (5)(5)=25}.

Thus, you can factor 9x2−25{\displaystyle 9x^{2}-25} using the difference of squares formula. , You know you are finding the difference if you have a polynomial that subtracts one term from another.

The difference of squares only applies to these polynomials, and not those in which addition is used.

For example, you cannot factor 9x2+25{\displaystyle 9x^{2}+25} using the difference of squares formula, because in this polynomial you are finding a sum, not a difference. , The formula is a2−b2=(a−b)(a+b){\displaystyle a^{2}-b^{2}=(a-b)(a+b)}.

The terms a2{\displaystyle a^{2}} and b2{\displaystyle b^{2}} are the perfect squares in your polynomial, and a{\displaystyle a} and b{\displaystyle b} are the roots of the perfect squares., This is the value for a{\displaystyle a}.

To find this value, take the square root of the first perfect square in the polynomial.

Remember that a square root of a number is a factor you multiply by itself to get that number.

For example, since (3x)(3x)=9x2{\displaystyle (3x)(3x)=9x^{2}}, the square root of 9x2{\displaystyle 9x^{2}} is 3x{\displaystyle 3x}.

So you should substitute this value for a{\displaystyle a} in the difference of squares formula: 9x2−25=(3x−b)(3x+b){\displaystyle 9x^{2}-25=(3x-b)(3x+b)}. , This is the value for b{\displaystyle b}, which is the square root of the second term in the polynomial.

For example, since (5)(5)=25{\displaystyle (5)(5)=25}, the square root of 25{\displaystyle 25} is 5{\displaystyle 5}.

So you should substitute this value for b{\displaystyle b} in the difference of squares formula: 9x2−25=(3x−5)(3x+5){\displaystyle 9x^{2}-25=(3x-5)(3x+5)}. , Use the FOIL method to multiply the two factors.

If your result is your original polynomial, you know you have factored correctly.

For example:(3x−5)(3x+5){\displaystyle (3x-5)(3x+5)}=9x2+15x−15x−25{\displaystyle =9x^{2}+15x-15x-25}=9x2−25{\displaystyle =9x^{2}-25}. , Use the difference of two squares formula: 36x4−9{\displaystyle 36x^{4}-9}.

The terms have no greatest common factor, so there is no need to factor anything out of the polynomial.

The term 36x4{\displaystyle 36x^{4}} is a perfect square, since (6x2)(6x2)=36x4{\displaystyle (6x^{2})(6x^{2})=36x^{4}}.

The term 9{\displaystyle 9} is a perfect square, since (3)(3)=9{\displaystyle (3)(3)=9}.

The difference of squares formula is a2−b2=(a−b)(a+b){\displaystyle a^{2}-b^{2}=(a-b)(a+b)}.

Thus, 36x4−9=(a−b)(a+b){\displaystyle 36x^{4}-9=(a-b)(a+b)}, where a{\displaystyle a} and b{\displaystyle b} are the square roots of the perfect squares.

The square root of 36x4{\displaystyle 36x^{4}} is 6x2{\displaystyle 6x^{2}}.

Plugging in for a{\displaystyle a} you have 36x4−9=(6x2−b)(6x2+b){\displaystyle 36x^{4}-9=(6x^{2}-b)(6x^{2}+b)}.

The square root of 9{\displaystyle 9} is 3{\displaystyle 3}.

So plugging in for b{\displaystyle b}, you have 36x4−9=(6x2−3)(6x2+3){\displaystyle 36x^{4}-9=(6x^{2}-3)(6x^{2}+3)}. , Make sure you factor out a greatest common factor, and use the difference of two squares: 48x3−27x{\displaystyle 48x^{3}-27x}.

Find the greatest common factor of each term.

This term is 3x{\displaystyle 3x}, so factor this out of the polynomial: 3x(16x2−9){\displaystyle 3x(16x^{2}-9)}.

The term 16x2{\displaystyle 16x^{2}} is a perfect square, since (4x)(4x)=16x2{\displaystyle (4x)(4x)=16x^{2}}.

The term 9{\displaystyle 9} is a perfect square, since (3)(3)=9{\displaystyle (3)(3)=9}.

The difference of squares formula is a2−b2=(a−b)(a+b){\displaystyle a^{2}-b^{2}=(a-b)(a+b)}.

Thus, 48x3−27x=3x(a−b)(a+b){\displaystyle 48x^{3}-27x=3x(a-b)(a+b)}, where a{\displaystyle a} and b{\displaystyle b} are the square roots of the perfect squares.

The square root of 16x2{\displaystyle 16x^{2}} is 4x{\displaystyle 4x}.

Plugging in for a{\displaystyle a} you have 48x3−27x=3x(4x−b)(4x+b){\displaystyle 48x^{3}-27x=3x(4x-b)(4x+b)}.

The square root of 9{\displaystyle 9} is 3{\displaystyle 3}.

So plugging in for b{\displaystyle b}, you have 48x3−27x=3x(4x−3)(4x+3){\displaystyle 48x^{3}-27x=3x(4x-3)(4x+3)}. , It has two variables, but it still follows the rules for the difference of squares method: 4x2−81y2{\displaystyle 4x^{2}-81y^{2}}.

No factor is common to each term in this polynomial, so there is nothing to factor out before you begin factoring the difference of squares.

The term 4x2{\displaystyle 4x^{2}} is a perfect square, since (2x)(2x)=4x2{\displaystyle (2x)(2x)=4x^{2}}.

The term 81y2{\displaystyle 81y^{2}} is a perfect square, since (9y)(9y)=81y2{\displaystyle (9y)(9y)=81y^{2}}.

The difference of squares formula is a2−b2=(a−b)(a+b){\displaystyle a^{2}-b^{2}=(a-b)(a+b)}.

Thus, 4x2−81y2=(a−b)(a+b){\displaystyle 4x^{2}-81y^{2}=(a-b)(a+b)}, where a{\displaystyle a} and b{\displaystyle b} are the square roots of the perfect squares.

The square root of 4x2{\displaystyle 4x^{2}} is 2x{\displaystyle 2x}.

Plugging in for a{\displaystyle a} you have 4x2−81y2=(2x−b)(2x+b){\displaystyle 4x^{2}-81y^{2}=(2x-b)(2x+b)}.

The square root of 81y2{\displaystyle 81y^{2}} is 9y{\displaystyle 9y}.

So plugging in for b{\displaystyle b}, you have 4x2−81y2=(2x−9y)(2x+9y){\displaystyle 4x^{2}-81y^{2}=(2x-9y)(2x+9y)}.

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