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.
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)} .
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.
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.
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)}.
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)}.
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}.
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)}.
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)}.
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)}.
