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Factor this polynomial. Try factoring this polynomial. Factor the following polynomial.

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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)}.