How do you simplify (64x^3+1)/( 4x^2-100) * (4x+20)/(64x^2-16x+4)64x3+14x2−100⋅4x+2064x2−16x+4?

1 Answer
George C.
Jul 18, 2015

(64x^3+1)/(4x^2-100)*(4x+20)/(64x^2-16x+4)64x3+14x2−100⋅4x+2064x2−16x+4

=(4x+1)/(4(x-5))=4x+14(x−5)

with exclusion x != -5x≠−5

Explanation:

Use sum of cubes identity: a^3+b^3 = (a+b)(a^2-ab+b^2)a3+b3=(a+b)(a2−ab+b2)

Use difference of squares identity: a^2-b^2 = (a-b)(a+b)a2−b2=(a−b)(a+b)

(64x^3+1)/(4x^2-100)*(4x+20)/(64x^2-16x+4)64x3+14x2−100⋅4x+2064x2−16x+4

=(((4x)^3+1^3)*4(x+5))/(4(x^2-5^2)*4((4x)^2-4x+1))=((4x)3+13)⋅4(x+5)4(x2−52)⋅4((4x)2−4x+1)

=(4(4x+1)((4x)^2-4x+1)(x+5))/(16(x-5)(x+5)((4x)^2-4x+1))=4(4x+1)((4x)2−4x+1)(x+5)16(x−5)(x+5)((4x)2−4x+1)

=(4x+1)/(4(x-5))=4x+14(x−5)

with exclusion x != -5x≠−5

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