How do you subtract $\frac { 2x ^ { 2} + 7x + 3} { 2x ^ { 2} - 5x - 3} - \frac { ( x - 4) ^ { 3} - 1} { x ^ { 2} - 2x - 15}$ ?

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Answer 1 · 2017-07-20T06:37:55

$((5-x)(x^3-11x^2+28x-48))/((x-3)(x^2-2x-14))$

You must have the same denominator : So you multiply and divide each fraction by the denominator of the other fraction.

$(2x^2+7x+3)/(2x^2-5x-3)-((x-4)^3-1)/(x^2-2x-15)=$

$(2x^2+7x+3)/(2x^2-5x-3)*(x^2-2x-15)/(x^2-2x-15)-$

$((x-4)^3-1)/(x^2-2x-15)*(2x^2-5x-3)/(2x^2-5x-3)=$

$((2x^2+7x+3)(x^2-2x-15)-((x-4)^3-1)(2x^2-5x-3))/((2x^2-5x-3)(x^2-2x-15))$

You do the multiplicantions, expansions, the factoring and get :
(It gets messy so I won't do it here but if you want me to just comment so.)

$-((x-5)cancel((2x+1))(x^3-11x^2+28x-48))/((x-3)cancel((2x+1))(x^2-2x-14))=$

$((5-x)(x^3-11x^2+28x-48))/((x-3)(x^2-2x-14))$

How do you subtract \frac { 2x ^ { 2} + 7x + 3} { 2x ^ { 2} - 5x - 3} - \frac { ( x - 4) ^ { 3} - 1} { x ^ { 2} - 2x - 15}\frac { 2x ^ { 2} + 7x + 3} { 2x ^ { 2} - 5x - 3} - \frac { ( x - 4) ^ { 3} - 1} { x ^ { 2} - 2x - 15}?