How do you divide $\frac { x ^ { 2} + 2x - 24} { 20- 5x } + \frac { x ^ { 2} + 4x - 12} { 2x + 4}$ ?

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Answer 1 · 2018-06-22T04:17:32

$(3x^2+4x-84)/(10x+20)$

$(x.(x+6)-4(x+6))/(-5(x-4))+(x^2+4x-12)/(2x+4)$

$((x-4)(x+6))/(-5(-4))+(x^2+4x-12)/(2x+4)$

$(x+6)/(-5)+(x^2+4x-12)/(2x+4)$

$-(x+6)/(5)+(x^2+4x-12)/(2x+4)$

$(-2(2x+4)(x+6)+5(x^2+4x-12))/(5(2x+4))$

$(-(2x^2+12x+4x+24)+5x^2+20x-60)/(10x+20)$

$(-2x^2-16x-24+5x^2+20x-60)/(10x+20)$

$(3x^2+4x-84)/(10x+20)$

How do you divide \frac { x ^ { 2} + 2x - 24} { 20- 5x } + \frac { x ^ { 2} + 4x - 12} { 2x + 4}\frac { x ^ { 2} + 2x - 24} { 20- 5x } + \frac { x ^ { 2} + 4x - 12} { 2x + 4}?