osula9a

2021-12-14

Perform the multiplication or division and simplify.

$\frac{x+3}{4{x}^{2}-9}\xf7\frac{{x}^{2}+7x+12}{2{x}^{2}+7x-15}$

zesponderyd

Beginner2021-12-15Added 41 answers

Step 1

Simplify:

Step 2

Inverting and changing to multiplication:

factorizing the quadratics:

Step 3

cancelling the same terms, we get:

Lindsey Gamble

Beginner2021-12-16Added 38 answers

Step 1

Given:$\frac{x+3}{4{x}^{2}-9}\xf7\frac{{x}^{2}+7x+12}{2{x}^{2}+7x-15}$

Divide$\frac{x+3}{4{x}^{2}-9}$ by $\frac{{x}^{2}+7x+12}{2{x}^{2}+7x-15}$ by multiplying $\frac{x+3}{4{x}^{2}-9}$ by the reciprocal of $\frac{{x}^{2}+7x+12}{2{x}^{2}+7x-15}$

$\frac{(x+3)(2{x}^{2}+7x-15)}{(4{x}^{2}-9)({x}^{2}+7x+12)}$

Factor the expressions that are not already factored.

$\frac{(2x-3)(x+3)(x+5)}{(2x-3)(x+3)(x+4)(2x+3)}$

Cancel out$(2x-3)(x+3)$ in both numerator and denominator.

$\frac{x+5}{(x+4)(2x+3)}$

Expand the expression.

$\frac{x+5}{2{x}^{2}+11x+12}$

Given:

Divide

Factor the expressions that are not already factored.

Cancel out

Expand the expression.

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?

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