Solve E(x)=(2x+3)^2-x^2/4x+4:x^2-9/8x-24-5

$$E ( x ) = \frac { ( 2 x + 3 ) ^ { 2 } - x ^ { 2 } } { 4 x + 4 } : \frac { x ^ { 2 } - 9 } { 8 x - 24 } - 5$$

Evaluate

$$1$$

Short Solution Steps

Divide \frac{\left(2x+3\right)^{2}-x^{2}}{4x+4} by \frac{x^{2}-9}{8x-24} by multiplying \frac{\left(2x+3\right)^{2}-x^{2}}{4x+4} by the reciprocal of \frac{x^{2}-9}{8x-24}.

$$\frac{\left(\left(2x+3\right)^{2}-x^{2}\right)\left(8x-24\right)}{\left(4x+4\right)\left(x^{2}-9\right)}-5$$

Factor the expressions that are not already factored in \frac{\left(\left(2x+3\right)^{2}-x^{2}\right)\left(8x-24\right)}{\left(4x+4\right)\left(x^{2}-9\right)}.

$$\frac{3\times 8\left(x-3\right)\left(x+1\right)\left(x+3\right)}{4\left(x-3\right)\left(x+1\right)\left(x+3\right)}-5$$

Cancel out 4\left(x-3\right)\left(x+1\right)\left(x+3\right) in both numerator and denominator.

$$2\times 3-5$$

Multiply 2 and 3 to get 6.

$$6-5$$

Subtract 5 from 6 to get 1.

$$1$$

Factor

$$1$$