# Simplify. `(x^3+4x^2-5)/(-4x^2-x+5)`

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### 1 Answer

`(x^3 + 4x^2 - 5)/(- 4x^2 - x + 5)`

To have a positive leading coefficient at the denominator, factor out the negative.

`= (x^3 + 4x^2 - 5)/(- (4x^2+x-5)) = -(x^3+4x^2-5)/(4x^2+x-5)`

Then, factor the polynomials.

For the numerator, apply factor by grouping. So re-write `4x^2` with the pair factor of -5. And group the terms into two.

`x^3 + 4x^2 - 5 = x^3 - x^2 + 5x^2 - 5 = (x^3 - x^2 ) + (5x^2 -5)`

Factor out the GCF in each group.

`x^3 + 4x^2 - 5 = x^2(x-1) + 5(x^2 -1)`

`x^3 + 4x^2 - 5 = x^2(x-1) + 5(x+1)(x-1)`

`x^3 + 4x^2 - 5 = (x - 1)(x^2 +5(x+1))`

`x^3 + 4x^2 - 5 = (x-1)(x^2 + 5x+ 5)`

For the denominator, let's use factor by grouping too. Multiply the

coefficient of `x^2` with the constant.

`4 * (-5) = -20`

And, re-write x with the pair factor of -20. Then, group the terms into two.

`4x^2 + x - 5 = 4x^2 -4x + 5x - 5 = (4x^2 - 4x) + (5x - 5)`

Factor out the GCF in each group.

`4x^2 + x - 5 = 4x (x - 1) + 5(x-1)`

`4x^2 + x - 5 = (x-1)(4x + 5)`

Then, we have:

`-(x^3+4x^2-5)/(4x^2+x-5) = ((x-1)(x^2+5x+5))/((x-1)(4x+5))`

From here. cancel common factor between numerator and denominator.

`= (x^2+5x+ 5)/(4x+5)`

**Hence, `(x^3 + 4x^2 - 5)/(- 4x^2 - x + 5)= (x^2+5x+ 5)/(4x+5)` .**