# Question #a6708

Jul 19, 2016

${x}^{2} + 5 x - 24$.

#### Explanation:

Expression$= \frac{{x}^{3} + 6 {x}^{2} - 19 x - 24}{x + 1}$

Let us note that for the poly. in $N r$,

the sum of the co-effs. of odd-powered terms$= 1 - 19 = - 18$, &,

the sum the the co-effs. of even-powered terms$6 - 24 = - 18$.

Therefore, $\left(x + 1\right)$ is a factor of the poly. in $N r .$

Now, $N r . = {x}^{3} + 6 {x}^{2} - 19 x - 24$,

$= {x}^{3} + {x}^{2} + 5 {x}^{2} + 5 x - 24 x - 24$,

$= {x}^{2} \left(x + 1\right) + 5 x \left(x + 1\right) - 24 \left(x + 1\right)$,

$= \left(x + 1\right) \left({x}^{2} + 5 x - 24\right)$

The Exp.$= \frac{\cancel{\left(x + 1\right)} \left({x}^{2} + 5 x - 24\right)}{\cancel{x + 1}}$

$= {x}^{2} + 5 x - 24$.

Jul 19, 2016

${x}^{2} + 5 x - 24$

#### Explanation:

Calling

$p \left(x\right) = {x}^{3} + 6 {x}^{2} - 19 x - 24$

we can verify that

$p \left(- 1\right) = 0$

so

$p \left(x\right) = \left(x + 1\right) \left({x}^{2} + a x + b\right)$

or

${x}^{3} + 6 {x}^{2} - 19 x - 24 = {x}^{3} + \left(a + 1\right) {x}^{2} + \left(a + b\right) x + b$

or

$b = - 24$
$a + b = - 19 \to a = 5$

Finally

$\frac{p \left(x\right)}{x + 1} = {x}^{2} + 5 x - 24$