Question #a6708

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Answer 1 · 2016-07-19T17:44:16

$x^2+5x-24$ .

Expression $=(x^3+6x^2-19x-24)/(x+1)$

Let us note that for the poly. in $Nr$ ,

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, $(x+1)$ is a factor of the poly. in $Nr.$

Now, $Nr.=x^3+6x^2-19x-24$ ,

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

$=x^2(x+1)+5x(x+1)-24(x+1)$ ,

$=(x+1)(x^2+5x-24)$

The Exp. $=(cancel((x+1))(x^2+5x-24))/cancel(x+1)$

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

Answer 2 · 2016-07-19T18:27:28

$x^2+5x-24$

Calling

$p(x) = x^3+6x^2-19x-24$

we can verify that

$p(-1)=0$

so

$p(x) = (x+1)(x^2+ax+b)$

or

$x^3+6x^2-19x-24 = x^3+(a+1)x^2+(a+b)x+b$

or

$b=-24$
$a+b=-19->a=5$

Finally

$(p(x))/(x+1) = x^2+5x-24$