Question

Rewrite the rational expression x^3+5x^2+3x-10/x+4 in the form q(x)+r(x)/b(x).

Answer 1

I have attached an image of a long division method that can be used to find the quotient and remainder.

Hope I helped :)

Answer 2

Answer:

$x^2+x-1-\frac{6}{x+4}$

Step-by-step explanation:

Rational expression form : $\frac{a(x)}{b(x)}=q(x)+\frac{r(x)}{b(x)}$

a(x) = dividend

b(x) = divisor

q(x) = quotient

r(x) = remainder

Given expression : $\frac{x^3+5x^2+3x-10}{x+4}$

We know that $Dividend = Divisor \times Quotient+Remainder$

So, $x^3+5x^2+3x-10 =x+4 \times (x^2+x-1)-6$

Thus a(x) = dividend =  $x^3+5x^2+3x-10$

b(x) = divisor= $x+4$

q(x) = quotient=$x^2+x-1$

r(x) = remainder=-6

Substitute the value in the rational form.

$\frac{x^3+5x^2+3x-10}{x+4}=x^2+x-1-\frac{6}{x+4}$

Hence the rational expression $\frac{x^3+5x^2+3x-10}{x+4}$ in the form of $q(x)+\frac{r(x)}{b(x)}$ is  $x^2+x-1-\frac{6}{x+4}$