Answer :
I have attached an image of a long division method that can be used to find the quotient and remainder.
Hope I helped :)
Hope I helped :)
Answer:
[tex]x^2+x-1-\frac{6}{x+4}[/tex]
Step-by-step explanation:
Rational expression form : [tex]\frac{a(x)}{b(x)}=q(x)+\frac{r(x)}{b(x)}[/tex]
a(x) = dividend
b(x) = divisor
q(x) = quotient
r(x) = remainder
Given expression : [tex]\frac{x^3+5x^2+3x-10}{x+4}[/tex]
We know that [tex]Dividend = Divisor \times Quotient+Remainder[/tex]
So, [tex]x^3+5x^2+3x-10 =x+4 \times (x^2+x-1)-6[/tex]
Thus a(x) = dividend = [tex]x^3+5x^2+3x-10 [/tex]
b(x) = divisor= [tex]x+4[/tex]
q(x) = quotient=[tex]x^2+x-1[/tex]
r(x) = remainder=-6
Substitute the value in the rational form.
[tex]\frac{x^3+5x^2+3x-10}{x+4}=x^2+x-1-\frac{6}{x+4}[/tex]
Hence the rational expression [tex]\frac{x^3+5x^2+3x-10}{x+4}[/tex] in the form of [tex]q(x)+\frac{r(x)}{b(x)}[/tex] is [tex]x^2+x-1-\frac{6}{x+4}[/tex]