Question

$p\left(x\right)=x^3+8x^2-7x+12$ and $g\left(x\right)=x-1$. If $p\left(x\right)$ is divided by $g\left(x\right)$, it gives $q\left(x\right)$ and $r\left(x\right)$ as quotient and remainder respectively. If $a$ is the degree of $q\left(x\right)$ and $b$ is the degree of $r\left(x\right)$, then, $(a-b)=?$.

• A. 2
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B

$2$

Remainder $=r\left(x\right)=p\left(1\right)=1^3+8\times 1^2-7\times 1+12=14$
The degree of a polynomial is the highest degree of its terms when the polynomial is expressed in its canonical form consisting of a linear combination of monomials.
So, Degree of $r\left(x\right)=b=0$
$q\left(x\right)=\frac{p\left(x\right)}{g\left(x\right)}$
Now , $p\left(x\right)=x^3+8x^2-7x+12$
$g\left(x\right)=(x-1)$
Performing long division
$x-1\stackrel{x^2+9x+2}{\sqrt{x^3+8x^2-7x+12}}{x}^3-x^2-+\overline{}9x^2-7x9x^2-9x-+\overline{}+2x+122x-2-+\overline{}14$
$\therefore q\left(x\right)=\frac{p\left(x\right)}{g\left(x\right)}=x^2+9x+2$
So, degree of $q\left(x\right)=a=2$
Hence, $a-b=2-0=2$