Question

The quotient and the remainder when $3x^4-x^3+2x^2-2x-4$ is divided by $(x+2)$ are respectively

• A. $3x^3-7x^2+16x-34;64$
• B. $3x^3+7x^2-16x-34;64$
• C. $3x^3-7x^2-16x-34;64$
• D. $3x^3+7x^2+16x-34;64$

Answer 1

The correct option is A $3x^3-7x^2+16x-34;64$

Let $a{x}^3+b{x}^2+cx+d$ and $r$ be the quotient and reminder.
Hence, we get $3x^4-x^3+2x^2-2x-4=$$(x+2)$$(a{x}^3+b{x}^2+cx+d)$$+$$r$
$\Rightarrow$ $3x^4-x^3+2x^2-2x-4=a{x}^4+b{x}^3+c{x}^2+dx+2a{x}^3+2b{x}^2+2cx+2d+r$
$\Rightarrow$ $3x^4-x^3+2x^2-2x-4=a{x}^4+(b+2a)x^3+(c+2b)x^2+(d+2c)x+2d+r$

Comparing the coefficients of $x^4,x^3,x^2,x$ and $x^0$, we get
$a=3$ .... (1)
$b+2a=-1$ .... (2)
$c+2b=2$ .... (3)
$d+2c=-2$ .... (4)
$r+2d=-4$ .... (5)
Putting $a$ from equation (1) in (2), we get: $b=-1-2a$$b=-1-2\left(3\right)=-7$
Putting $b$ in (3), we get: $c=2-2b=2-2(-7)=16$
Putting $c$ in (4), we get: $d=-2-2c=-2-2\left(16\right)=-34$
Putting $d$ in (5), we get: $r=-4-2d=-4-2(-34)=64$
Hence, the quotient and remainder are $3x^3-7x^2+16x-34$ and $64$.