Question

If $2x^3+3x^2+ax+b$ is divided by $(x-2)$ leaves the remainder $2$ and divided by $(x+2)$ leaves the remainder $-2$, then the values of $a,b$ are respectively

• A. $7,12$
• B. $-7,12$
• C. $7,-12$
• D. $-7,-12$

Answer 1

Answer: (D)

$-7,-12$

$f\left(x\right)=2x^3+3x^2+ax+b$

when divided by $(x-2)$

$f\left(2\right)=2\ast 2^3+3\ast 2^2+a\ast 2+b$

$f\left(2\right)=16+12+2a+b=R1=2$

$2a+b=2-28=-26$

$2a+b=-26$...........(equation 1)

when divided by $(x+2)$

$f(-2)=2\ast (-2{)}^3+3\ast (-2{)}^2+a\ast (-2)+b$

$f(-2)=-16+12-2a+b=R2=-2$

$-2a+b=-2+4$

$-2a+b=2$...........(equation 2)

Now solve equation $1$ and $2$

$2a+b=-26$...........(equation 1)

$-2a+b=2$...........(equation 2)

By adding equation $1$ and $2$ we get $b=-12$ and by subtracting we get $a=-7$