Question

If $x^3+a{x}^2+bx+6$ divided by $x-2$ as factor then remainder becomes O, and leaves remainder $3$ when divided by $x-3$ find the values of a and b.

Answer 1

We are given,

function $F\left(x\right)=x^3+a{x}^2+bx+6$

at $x=2$, we get $y=remainder\Rightarrow y=0$

$f\left(x\right)=(2{)}^3+a(2{)}^2+b\left(2\right)+6=0$

$4a+2b+14=0-------\left(1\right)$

at $x=3$, we get $y=remainder\Rightarrow y=3$

$f\left(3\right)=(3{)}^3+4(3{)}^2+b\left(3\right)+6=3$

$27+9a+3b+6=3$

$9a+3b+30=0----\left(2\right)$

Let, $3\times \left(equation\right(1\left)\right)-2\times \left(equation\right(2\left)\right)$

we get,

$12a+6b+42=018a+6b+60=0-\_\_\_\_\_\_\_\_\_\_\_\_\_-6a-18=0$

$a=-3$

From eqn $\left(1\right)$ $b=\frac{-14-4a}{2}$

$b=-7-2a=-7-2(-3)=-7+6=-1$

$b=-1$

So, $a=-3$ & $b=-1$