$x^3+ax^2+bx+6$ has $(x-2)$ as a factor and leaves remainder $3$ when divided by $(x-3)$. Find the values of a and b

Question

$$F(x)=x^3+ax^2+bx+6$$ If $x-2=0$ is solution,then for $x=2$ $$F(2)=(2)^3+a(2)^2+b(2)+6=14+4a+2b$$ What will be the value of $a$ and $b$?

Answer 1

hint: You have: $f(3) = 3, f(2) = 0$.

Answer 2

You should have two equations in $a,b$.First, $8+4a+2b+6=0 \implies 2a+b=-7$

and $27+9a+3b+6=3\implies 9a+3b=-30 \implies 3a+b=-10$

So, we have $a=-3$, $b=-1$

Answer 3

$f(2)=0\ $ so $\ f = (x\!-\!2)(x^2\!+\!ax\!-\!3).\ $ $\,3=f(3)= 6+3a\,\Rightarrow\, a=-1$