Question

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

Answer 1

ANSWER:
Let:
$f\left(x\right)=x^3+a{x}^2+bx+6$
x-2 is a factor of $f\left(x\right)=x^3+a{x}^2+bx+6$.
⇒$f\left(2\right)=0\Rightarrow 2^3+a\times 2^2+b\times 2+6=0\Rightarrow 14+4a+2b=0\Rightarrow 4a+2b=-14\Rightarrow 2a+b=-7....\left(1\right)$
Now,
x-3=0⇒x=3
By the factor theorem, we can say:
When f(x) will be divided by x-3, 3 will be its remainder ⇒f(3)=3
Now,
$f\left(3\right)=3^3+a\times 3^2+b\times 3+6=27+9a+3b+6=33+9a+3b$
Thus, we have:
f(3)=3

⇒33+9a+3b=3

⇒9a+3b=-30

⇒3a+b=-10 ...(2)
Subtracting (1) from (2), we get:
a = -3
By putting the value of a in (1), we get the value of b, i.e., -1.
∴ a = -3 and b = -1