Question
 $(x-2)$ is a factor of the expression $x^3+a x^2+b x+6$. When this expression is divided by $(x-3)$, it leaves the remainder 3 . Find the values of $a$ and $b$.

Answer

As $x – 2$ is a factor of
$f(x) = x^3 + ax^2 + bx + 6$
$\therefore f(2) = 0$
$\therefore (2)^3 + a(2)^2 + b(2) + 6 = 0$
$\Rightarrow 8 + 4a + 2b + 6 = 0$
$\Rightarrow 4a + 2b = –14$
$\Rightarrow 2a + b = –7 ....(i)$
as on dividing f(x) by$ x – 3$
remainder $= 3$
$\therefore f(3) = 3$
$\therefore (3)^3 + a(3)^2 + b(3) + 6 = 3$
$\Rightarrow 27 + 9a + 3b + 6 = 3$
$\Rightarrow 9a + 3b = –30$
$\Rightarrow 3a + b = –10 ....(ii)$
Solving simultaneously equation (i) and (ii),
$\therefore 2a + b = –7$
$3a + b – 10$
Subtracting $– – +$
$–a = 3 $
$a = –3$
Subtracting value of a in equation (i)
$2(–3) + b = –7$
$\therefore –6 + b = –7$
$\therefore b = –1$
$\therefore a = –3, b = –1.$

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