Question

If $x^3-{ax}^2+bx-6$ is exactly divisible by $x^2-5x+6$, then $\frac{a}{b}$ is:

• A. an integer
• B. an irrational number
• C. $\frac{6}{11}$
• D. $\frac{1}{3}$

Answer 1

The correct option is A an integer

Let the given polynomial be $p\left(x\right)=x^3+a{x}^2+bx+6$

Given $p\left(x\right)$ is divisible by $(x–2)$, hence $p\left(2\right)=0$

Put $x=2$ in $p\left(x\right)$

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

$\Rightarrow 0=8+4a+2b+6$

$\Rightarrow 4a+2b=–14$

$\Rightarrow 2a+b=–7$ (1)

It is also given that when $p\left(x\right)$ is divided by $(x–3)$ the remainder is $3$

That $p\left(3\right)=3$

Put $x=3$ in $p\left(x\right)$

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

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

$\Rightarrow 9a+3b=–30$

$\Rightarrow 3a+b=–10$ (2)

Subtract (2) from (1)

$2a+b=–7$

$3a+b=–10$

-------------------

$–a=3$

$\therefore a=–3$

Substitute $a=–3$ in (1), we get

$2(–3)+b=–7$

$\therefore b=–1$

$\frac{a}{b}=\frac{-3}{-1}=3$