Question

If $f\left(x\right)=2x^4-13x^2+ax+b$ is divisible by $x^2-3x+2$, then $(a,b)$ is equal to:

• A. $\left(-9,-2\right)$
• B. $\left(6,4\right)$
• C. $\left(9,2\right)$
• D. $\left(2,9\right)$

Answer 1

The correct option is C

$\left(9,2\right)$

Explanation for the correct option.

Step 1: Find the value of $f\left(x\right)$, for the given factors.

$\begin{array}{rcl}{x}^2-3x+2& =& x^2-x-2x+2\\ & =& \left(x-1\right)\left(x-2\right)\end{array}$

Now, as it is given that $f\left(x\right)=2x^4-13x^2+ax+b$ is divisible by $x^2-3x+2$, then for $x=1\mathrm{and}x=2$, $f\left(x\right)$ will be equal to $0$.

So,

$\begin{array}{rcl}f\left(1\right)& =& 0\\ & \Rightarrow & 2{\left(1\right)}^4-13{\left(1\right)}^2+a\left(1\right)+b=0\\ \Rightarrow a+b& =& 11......\left(1\right)\end{array}$

and

$\begin{array}{rcl}f\left(2\right)& =& 0\\ & \Rightarrow & 2{\left(2\right)}^4-13{\left(2\right)}^2+a\left(2\right)+b=0\\ \Rightarrow 2a+b& =& 20......\left(2\right)\end{array}$

Step 2: Find the value of $a\mathrm{and}b$.

On subtracting $\left(1\right)$ from $\left(2\right)$, we get

$a=9$

Substituting the value of $a$ in $\left(1\right)$, we get

$b=2$

So, $(a,b)$ is equal to $(9,2)$.

Hence, option C is correct.