Question

If $2x^3+a{x}^2+bx-6$ has $(x-1)$ as a factor and leaves a remainder $2$ when divided by $(x-2)$, find the value of '$a$' and '$b$' respectively.

• A. $-8,12$
• B. $8,-12$
• C. $-4,10$
• D. $4,-10$

Answer 1

The correct option is A $-8,12$

Let $f\left(x\right)=2x^3+a{x}^2+bx-6$

Since $(x-1)$ is the factor of $f\left(x\right)$, therefore, by factor theorem, $f\left(1\right)=0$ that is:

$f\left(1\right)=2(1{)}^3+a(1{)}^2+(b\times 1)-6\Rightarrow 0=(2\times 1)+(a\times 1)+b-6\Rightarrow 0=2+a+b-6\Rightarrow a+b=6-2\Rightarrow a+b=4.......\left(1\right)$

Also, it is given that when $f\left(x\right)$ is divided by $(x-2)$, it leaves remainder $2$, therefore, by remainder theorem, $f\left(2\right)=2$ that is:

$f\left(2\right)=2(2{)}^3+a(2{)}^2+(b\times 2)-6\Rightarrow 2=(2\times 8)+(a\times 4)+2b-6\Rightarrow 2=16+4a+2b-6\Rightarrow 4a+2b=2+6-16\Rightarrow 4a+2b=-8.......\left(2\right)$

Multiply equation 1 by $2$ as follows:

$2(a+b)=2\times 4\Rightarrow 2a+2b=8.....\left(3\right)$

Now, subtract equation 3 from equation 2 as shown below:

$(4a-2a)+(2b-2b)=-8-8\Rightarrow 2a=-16\Rightarrow a=-\frac{16}{2}\Rightarrow a=-8$

Substituting the value of $a$ in equation 1, we have:

$a+b=4\Rightarrow -8+b=4\Rightarrow b=4+8\Rightarrow b=12$

Hence, $a=-8$ and $b=12$.