Question

The polynomials $P\left(x\right)=a{x}^3+3x^2-13$ and $g\left(x\right)=2x^3-4x+a$ are divided by $(x-3)$. If the remainder in each case is the same, find the value of $a$.

Answer 1

Solution:

$P\left(x\right)=a{x}^3+3x^2-13$

Put $x=3$

$P\left(3\right)=a\times 3^3+3\times 3^2-13$

$P\left(3\right)=27a+27-13$

$P\left(3\right)=27a+14$

$P\left(3\right)=27a+14$ $..............\left(i\right)$

$g\left(x\right)=2x^3-4x+a$

Put $x=3$

$g\left(3\right)=2\times 3^3-4\times 3+a$

$g\left(3\right)=2\times 27-12+a$

$g\left(3\right)=54-12+a$

$g\left(3\right)=42+a$ $...............\left(ii\right)$

Remainders of $P\left(x\right)$ and $g\left(x\right)$ are same.

$\therefore$ $27a+14=42+a$ (from eqn.(i) & (ii))

or, $26a=28$

or, $a=\frac{28}{26}=\frac{14}{13}$

Therefore, value of $a=\frac{14}{13}$