The values of ${}^{\prime }{a}^{\prime }$ & ${}^{\prime }{b}^{\prime }$ so that the polynomial $x^3-a{x}^2-13x+b$ is divisible by $(x-1)$ & $(x+3)$ are

The correct option is A: $a=3,b=15$
Given, $f\left(x\right)=x^3-a{x}^2-13x+b$

$f\left(x\right)$ is divisible by $(x-1)$ and $(x+3)$

$\therefore f\left(1\right)=0$ and $f(-3)=0$ [Using Factor theorem]

Now, $f\left(1\right)=0$

$(1{)}^3-a(1{)}^2-13\left(1\right)+b=0$

$\Rightarrow 1-a-13+b=0$

$\Rightarrow -a+b=12\dots \dots \left(i\right)$

And, $f(-3)=0$

$(-3{)}^3-a(-3{)}^2-13(-3)+b=0$

$\Rightarrow -27-9a+39+b=0$

$\Rightarrow -9a+b=-12\dots \dots \left(ii\right)$

Now subtracting $eq.\left(ii\right)$ from $\left(i\right),$ we get
$-a+b-(-9a+b)=12-(-12)$

$\Rightarrow 8a=24$

$\Rightarrow a=3$

By putting $a=3$ in equation $\left(i\right)$, we get
$-3+b=12$

$\Rightarrow b=15$

So, $a=3,b=15$