Question

If the zeroes of monic cubic polynomial are $3,5$ and $6,$ then the cubic polynomial is

• A. $3x^3-42x^2+189x-270$
• B. $2x^3-14x^2+60x-78$
• C. $2x^3-28x^2+126x-180$
• D. $x^3-14x^2+63x-90$

Answer 1

The correct option is D $x^3-14x^2+63x-90$

As the zereos are $3,5$ and $6,$ so the polynomial is
$f\left(x\right)=a(x-3)(x-5)(x-6)$
Since, it is a monic cubic polynomial $\therefore a=1$
$\Rightarrow f\left(x\right)=(x-3)(x-5)(x-6)\Rightarrow f\left(x\right)=(x^2-8x+15)(x-6)\Rightarrow f\left(x\right)=x^3-8x^2+15x-6x^2+48x-90\therefore f\left(x\right)=x^3-14x^2+63x-90$