Question

If a polynomial $3x^4-15x^3+14x^2+2px-q$ is exactly divisible by $x^2-5x+6$ then find the value of $p$ & $q.$

• A. $p=24,q=10$
• B. $p=10,q=24$
• C. $p=12,q=20$
• D. $p=20,q=10$

Answer 1

Answer: (B)

$p=10,q=24$

${3x}^4-15x^3+14x^2+2px-q$ is divisible by $x^2-5x+16$ i.e. $(x-2)(x-3)$

$\therefore$ $x=2$ & $x=3$, are the zeroes of polynomial

So, at $x=2$ & $x=3$ , the value of polynomial is zero.

$\therefore 3{\left(2\right)}^4-15{\left(2\right)}^3+14{\left(2\right)}^2+2p\left(2\right)-q=0$

$\Rightarrow 48-120+56+4p-q=0$

$\Rightarrow 4p-q=16⟶\left(1\right)$

& $3{\left(3\right)}^4-15{\left(3\right)}^3+14{\left(3\right)}^2+2p\left(3\right)-q$

$\Rightarrow 243-405+126+6p-q=0$

$\Rightarrow 6p-q=36⟶\left(2\right)$

Solving both equ (1) & equ(2)

$2p=20\Rightarrow p=10$

& $q=4\left(10\right)-16=24$