Question

Polynomial x⁴+7x³+px+q is exactly divisible by x²+7x+12, then find the value of p.

A)-35
B)-94
C)-37
D)35
given solution step by step and explain it after giving a solution​

Answer 1

Answer:

$\huge\bf \blue{Question}$

Polynomial x⁴+7x³+px+q is exactly divisible by x²+7x+12, then find the value of p.

$\huge \bf \blue{Answer}$

Factors of x²+7x+12

x²+7x+12=0

x²+4x+3x+12=0

x(x+4)+3(x+4)=0

(x+4)(x+3)=0

⇒x=-4,-3

Since,p`(x)=x⁴+7x³+7x²+px+q

If p(x) is exactly divisible by x²+7x+12, then x=-4 and x=-3 are its zeroes.So putting x=-4 and x=-3.

p(-4)=(-4)⁴+7(-4)³+7(-4)²+p(-4)+q but p(-4)=0.

0=256-448+112-4p+q

0=-4p+q-80

4p-q=-80

and, p(-3)=(-3)⁴+7(-3)³+7(-3)²+p(-3)+q

but, p(-3)=0

0=81-189+63-3p+q

0=-3p+q-45

3p-q=-45

$\sf{4p - q = - 80} \\ \sf{3p - q = - 45}$

(By changing signs)

$\sf{ \: \: \: \: 4p - q = - 80 } \\ \sf{ - 3p - q = - 45} \\ \sf\frac{}{ \: \: \: \: \: \: p = - 35 \: \: \: \: \: \: \: }$

On putting value of p in equation,

4(-35)-q=-80

-140-q=-80

-q=140-80

-q=60

or

q=-60

Hence,p=-35 and q=-60.

So the option (a)-35 is correct.