Question

If the polynomial ${x^4} + 7{x^3} + 7{x^2} + px + q$ is exactly divisible by ${x^2} + 7x + 12$, then find the values of p and q.

Answer 1

Hint: In the given question, we are given that the polynomial ${x^4} + 7{x^3} + 7{x^2} + px + q$ is exactly divisible by another polynomial ${x^2} + 7x + 12$. This means that the remainder obtained on dividing ${x^4} + 7{x^3} + 7{x^2} + px + q$ by ${x^2} + 7x + 12$ is equal to zero.
So, we do the long division and equate the remainder to zero to find the values of x and y.

Complete answer:
So, in the problem, dividend $ = p\left( x \right) = {x^4} + 7{x^3} + 7{x^2} + px + q$
Divisor $ = g\left( x \right) = {x^2} + 7x + 12$
Now, we will follow the long division method to divide one polynomial by another and then equate the remainder as zero. So, we have,
\[{x}^{2}+7x+12\overset{x^{2}+7}{\overline{\left){\begin{align} & {{x}^{4}}+7{{x}^{3}}+7{{x}^{2}}+px+q \\ & \underline{{{x}^{4}}+7{x^{3}}} \\ & 7{{x}^{2}}+px+q \\ & \underline{7{{x}^{2}}+49x+84} \\ & (p-49)x+(q-84)\\ \end{align}}\right.}}\]
Hence, we have divided the polynomial $p\left( x \right) = {x^4} + 7{x^3} + 7{x^2} + px + q$ by $g\left( x \right) = {x^2} + 7x + 12$ using the long division method and obtained the quotient polynomial as $q\left( x \right) = {x^2} + 7$ and remainder polynomial as $r\left( x \right) = \left( {p - 49} \right)x + \left( {q - 84} \right)$.
Now, we equate the remainder to zero to find the values of p and q. Hence, we get,
$r\left( x \right) = \left( {p - 49} \right)x + \left( {q - 84} \right) = 0$
So, we equate the coefficient of x and constant term as zero. So, we get,
$\left( {p - 49} \right) = 0$ and $\left( {q - 84} \right) = 0$
So, we get the values of p and q as,
$p = 49$ and $q = 84$

Note:
The degree of dividend given to us is $4$ and the degree of the divisor is $2$. So, we must remember that the degree of the remainder must be less than that of the divisor. So, in this case, the degree of remainder obtained is $0$ as zero is a constant polynomial.