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Question: If the polynomial \({x^4} + 7{x^3} + 7{x^2} + px + q\) is exactly divisible by \({x^2} + 7x + 12\), ...

If the polynomial x4+7x3+7x2+px+q{x^4} + 7{x^3} + 7{x^2} + px + q is exactly divisible by x2+7x+12{x^2} + 7x + 12, then find the values of p and q.

Explanation

Solution

In the given question, we are given that the polynomial x4+7x3+7x2+px+q{x^4} + 7{x^3} + 7{x^2} + px + q is exactly divisible by another polynomial x2+7x+12{x^2} + 7x + 12. This means that the remainder obtained on dividing x4+7x3+7x2+px+q{x^4} + 7{x^3} + 7{x^2} + px + q by x2+7x+12{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(x)=x4+7x3+7x2+px+q = p\left( x \right) = {x^4} + 7{x^3} + 7{x^2} + px + q
Divisor =g(x)=x2+7x+12 = 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,
x2+7x+12)x4+7x3+7x2+px+q x4+7x3‾ 7x2+px+q 7x2+49x+84‾ (p−49)x+(q−84) ‾x2+7{x}^{2}+7x+12\overset{x^{2}+7}{\overline{\left){\begin{aligned} & {{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{aligned}}\right.}}
Hence, we have divided the polynomial p(x)=x4+7x3+7x2+px+qp\left( x \right) = {x^4} + 7{x^3} + 7{x^2} + px + q by g(x)=x2+7x+12g\left( x \right) = {x^2} + 7x + 12 using the long division method and obtained the quotient polynomial as q(x)=x2+7q\left( x \right) = {x^2} + 7 and remainder polynomial as r(x)=(p−49)x+(q−84)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(x)=(p−49)x+(q−84)=0r\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,
(p−49)=0\left( {p - 49} \right) = 0 and (q−84)=0\left( {q - 84} \right) = 0
So, we get the values of p and q as,
p=49p = 49 and q=84q = 84

Note:
The degree of dividend given to us is 44 and the degree of the divisor is 22. 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 00 as zero is a constant polynomial.