Question

x^4+px^3+rx+s^2 is a perfect square then find p^3+8r
Pls solve ​

Answer 1

$\textbf{Given:}$

$\text{ x^4+px^3+rx+s^2 is a perfect square}$

$\textbf{To find:}$

$\text{The value of p^3+8r }$

$\textbf{Solution:}$

$\text{We apply long division to solve this problem}$

$\begin{array}{r|lllllllll}&x^2&+&\frac{p}{2}x&-&\frac{p^2}{8}&&&&\\\cline{2-10}x^2&x^4&+&px^3&+&0x^2&+&r\,x&+&s^2\\&x^4&&&&&&&&\\\cline{2-10}\\2x^2+\frac{p}{2}x&&&px^3&+&0x^2&&&&\\&&&px^3&+&\frac{p^2}{4}x^2&&&&\\\cline{2-10}2x^2+px-\frac{p^2}{8}&&&&&\frac{-p^2}{4}x^2&+&r\,x&+&s^2\\&&&&&\frac{-p^2}{4}x^2&-&\frac{p^3}{8}x&+&\frac{p^3}{64}\\\cline{2-10}&&&&&&&(\frac{p^3}{8}+r)x&+&(s^2-\frac{p^3}{64})\\\cline{2-10}\end{array}$

$\text{Since x^4+px^3+rx+s^2 is a perfect square, the remainder is zero}$

$\implies\,\frac{p^3}{8}+r=0\;\text{and}\;s^2-\frac{p^3}{64}=0$

$\implies\,\frac{p^3+8r}{8}=0$

$\implies\bf\,p^3+8r=0$

$\therefore\textbf{The value of \bf\,p^3+8r is 0}$

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