Question

the value of n for which the expression x^4+4x^3+nx^2+4x+1 becomes a perfect square

Answer 1

$\textbf{Given:}$

$\text{ x^4+4x^3+nx^2+4x+1 is perfect square}$

$\textbf{To find:}$

$\text{The value of 'n'}$

$\textbf{Solution:}$

$\text{Since x^4+4x^3+nx^2+4x+1 is perfect square, we can write}$

$\bf\,x^4+4x^3+nx^2+4x+1=(x^2+px+1)(x^2+px+1)$

$\text{Equating corresponding coefficient of x^2 and x on bothsides, we get}$

$n=1+1+p^2$ and $4=p+p$

$n=2+p^2$ and $2p=4$

$n=2+p^2$ and $p=2$

$\implies\,n=2+4$

$\implies\bf\,n=6$

$\therefore\textbf{The value of n is 6}$

Find more:

X^4+px^3+rx+s^2 is a perfect square then find p^3+8r

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