Question

Find the values of m and n if x^4-8x^3+(m +16)x^2+ +(n-16)X +16 is a perfect square.​

Answer 1

Given,

$\longrightarrow\sf{p(x)=x^4-8x^3+(m+16)x^2+(n-16)x+16\quad\quad\dots(1)}$

Since $\sf{p(x)}$ is a perfect square, the square root of $\sf{p(x)}$ should be a quadratic polynomial, since $\sf{p(x)}$ has a degree 4.

We see the coefficient of $\sf{x^4}$ in $\sf{p(x)}$ is 1, so is that of $\sf{x^2}$ in square root of $\sf{p(x).}$ Then let,

$\longrightarrow\sf{p(x)=(x^2+ax+b)^2\quad\quad\dots(2)}$

Equating right hand sides of (1) and (2),

$\longrightarrow\sf{(x^2+ax+b)^2=x^4-8x^3+(m+16)x^2+(n-16)x+16}$

$\longrightarrow\sf{x^4+a^2x^2+b^2+2\left(ax^3+bx^2+abx\right)=x^4-8x^3+(m+16)x^2+(n-16)x+16}$

$\longrightarrow\sf{x^4+2ax^3+\left(a^2+2b\right)x^2+2abx+b^2=x^4-8x^3+(m+16)x^2+(n-16)x+16}$

By equating like terms, we get,

$\longrightarrow\sf{2a=-8\quad\quad\dots(3)}$

$\longrightarrow\sf{a^2+2b=m+16\quad\quad\dots(4)}$

$\longrightarrow\sf{2ab=n-16\quad\quad\dots(5)}$

$\longrightarrow\sf{b^2=16\quad\quad\dots(6)}$

From (3),

$\longrightarrow\sf{a=-4}$

And from (6),

$\longrightarrow\sf{b=\pm4}$

Let $\sf{b=4.}$ Then from (4),

$\longrightarrow\sf{(-4)^2+2\times4=m+16}$

$\longrightarrow\sf{m+16=24}$

$\longrightarrow\sf{\underline{\underline{m=8}}}$

And from (5),

$\longrightarrow\sf{2\times-4\times4=n-16}$

$\longrightarrow\sf{n-16=-32}$

$\longrightarrow\sf{\underline{\underline{n=-16}}}$

Let $\sf{b=-4.}$ Then from (4),

$\longrightarrow\sf{(-4)^2+2\times-4=m+16}$

$\longrightarrow\sf{m+16=8}$

$\longrightarrow\sf{\underline{\underline{m=-8}}}$

And from (5),

$\longrightarrow\sf{2\times-4\times-4=n-16}$

$\longrightarrow\sf{n-16=32}$

$\longrightarrow\sf{\underline{\underline{n=48}}}$