Question

If the polynomial $16x^4-24x^3+41x^2-mx+16$ be a perfect square, then the value of "m" is

• A. $12$
• B. $-12$
• C. $24$
• D. $-24$

Answer 1

The correct option is C $24$

It is given that the polynomial $16x^4-24x^3+41x^2-mx+16$ to be a perfect square, therefore, we have:

$16x^4-24x^3+41x^2-mx+16=(a{x}^2+bx+c{)}^2\Rightarrow 16x^4-24x^3+41x^2-mx+16=a^2{x}^4+b^2{x}^2+c^2+2ab{x}^3+2ca{x}^2+2bcx(\because ({a+b+c)}^2=a^2+b^2+c^2+2ab+2bc+2ca)\Rightarrow 16x^4-24x^3+41x^2-mx+16=a^2{x}^4+2ab{x}^3+(2ca+b^2)x^2+2bcx+c^2$

Comparing the coefficient of $x$ on both sides, we get

$a^2=16\Rightarrow a=4c^2=16\Rightarrow c=4$

$2ab=-24\Rightarrow 2\times 4\times b=-24\Rightarrow 8b=-24\Rightarrow b=-\frac{24}{8}\Rightarrow b=-3$

$2bc=-m\Rightarrow 2\times -3\times 4=-m\Rightarrow -24=-m\Rightarrow m=24$

Hence, $m=24$.