Question

If the polynomial $16x^4-24x^3+41x^2-mx+16$ be a perfect square, then the value of "m" is _____.
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• A. 12
• B. -12
• C. 24
• D. -24

Answer 1

Answer: (C)

24

$16x^4-24x^3+41x^2-mx+16=(a{x}^2+bx+c{)}^2$
$\Rightarrow 16x^4-24x^3+41x^2-mx+16$= $a^2{x}^4+b^2{x}^2+c^2+2ab{x}^3+2ac{x}^2+2bcx$
$\Rightarrow 16x^4-24x^3+41x^2-mx+16$= $a^2{x}^4+2ab{x}^3+(2ac+b^2)x^2+2bcx+c^2$
comparing coeffient of x on both side ,we have
$a^2=16\Rightarrow a=4$
$c^2=16\Rightarrow c=4$
2ab=-24
$\Rightarrow 2\times 4b=-24\Rightarrow b=-3$
$\Rightarrow 2\times bc=-m\Rightarrow 2\times (-3)\left(4\right)=-m\Rightarrow m=24$