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Expansion of (x^2 - y^2)

If √x4-10x3...

Question

If $\sqrt{x^{4} - 10 x^{3} + 37 x^{2} - p x + 36}$ is a perfect square, then find the value of $p$ .

A

60

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B

10

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C

- 60

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D

- 10

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Solution

The correct option is C $- 60$
$\div$ $x^{2} - 5 x + 6$ <= Quotient
$x^{2}$
$x^{4} - 10 x^{3} + 37 x^{2} - p x + 36$
$x^{4}$
$2 x^{2} - 5 x$ $- 10 x^{3} + 37 x^{2}$
$- 10 x^{3} + 25 x^{2}$
$2 x^{2} - 10 x + 6$ $12 x^{2} - p x + 36$
$12 x^{2} - 60 x + 36$
Remainder $\Rightarrow 0$
Since the given polynomial is a perfect square, we must have $p = - 60$ .

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