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Factorize: ...

Question

Factorize: $28 p^{3} q^{2} - 56 p^{2} q^{3} + 16 p^{3} q^{2} - 32 p^{2} q^{3}$

A

44 p^{2} q^{2} (p + 2 q)

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B

44 p^{2} q^{2} (p - 2 q)

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C

44 p^{2} q^{2} (- p - 2 q)

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D

44 p^{2} q^{2} (p^{2} + 2 q)

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Solution

The correct option is B $44 p^{2} q^{2} (p - 2 q)$
$28 p^{3} q^{2} - 56 p^{2} q^{3} + 16 p^{3} q^{2} - 32 p^{2} q^{3}$

$= 28 p^{2} q^{2} (p - 2 q) + 16 p^{2} q^{2} (p - 2 q)$

$= 44 p^{2} q^{2} (p - 2 q)$

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Similar questions

| x | < 1

\frac{d}{d x} [1 + \frac{p}{q} x + \frac{p (p + q)}{2!} {(\frac{x}{q})}^{2} + \frac{p (p + q) (p + 2 q)}{3!} {(\frac{x}{q})}^{3} . . . . \infty] =

q² + 2q + 1

Q. Let P and Q be

3 α 3

P \neq Q

P^{3} = Q^{3}

P^{2} Q = Q^{2} P

P^{2} + Q^{2} | =

P

Q

3 \times 3

P \neq Q

P^{3} = Q^{3}

P^{2} Q = Q^{2} P

, then determinant of

(P^{2} + Q^{2})

Q. Let P and Q be

3 \times 3

matrices such that

P \neq Q

P^{3} = Q^{3}

P^{2} Q = Q^{2} P

then determinant of

(P^{2} + Q^{2})

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