Factorisation of the expression : $- 2 x^{2} y^{3} + 6 x^{3} y^{2} - 8 x^{2} y^{2}$ results in :

- 2 x^{2} y^{2} (- y - 3 x - 4)

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- 2 x^{2} y^{2} (y - 3 x + 4)

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2 x^{2} y^{2} (3 x + y - 4)

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2 x^{2} y^{2} (3 x - y + 4)

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Solution

The correct option is A $- 2 x^{2} y^{2} (y - 3 x + 4)$
$- 2 x^{2} y^{3} + 6 x^{3} y^{2} - 8 x^{2} y^{2}$

$= [(- 2) \times x^{2} \times y^{2} \times y] + [(- 2) \times (- 3) \times x^{2} \times x \times y^{2}] + [(- 2) \times 4 \times x^{2} \times y^{2}]$

$= - 2 x^{2} y^{2} (y - 3 x + 4)$

$= - 2 x^{2} y^{2} (y - 3 x - 4)$

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Similar questions

If \frac{4 x - 5 y}{3 x + y} = 1, then \frac{2 x^{2} + y^{2}}{2 x^{2} - y^{2}} =

y^{2} = 16 \sqrt{3} x

2 x^{2} + y^{2} = 4

y = 2 x + 2 \sqrt{3}

y = m x + \frac{4 \sqrt{3}}{m}, (m \neq 0)

y^{2} = 16 \sqrt{3} x

2 x^{2} + y^{2} = 4

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m^{4} + 2 m^{2} = 24

2 x^{2} + y^{2} + 8 z^{2} - 2 \sqrt{2} x y + 4 \sqrt{2} y z - 8 x z

Factorise : [4 MARKS]
$1) x^{3} - 2 x^{2} - x + 2$
$2) 2 y^{3} + y^{2} - 2 y - 1$

- 2 x^{2} + 3 x - 4 = 0

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