Factorise m 2-7 m -60-16 n 2A- m -3-4 n m -3-4 n B- m-3-4 nm-3-4 nC- m -3-4 n m -3-4 n D- m -2-4 n m -2-4 n

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Standard VIII

Mathematics

Method of Common Factors

Factorise m 2...

Question

Factorise $m^{2} + 7 m - 60 - 16 n^{2}$

$(m + 3 + 4 n) (m + 3 - 4 n)$

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$(m + 3 - 4 n) (m + 3 - 4 n)$

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$(m - 3 + 4 n) (m - 3 - 4 n)$

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$(m + 2 + 4 n) (m + 2 - 4 n)$

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Solution

The correct option is A
$(m + 3 + 4 n) (m + 3 - 4 n)$

Given that $m^{2} + 6 m + 9 - 16 n^{2}$

$m^{2} + 6 m + 9 = (m + 3)^{2}$

$16 n^{2} = (4 n)^{2}$

Hence using difference of squares $x^{2} - y^{2} = (x + y) (x - y)$

Hence $x = (m + 3) a n d y = 4 n$

= $(m + 3 + 4 n) (m + 3 - 4 n)$

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Factorise m2+7m−60−16n2

The correct option is A (m+3+4n)(m+3−4n) Given that m2+6m+9−16n2 m2+6m+9=(m+3)2 16n2=(4n)2 Hence using difference of squaresx2−y2=(x+y)(x−y) Hence x=(m+3) and y=4n = (m+3+4n)(m+3−4n)

Factorise $m^{2} + 7 m - 60 - 16 n^{2}$

The correct option is A
$(m + 3 + 4 n) (m + 3 - 4 n)$

Given that $m^{2} + 6 m + 9 - 16 n^{2}$

$m^{2} + 6 m + 9 = (m + 3)^{2}$

$16 n^{2} = (4 n)^{2}$

Hence using difference of squares $x^{2} - y^{2} = (x + y) (x - y)$

Hence $x = (m + 3) a n d y = 4 n$

= $(m + 3 + 4 n) (m + 3 - 4 n)$