Factorise:
$9 m^{2} + 24 m + 16$

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Solution

$9 m^{2} + 24 m + 16 = (3 m)^{2} + 2 \times 3 m \times 4 + (4)^{2} = (3 m + 4)^{2} {a^{2} + 2 a b + b^{2} = (a + b)^{2}}$

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Similar questions

9 a^{2} - 12 a b + 4 b^{2} - 9 m^{2} + 6 m n - n^{2}

a^{2} + 4 a b + 4 b^{2} - 9 m^{2} + 6 m n - n^{2}

Factorise:

$3 m^{2} + 24 m + 36$

{(\frac{4}{3} m - \frac{3}{4} n)}^{2} + 2 m n = \frac{16}{9} m^{2} + \frac{9}{16} n^{2}

{(\frac{4}{3} m - \frac{3}{4} n)}^{2} + 2 m n = \frac{16}{9} m^{2} + \frac{9}{16} n^{2}

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