Factorise- 63a-2 - 112b-2 - Maths Q-A

Factorise the polynomial63a^2 112b^20ex0ex=7(9a^2 16b^2)0ex0ex=7[(3a)^2 (4b)^2]Using the identity 𝐚^2 𝐛^2=(𝐚+𝐛)(𝐚 𝐛)=7(3a+4b)(3a 4b)Hence, the factors of 63a^2 ...

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Standard VIII

Mathematics

Factorisation by Regrouping Terms

Factorise: 63...

Question

Factorise: $63 a^{2} - 112 b^{2}$ .

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Solution

Factorise the polynomial
$63 a^{2} - 112 b^{2} = 7 (9 a^{2} - 16 b^{2}) = 7 [{(3 a)}^{2} - {(4 b)}^{2}]$
Using the identity $a^{2} - b^{2} = (a + b) (a - b)$
$= 7 (3 a + 4 b) (3 a - 4 b)$
Hence, the factors of $63 a^{2} - 112 b^{2}$ are $7, (3 a + 4 b)$ and $(3 a - 4 b)$ .

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$63 a^{2} - 112 b^{2}$

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Q. Factorise

63 a^{2} - 112 b^{2}

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Factorise: 63a2-112b2.

Factorise the polynomial63a2-112b2=79a2-16b2=7(3a)2-(4b)2Using the identity a2-b2=(a+b)(a-b)=7(3a+4b)(3a-4b)Hence, the factors of 63a2-112b2are 7,(3a+4b) and (3a-4b).

Factorise: $63 a^{2} - 112 b^{2}$ .

Factorise the polynomial
$63 a^{2} - 112 b^{2} = 7 (9 a^{2} - 16 b^{2}) = 7 [{(3 a)}^{2} - {(4 b)}^{2}]$
Using the identity $a^{2} - b^{2} = (a + b) (a - b)$
$= 7 (3 a + 4 b) (3 a - 4 b)$
Hence, the factors of $63 a^{2} - 112 b^{2}$ are $7, (3 a + 4 b)$ and $(3 a - 4 b)$ .