Question

Subject: Factorization

1. Factorise:
(a) 14m5n4p² - 42m7n3p⁷ - 70m6n4p³
(b) 2a²(b² - c²) + b² (2c² - 2a²) + 2c²(a² - b²)

2. Express the following as in the form of (a + b)(a - b):
(i) a² - 64
(ii) 20a² - 45b²
(iii) 32x²y² - 8
(iv) x² - 2xy + y² - z²
(v) 49x² - 1

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Answer 1

Question no. 1 :

Equation 1 :

We can factorize the first equation (a) 14m5n4p² - 42m7n3p⁷ - 70m6n4p³ as follows;

$\\\\\sf{14m5n4 {p}^{2} - 42m7n3 {p}^{7} - 70m6n4 {p}^{3} }$

${ : \implies \sf{14\left(m_{5}n_{4}p^{2}-3m_{7}n_{3}p^{7}-5m_{6}n_{4}p^{3}\right) }}$

${ : \implies \sf{p^{2}\left(m_{5}n_{4}-3m_{7}n_{3}p^{5}-5m_{6}n_{4}p\right) }}$

${ : \implies { \pmb{ \sf{ \red{14p^{2}\left(m_{5}n_{4}-3m_{7}n_{3}p^{5}-5m_{6}n_{4}p\right) }}}}}$

∴ Hence, the factorisation for $\bf{14m5n4 {p}^{2} - 42m7n3 {p}^{7} - 70m6n4 {p}^{3}}$will be$\sf{14p^{2}\left(m_{5}n_{4}-3m_{7}n_{3}p^{5}-5m_{6}n_{4}p\right) }$

Equation 2 :

Now, we can factorize the second equation (b) 2a²(b² - c²) + b² (2c² - 2a²) + 2c²(a² - b²) as follows;

$\\\\\sf{2 {a}^{2} ( {b}^{2} - {c}^{2} ) + {b}^{2} (2 {c}^{2} - 2 {a}^{2} ) + 2 {c}^{2} ( {a}^{2} - {b}^{2} )}$

${ : \implies \sf{2a^{2}b^{2}-2a^{2}c^{2}+b^{2}\left(2c^{2}-2a^{2}\right)+2c^{2}\left(a^{2}-b^{2}\right) }}$

${ :\implies \sf{2a^{2}b^{2}-2a^{2}c^{2}+2b^{2}c^{2}-2b^{2}a^{2}+2c^{2}\left(a^{2}-b^{2}\right) }}$

${ :\implies \sf{-2a^{2}c^{2}+2b^{2}c^{2}+2c^{2}\left(a^{2}-b^{2}\right) }}$

${ :\implies \sf{-2a^{2}c^{2}+2b^{2}c^{2}+2c^{2}a^{2}-2b^{2}c^{2} }}$

${ : \implies \sf{2b^{2}c^{2}-2b^{2}c^{2} }}$

$: \implies { \pmb{ \sf{ \red{0 }}}}$

∴ Hence, the value for$\sf{2 {a}^{2} ( {b}^{2} - {c}^{2} ) + {b}^{2} (2 {c}^{2} - 2 {a}^{2} ) + 2 {c}^{2} ( {a}^{2} - {b}^{2} )}$will be 0.

Question 2

${ \underline{\boxed{ \bigstar \: \bold{ ( {a}^{2} - {b}^{2}) = (a + b)(a - b) }}}}$

$❶ \: \: \bold { {a}^{2} - 64 }$

$:\implies \tt{ ({a})^{2} - ( {8})^{2} }$

$:\implies \tt{(a + 8)(a - 8)}$

$❷ \: \: \bold{20 {a}^{2} - 45b }$

$:\implies \tt{(2 \sqrt{5}a)^{2} - (3 \sqrt{5b})^{2} }$

${ :\implies \tt{(2 \sqrt{5}a + 3 \sqrt{5b})(2 \sqrt{5}a - 3 \sqrt{5b} ) }}$

$❸\: \: \bold{36 {x}^{2} {y}^{2} - 8 }$

$:\implies \tt{( {6xy})^{2} - ({2 \sqrt{2} })^{2} }$

${: \implies \tt{(6xy + 2 \sqrt{2})(6xy - 2 \sqrt{2} ) }}$

$❹\: \: \bold{ {x}^{2} - 2xy + {y}^{2} - 2 }$

$:\implies \tt{ {(x - y)}^{2} - ({ \sqrt{ 2} })^{2} }$

$:{ \implies \tt{(x - y + \sqrt{2})(x - y - \sqrt{2}) }}$

$❺ \: \: \bold{49 {x}^{2} - 1}$

$:\implies \tt{ ({7x})^{2} - ({1})^{2} }$

$:{ \implies \tt{(7x + 1)(7x - 1)}}$