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Maharashtra State BoardSSC (English Medium) 8th Standard

Simplify: a2+10a+21a2+6a−7×a2−1a+3 - Marathi (Second Language) [मराठी (द्वितीय भाषा)]

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Question

Simplify:

\[\frac{a^2 + 10a + 21}{a^2 + 6a - 7} \times \frac{a^2 - 1}{a + 3}\]

Simplify

Solution

It is known that,

a2 − b= (a + b) (a − b)

a3 − b3 = (a − b)(a2 + ab + b2)

\[\ \frac{a^2 + 10a + 21}{a^2 + 6a - 7} \times \frac{a^2 - 1}{a + 3}\] 

\[= \frac{a^2 + 7a + 3a + 21}{a^2 + 7a - a - 7} \times   \frac{\left(a + 1 \right)\left(a - 1 \right)}{\left(a + 3 \right)}\] 

\[= \frac{a\left(a + 7 \right) + 3\left(a + 7 \right)}{a \left(a + 7 \right) - 1\left(a + 7 \right)} \times \frac{\left(a + 1 \right)\left(a - 1 \right)}{\left(a + 3 \right)}\] 

\[ = \frac{\left(a + 7 \right)\left(a + 3 \right)}{\left(a - 1 \right)\left(a + 7 \right)} \times \frac{\left(a + 1 \right)\left(a - 1 \right)}{\left(a + 3 \right)}\] 

= (a + 1)

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Chapter 6: Factorisation of Algebraic expressions - Practice Set 6.4 [Page 33]

APPEARS IN

Balbharati Mathematics [English] 8 Standard Maharashtra State Board
Chapter 6 Factorisation of Algebraic expressions
Practice Set 6.4 | Q 2 | Page 33
Balbharati Integrated 8 Standard Part 2 [English Medium] Maharashtra State Board
Chapter 3.1 Factorisation of Algebraic expressions
Practice Set 6.4 | Q 1 (2) | Page 48
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