Question

7.Factorise : 121b2 – 88bc + 16c2​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

The factorization of 121b^2 - 88bc + 16c^2121b

2

−88bc+16c

2

= (11b - 4c)(11b - 4c)(11b−4c)(11b−4c)

Step-by-step explanation:

We have,

121b^2 - 88bc + 16c^2121b

2

−88bc+16c

2

To find, the factorization of 121b^2 - 88bc + 16c^2121b

2

−88bc+16c

2

= ?

∴ 121b^2 - 88bc + 16c^2121b

2

−88bc+16c

2

= (11 b)^2 - 2(11 b)(4c) + (4 c)^2(11b)

2

−2(11b)(4c)+(4c)

2

Using the algebric identity,

(a - b)^2 = a^2 - 2ab + b^2(a−b)

2

=a

2

−2ab+b

2

= (11b - 4c)^2(11b−4c)

2

= (11b - 4c)(11b - 4c)(11b−4c)(11b−4c)

∴ The factorization of 121b^2 - 88bc + 16c^2121b

2

−88bc+16c

2

= (11b - 4c)(11b - 4c)(11b−4c)(11b−4c)

Thus, the factorization of 121b^2 - 88bc + 16c^2121b

2

−88bc+16c

2

is equal to (11b - 4c)(11b - 4c)(11b−4c)(11b−4c) .