Question

Factorise 121b2-88bc+16c2

Answer 1

The factorization of $121b^2 - 88bc + 16c^2$ = $(11b - 4c)(11b - 4c)$

Step-by-step explanation:

We have,

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

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

∴ $121b^2 - 88bc + 16c^2$

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

Using the algebric identity,

$(a - b)^2 = a^2 - 2ab + b^2$

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

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

∴ The factorization of $121b^2 - 88bc + 16c^2$ = $(11b - 4c)(11b - 4c)$

Thus, the factorization of $121b^2 - 88bc + 16c^2$ is equal to $(11b - 4c)(11b - 4c)$.

Answer 2

Aɴsᴡᴇ :-

121b2 – 88bc + 16c2

Here, 44 + 44 = 88 and 44 × 44 = 121 × 16 = 1936

121b2 – 88bc + 16c2

= 121b2 – 44bc – 44bc + 16c2

= 11b(11b – 4c) – 4c(11b – 4c)

= (11b – 4c) (11b – 4c)

= (11b – 4c)2