Question

Factorise+121b2-88bc+16c2

Answer 1

HERE IS YOUR ANSWER MATE

PLZZZ MARK BRAINLIEST

Answer 2

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

Solution:

Given that,

We have to factorize:

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

$\mathrm{Rewrite\:}121\mathrm{\:as\:}11^2\\\\11^2b^2-88bc+16c^2\\\\\mathrm{Rewrite\:}16\mathrm{\:as\:}4^2\\\\ 11^2b^2-88bc+4^2c^2\\\\ \left(11b\right)^2-88bc+4^2c^2\\\\\mathrm{Rewrite\:}88bc\mathrm{\:as\:}2\cdot \:11b\cdot \:4c\\\\\left(11b\right)^2-2\cdot \:11b\cdot \:4c+\left(4c\right)^2\\\\\mathrm{Apply\:Perfect\:Square\:Formula}:\quad \left(a-b\right)^2=a^2-2ab+b^2\\\\a=11b,\:b=4c\\\\\left(11b-4c\right)^2\\\\(11b - 4c)(11b + 4c)$

Thus the given expression is factored

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