Question

how to factorise 121 a square + 154 AB + 49 b square

Answer 1

$121 {a}^{2} + 154ab + 49 {b}^{2} \\ \\ = > {(11a)}^{2} + 2(11a)(7b) + {(7b)}^{2} \\ \\ = > {(11a + 7b)}^{2} \\ \\ As \: we \: know \: that :\: {(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2}$

Answer 2

Answer:

Step-by-step explanation:

121 a^2 + 154 ab + 49 b^2

It is similar to a^2 + 2ab+ b^2

So (11 a + 7 b) whole square = 11 a whole square + 7 b whole square + 2 × 11 a × 7 b.

= 121 a^2 + 49 b^2 + 154 ab.

Hence the answer is (11a+7b) whole square.

^ this sign means power that is raised to the power of eg. 2^2 =4