Question:
Factorise:
121a2 − 88ab + 16b2
Solution:
We have:
$121 a^{2}-88 a b+16 b^{2}=(11 a)^{2}-2 \times 11 a \times 4 b+(4 b)^{2}$
$=(11 a-4 b)^{2}$
$\therefore 121 a^{2}-88 a b+16 b^{2}=(11 a-4 b)^{2}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.