Factorise: 49 a^{2}+70 a b+25 b^{2} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) To factorise the expression $49 a^{2}+70 a b+25 b^{2}$,we observe the terms:$49 a^{2} = (7 a)^{2}$$25 b^{2} = (5 b)^{2}$$70 a b = 2(7 a)(5 b)$Comp

Vedclass · Accepted Answer

$(7a+5b)^{2}$

Vedclass · Answer

(B) To factorise the expression $49 a^{2}+70 a b+25 b^{2}$,we observe the terms:$49 a^{2} = (7 a)^{2}$$25 b^{2} = (5 b)^{2}$$70 a b = 2(7 a)(5 b)$Comparing this with the algebraic identity $(x+y)^{2} = x^{2}+2 x y+y^{2}$,we identify $x = 7 a$ and $y = 5 b$.Substituting these values into the identity,we get:$49 a^{2}+70 a b+25 b^{2} = (7 a+5 b)^{2} = (7 a+5 b)(7 a+5 b)$.

Factorise: $49 a^{2}+70 a b+25 b^{2}$

Similar Questions

Factorise: $\frac{25}{4} x^{2}-\frac{y^{2}}{9}$

Factorise $6x^2 + 17x + 5$ by splitting the middle term,and by using the Factor Theorem.

Find $p(0)$,$p(1)$,and $p(2)$ for the following polynomial: $p(x) = x^{3}$

Factorise $4 x^{2}+y^{2}+z^{2}-4 x y-2 y z+4 x z$.

Evaluate the following using suitable identities: $(102)^{3}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module