Factorise 169 x^{2}-625. | vedclass

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

Question

(A) The given expression is $169 x^{2}-625$.We can write this as $(13 x)^{2}-(25)^{2}$.Using the algebraic identity $a^{2}-b^{2} = (a+b)(a-b)$,where $

Vedclass · Accepted Answer

$(13 x+25)(13 x-25)$

Vedclass · Answer

(A) The given expression is $169 x^{2}-625$.We can write this as $(13 x)^{2}-(25)^{2}$.Using the algebraic identity $a^{2}-b^{2} = (a+b)(a-b)$,where $a = 13 x$ and $b = 25$,we get:$(13 x)^{2}-(25)^{2} = (13 x+25)(13 x-25)$.

Factorise $169 x^{2}-625$.

Similar Questions

Factorise:
$2 \sqrt{2} a^{3} + 8 b^{3} - 27 c^{3} + 18 \sqrt{2} a b c$

Evaluate $(98)^{2}$.

Factorise the following:
$25 x^{2}+16 y^{2}+4 z^{2}-40 x y+16 y z-20 x z$

Expand $(11 x+18)(11 x-18)$.

Factorise $49 x^{2}-42 x+9$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module