One of the factors of (25 x^{2}-1)+(1+5 x)^{2 | vedclass

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

Question

(C) Given expression: $(25 x^{2}-1)+(1+5 x)^{2}$Using the identity $a^{2}-b^{2}=(a-b)(a+b)$,we can write $(25 x^{2}-1)$ as $(5 x)^{2}-1^{2}=(5 x-1)(5

Vedclass · Accepted Answer

$10x$

Vedclass · Answer

(C) Given expression: $(25 x^{2}-1)+(1+5 x)^{2}$Using the identity $a^{2}-b^{2}=(a-b)(a+b)$,we can write $(25 x^{2}-1)$ as $(5 x)^{2}-1^{2}=(5 x-1)(5 x+1)$.Substituting this into the expression:$(5 x-1)(5 x+1)+(5 x+1)^{2}$Taking $(5 x+1)$ as a common factor:$(5 x+1) 	imes [(5 x-1)+(5 x+1)]$Simplifying the terms inside the square bracket:$(5 x+1) 	imes (5 x-1+5 x+1) = (5 x+1)(10 x)$Thus,the factors are $10 x$ and $(5 x+1)$.Therefore,$(c)$ is the correct answer.

One of the factors of $(25 x^{2}-1)+(1+5 x)^{2}$ is

Similar Questions

Expand $(3x + 7)(-3x + 7)$.

Without finding the cubes,factorise $(x-2 y)^{3}+(2 y-3 z)^{3}+(3 z-x)^{3}$.

Expand $(2a + 3b)^2$.

Find the zeroes of the polynomial in each of the following:
$h(y) = 2y$

If the polynomials $ax^3 + 4x^2 + 3x - 4$ and $x^3 - 4x + a$ leave the same remainder when each is divided by $x - 3$,find the value of $a$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module