4x^{2} - 20x + 25 = (ldots ldots ldots)^{2} | vedclass

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

Question

(C) To factorize the quadratic expression $4x^{2} - 20x + 25$,we observe that it follows the algebraic identity $(a - b)^{2} = a^{2} - 2ab + b^{2}$.He

Vedclass · Accepted Answer

$2x - 5$

Vedclass · Answer

(C) To factorize the quadratic expression $4x^{2} - 20x + 25$,we observe that it follows the algebraic identity $(a - b)^{2} = a^{2} - 2ab + b^{2}$.Here,$a^{2} = 4x^{2} \implies a = 2x$.Also,$b^{2} = 25 \implies b = 5$.Checking the middle term: $-2ab = -2(2x)(5) = -20x$,which matches the given expression.Therefore,$4x^{2} - 20x + 25 = (2x - 5)^{2}$.

$4x^{2} - 20x + 25 = (\ldots \ldots \ldots)^{2}$

Similar Questions

Degree of the polynomial $4 x^{4}+0 x^{3}+0 x^{5}+5 x+7$ is

Determine whether $(x+1)$ is a factor of the polynomial $p(x) = x^{3} - 2x^{2} - 5x + 6$.

The coefficient of $x$ in the expansion of $(x+3)^{3}$ is

Find the quotient and the remainder when $x^{3}+x^{2}-10x+8$ is divided by $x+3$.

Factorise $x^{3}-8 y^{3}-27-18 x y$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module