The factors of (a^{2}+4 b^{2}+4 b-4 a b-2 a-8 | vedclass

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

Question

(A) Given expression: $a^{2}+4 b^{2}+4 b-4 a b-2 a-8$Rearranging the terms: $a^{2}+4 b^{2}-4 a b-2 a+4 b-8$Recognizing the perfect square: $(a-2 b)^{2

Vedclass · Accepted Answer

$(a-2 b-4)(a-2 b+2)$

Vedclass · Answer

(A) Given expression: $a^{2}+4 b^{2}+4 b-4 a b-2 a-8$Rearranging the terms: $a^{2}+4 b^{2}-4 a b-2 a+4 b-8$Recognizing the perfect square: $(a-2 b)^{2}-2(a-2 b)-8$Let $x = (a-2 b)$.Substituting $x$ into the expression: $x^{2}-2 x-8$Factoring the quadratic expression: $x^{2}-4 x+2 x-8 = x(x-4)+2(x-4) = (x-4)(x+2)$Substituting back $x = (a-2 b)$: $(a-2 b-4)(a-2 b+2)$

The factors of $(a^{2}+4 b^{2}+4 b-4 a b-2 a-8)$ are:

Similar Questions

Solve the given two equations and select the correct answer from the given options.
$I.$ $5x^2 + 3x - 14 = 0$
$II.$ $10y^2 - 3y - 27 = 0$

The expression $y = ax^2 + bx + c$ has always the same sign as $c$ if

The roots of the equation $a(x^2 + 1) - (a^2 + 1)x = 0$ are

The solutions of the quadratic equation $(3|x| - 3)^2 = |x| + 7$ which belong to the domain of definition of the function $y = \sqrt{x(x - 3)}$ are given by

The sum of the roots of the equation,${x^2} + |2x - 3| - 4 = 0$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module