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

  • A
    $(a-2 b-4)(a-2 b+2)$
  • B
    $(a-b+2)(a+4 b+4)$
  • C
    $(a+2 b-4)(a+2 b+2)$
  • D
    $(a+2 b-1)(a-2 b+1)$

Explore More

Similar Questions

If the roots of the equation $5x^2 - 7x + k = 0$ are reciprocals of each other,then the value of $k$ is:

$I. 3x^2 - 20x - 32 = 0$
$II. 2y^2 - 3y - 20 = 0$

Difficult
View Solution

If the expression $\left( mx - 1 + \frac{1}{x} \right)$ is always non-negative for $x > 0$,then the minimum value of $m$ must be

Difficult
View Solution

If the difference between the roots of the equation $x^2 + ax + 1 = 0$ is less than $\sqrt{5}$,then the set of possible values of $a$ is

If $8, 2$ are the roots of ${x^2} + ax + \beta = 0$ and $3, 3$ are the roots of ${x^2} + \alpha x + b = 0$,then the roots of ${x^2} + ax + b = 0$ are

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo