Expand $(4a - 2b - 3c)^2.$
(N/A) Using the algebraic identity $(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$,we have:
$(4a - 2b - 3c)^2 = [4a + (-2b) + (-3c)]^2$
$= (4a)^2 + (-2b)^2 + (-3c)^2 + 2(4a)(-2b) + 2(-2b)(-3c) + 2(-3c)(4a)$
$= 16a^2 + 4b^2 + 9c^2 - 16ab + 12bc - 24ac$
$(4a - 2b - 3c)^2 = [4a + (-2b) + (-3c)]^2$
$= (4a)^2 + (-2b)^2 + (-3c)^2 + 2(4a)(-2b) + 2(-2b)(-3c) + 2(-3c)(4a)$
$= 16a^2 + 4b^2 + 9c^2 - 16ab + 12bc - 24ac$
Explore More
Similar Questions
Factorise: $12 x^{2}-7 x+1$
Easy
View SolutionFactorise: $x^{3}-2x^{2}-x+2$
Medium
View SolutionFind the remainder when $x^{3}-ax^{2}+6x-a$ is divided by $x-a$.
Medium
View SolutionFind the value of the polynomial $p(x) = 5x - 4x^2 + 3$ at $x = 2$.
Easy
View SolutionDetermine which of the following polynomials has $(x + 1)$ as a factor: $x^{3} + x^{2} + x + 1$.
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo