Question

28v^3+16v^2-21v-12 factor by group please

Answer 1

The factor of $28v^3+16v^2-21v-12$ = $(7v+4)(2v+\sqrt{3})(2v+\sqrt{3})$ or $(7v+4)(4v^2-3)$

Step-by-step explanation:

The given equation:

$28v^3+16v^2-21v-12$

To find, the factors of $28v^3+16v^2-21v-12$ = ?

∴ $28v^3+16v^2-21v-12$

$=(4\times 7)v^3+(4\times 4)v^2-(3\times 7)v-(3\times 4)$

= $4v^2(7v+4)-3(7v+4)$

= $(7v+4)(4v^2-3)$

= $(7v+4)[(2v)^2-\sqrt{3}^2]$

Using the algebraic identity,

$a^{2}-b^{2}=(a+b)(a-b)$

= $(7v+4)(2v+\sqrt{3})(2v+\sqrt{3})$

∴ The factor of $28v^3+16v^2-21v-12$ = $(7v+4)(2v+\sqrt{3})(2v+\sqrt{3})$ or $(7v+4)(4v^2-3)$