Step-by-step explanation:
The given equation:
[tex]28v^3+16v^2-21v-12[/tex]
To find, the factors of [tex]28v^3+16v^2-21v-12[/tex] = ?
∴ [tex]28v^3+16v^2-21v-12[/tex]
[tex]=(4\times 7)v^3+(4\times 4)v^2-(3\times 7)v-(3\times 4)[/tex]
= [tex]4v^2(7v+4)-3(7v+4)[/tex]
= [tex](7v+4)(4v^2-3)[/tex]
= [tex](7v+4)[(2v)^2-\sqrt{3}^2][/tex]
Using the algebraic identity,
[tex]a^{2}-b^{2}=(a+b)(a-b)[/tex]
= [tex](7v+4)(2v+\sqrt{3})(2v+\sqrt{3})[/tex]
∴ The factor of [tex]28v^3+16v^2-21v-12[/tex] = [tex](7v+4)(2v+\sqrt{3})(2v+\sqrt{3})[/tex] or [tex](7v+4)(4v^2-3)[/tex]
