How do you find the factors of $f (z)$ $f(z)$ over C if $f (z) = 2 z^{3} + 3 z^{2} - 14 z - 15$ $f(z)=2z^3+3z^2-14z-15$ ?

Question

2779 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-17T19:14:22

$(z + 1) (z + 3) (2 z - 5)$ $(z+1)(z+3)(2z-5)$ .

We observe that,

The Sum of the co-effs. of odd powered terms of $z$ $z$ is

$2 - 14 = - 12$ $2-14=-12$ ,

and, that of even powered, $3 - 15 = - 12$ $3-15=-12$

We conclude that $(z + 1)$ $(z+1)$ is a factor of $f (z)$ $f(z)$ .

$Now, f (z) = 2 z^{3} + 3 z^{2} - 14 z - 15$ $"Now, "f(z)=2z^3+3z^2-14z-15$

$=ul(2z^3+2z^2)+ul(z^2+z)-ul(15z-15)$

$=2z^2(z+1)+z(z+1)-15(z+1)$

$=(z+1)(2z^2+z-15)$

$=(z+1){ul(2z^2+6z)-ul(5z-15)}$

$=(z+1){2z(z+3)-5(z+3)}$

$=(z+1)(z+3)(2z-5)$ .

How do you find the factors of f(z)f(z)f(z) over C if f(z)=2z^3+3z^2-14z-15f(z)=2z3+3z2−14z−15f(z)=2z^3+3z^2-14z-15?