Can anyone explain how this:
p(z) = 2 + 5z - 3z^2
factors into this:
-3(z + 1/3)(z-2)
?

This seems to be a similar problem to the following:

7x^2 -5x -2
becomes:
7(x - 1)(x + 2/7)

I don't understand the steps required to reach these solutions. :(

Question

Can anyone explain how this:
p(z) = 2 + 5z - 3z^2
factors into this:
-3(z + 1/3)(z-2)
?

This seems to be a similar problem to the following:

7x^2 -5x -2
becomes:
7(x - 1)(x + 2/7)

I don't understand the steps required to reach these solutions. :(

mayankdevnani · Answer

by applying factor theorm..

mayankdevnani · Answer

@math0101

anonymous · Answer

-3x^2 + 5x + 2 factor -1 first

mayankdevnani · Answer

first of all , p(z) converts into 0 i.e p(z)=0[remainder theorem]

mayankdevnani · Answer

so, p(-2)=2+5(-2)-3(2)^2
p(-2)=12-12=0

mayankdevnani · Answer

therefore , -2 is a zero of a polynomial

mayankdevnani · Answer

and facrtorize it

anonymous · Answer

nope, still not getting it. :)

Here's what I've got so far:

$$2+5z-3z^2$$
$$-3z^2+5z+2$$
$$3(z^2+5z/3+2/3)$$
$$-(5z/3)/2 +- \sqrt{(5z/3)/2-2/3}$$
$$x1 = \sqrt{2/3}$$

Well... somewhere here everything breaks down. :/

anonymous · Answer

Hmm, I think I'll rephrase the question.