Question

Rephrasing a question! Can someone tell me where I'm going wrong?

I'm supposed to factor this: −3z2+5z+2
And get this: -3(z+1/3)(z-2)

This is what I've got so far:

−3z2+5z+2

3(z2+5z/3+2/3)

−(5z/3)/2+−(5z/3)/2−2/3−−−−−−−−−−−−√

x1=2/3−−−√

Aaannd here it breaks down...

Answer 1

This step: −(5z/3)/2+−(5z/3)/2−2/3−−−−−−−−−−−−√
Is supposed to read:

\[-(5z/3)/2+-\sqrt{(5z/3)/2-2/3}\]

Answer 2

Nope, it's +2 at the end. :)

Answer 3

I realize there may be several correct answers, but what my book wants and I don't get how to arrive at is -3(z+1/3)(z-2)

Answer 4

Nope, it only gives a hint. Start by breaking out -3. So the first step seems to be -3(z2+5z/3+2/3)

Answer 5

polynomials in factored form

Answer 6

I've only got one identical problem.

This: \[7x^2 - 5x -2\]
Apparently (?) factors into this: \[7(x - 1)(x + 2/7)\]

Answer 7

I think part of the solution is the application of the following formula:

\[x = -p/2 +- \sqrt{(p/2)^2-q}\]

Answer 8

But, I don't know, haven't solved it...

Answer 9

This is actually fun, I just wish I didn't get stuck all the time!

Answer 10

Hmm, I get this:

\[7x^2-5x-2\]
\[-(5/7)/2 +- \sqrt{((5/7)/2)^2+2/7}\]
\[x1: \sqrt{2/7}\]
And stuck... the answers should be 2/7 and -1

Answer 11

well yeah, but x1 is already wrong. I should be getting 2/7, not sqrt(2/7)

Answer 12

There's a facepalm moment coming up soon, I can feel it.

Answer 13

I can't think of a way of factoring x^2 -5x/7-2/7 without it

Answer 14

thanks, I'll watch it

Answer 15

\[\Huge{-3z^2-6z+z+2}\]now take common from \(\Large{(-3z^2-6z)}\) & \(\Large{(z+2)}\) i.e.;\[\Huge{-3z(z+2)+1(z+2)}\]now take common \[\Huge{(-3z+1)(z+2)Ans.}\]

Answer 16

gt it @math0101 ???

Answer 17

@math0101 the answer i have founded is correct .


you can check http://www.wolframalpha.com/input/?i=factor+%3A+%28-3z^2%2B5z%2B2%29

Answer 18

click on this highlighted site :)

Answer 19

that may be, and I appreciate it, but I still don't know how to arrive at the answer the book wants :/

Answer 20

interesting site btw

Answer 21

Do u want me to explain it to you

Answer 22

yeah ! computing site :

Answer 23

can u do it @hba ???

Answer 24

hba: if you please ;)

Answer 25

Yeah it easy ill do it wait