Urgent help needed with polynomial factoring+finding zeros

[masterchief007]

Ok...this is my first time on these forums, which should tell you the urgency of the help I need currently.

FINDING RATIONAL ZEROS:

Find all zeroes of f(x) = 2x^3 - 3x^2 + 11x - 40 (IF ONE ZERO IS 5/2)

In that problem so far, I have done synthetic substitution, but after that I get stuck.

This is what I have so far after synthetic subst

f(x) = (x+5/2) (2x^2+2x+16)

what should I do next?! Please give me the procedure ASAP. Thanks in advance.

 

[soroban]

Re: Urgent help needed with polynomial factoring+finding zer

Hello, masterchief007!

Find all zeroes of \(\displaystyle f(x)\:=\: 2x^3\,-\,3x^2\,+\,11x\,-\,40\) (if one zero is \(\displaystyle \frac{5}{2}\))

This is what I have so far after synthetic subst:
. . . \(\displaystyle \,f(x)\:=\x\,+\,\frac{5}{2})(2x^2\,+\,2x\,+\,16)\;\) This is not correct . . . it ends in +40

If a zero is \(\displaystyle \frac{5}{2}\), then a factor is \(\displaystyle (x\,-\,\frac{5}{2})\) or \(\displaystyle (2x\,-\,5)\)

Try the division again . . .

 

[masterchief007]

Thanks for helping man, but I tried (x-5/2) and it doesnt work either.

I officially have no idea what i'm doing now...its taken me 4 pages (front and back) to try just THIS problem and i still havnt gotten it. I'm all out of ideas man...

 

[tkhunny]

2(5/2)^3 - 3(5/2)^2 + 11(5/2) - 40 =
2*(125/8) - 3*(25/4) + 11*(5/2) - 40 =
(125/4) - 3*(25/4) + 22*(5/4) - 160/4 =
(125 - 75 + 110 - 160)/4 = 0/4 = 0

What were you getting?

Have you met Synthetic Division?

Planning to move on to OCS, are we?

 

[Denis]

What is your problem?

Divide 2x^3 - 3x^2 + 11x - 40 by 2x - 5 (as Soroban told you)
and you get x^2 + x + 8

Use quadratic to get x = (-1 +- sqrt(-31)) / 2