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find a third degree polynomial with real coefficients and with zeros 1 and -4+i
the answer given is z^3+7z^2+9z-17

I need the formula to work this problem and then I think I am mostly getting hung up on how to multiply the last part of the problem...I used this formula that someone gave me z-z1*z-z2*z-z3

here is what I did:
(z-1) (z-(-4+i)) (z-(-4-i))
(z-1) (z+4-i) (z+4+i)
(z-1) (z+4)^2 +1
(z-1) (z^2+8z+16+1)
(z-1) (z^2+8z+17)
is this right so far?? if it is then this is where I start having problems, if not then what am I doing wrong??

2007-07-10 13:10:05 · 8 answers · asked by kimmeez 1 in Science & Mathematics Mathematics

thank you to those of you who helped and for the person with the smartaleck einstein reply, you really should not email rude replies........in fact if you have no answer you have no business replying to begin with, don't waste people's time with your rude behavior!! As for your einstein remark , as you can see several people knew how to do it so maybe you should go back and educate yourself so you may provide decent answers to people's genuine questions rather than rude and ignorant remarks. I came here for math help not petty childish remarks.....

2007-07-10 13:30:03 · update #1

8 answers

Looks correct so far.
You might try this to reduce confusion:
(z - 1) (z^2 + 8z + 17) =
(take advantage of the cut-and-paste features of the text editor)
z(z^2 + 8z + 17) - 1(z^2 + 8z + 17) =
z^3 + 8z^2 + 17z - z^2 - 8z - 17 =
z^3 + 7z^2 + 9z - 17

2007-07-10 13:25:06 · answer #1 · answered by Helmut 7 · 0 0

Yes, you are perfectly on the right course.
The next step is to distribute (z-1) over the second degree polynomial, returning:

(z^3+8z^2+17x) - (z^2 + 8z +17) which does reduce to z^3 +7z^2 +9z -17 which is exactly your target result.

2007-07-10 20:16:58 · answer #2 · answered by Vincent G 7 · 0 0

(z-1) (z^2+8z+17) =
z^3 -z^2 +8z^2 -8z + 17z - 17 =
z^3 +7z^2 +9z -17

So, you were right.

2007-07-10 20:18:34 · answer #3 · answered by Steve A 7 · 0 0

correct. then multiply the last.
Just distribute the first factor to the second.
z(z^2+8z+17) -(z^2+8z+17)

d:

2007-07-10 20:15:12 · answer #4 · answered by Alam Ko Iyan 7 · 0 0

z^3+8z^2+17z-z^2-8z-17
z^3+7z^2+9z-17

2007-07-10 20:18:43 · answer #5 · answered by liam_jones_10_10 2 · 0 0

z^3+8z^2+17z-z^2-8z-17
z^3+7z^2+9z-17

2007-07-10 20:15:08 · answer #6 · answered by leo 6 · 0 0

what are you einstein or something, no wonder nobody can help you, nobody knows how do do that.

2007-07-10 20:14:53 · answer #7 · answered by yowuzup 5 · 0 2

that is crack nasty

2007-07-10 20:12:35 · answer #8 · answered by Anonymous · 0 3

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