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# FP1 Complex numbers watch

1. Complete losing the plot.

2x^3 + ax^2 + bx - 10

has root (3 + i)

other root is therefore (3 - i)

or (x^2 - 6x + 10) together. Find the other root. Missing how I can find the real root and it blatantly obvious. Help?!
2. product of roots is .......
3. 2x^3 + ax^2 + bx - 10

= (x - 3+i)(x - 3-i)(px + q)
4. dude,
remember root1+root2+root3=-d/a?
6+rt3=-10/2=-5
rt3=-11

EDIT:
got it wrong its -1(i hope)
5. (Original post by mathz)
product of roots is .......
Errrrrr, I'm confused as to your question, the product of the complex conjugate roots is 10... how does that help me find the real root? Am I being dense?
6. (Original post by england_sucks)
dude,
remember root1+root2+root3=-d/a?
6+rt3=-10/2=-5
rt3=-11

EDIT:
got it wrong its -1(i hope)
not quite. -d/a is product of roots.
7. (Original post by AnonyMatt)
2x^3 + ax^2 + bx - 10

= (x - 3+i)(x - 3-i)(px + q)
Jaja, I get that but the cubic has two unknown variables so you can't use polynomial division, or well you can but you then have two unknowns, what's the method?
8. (Original post by Clarity Incognito)
Errrrrr, I'm confused as to your question, the product of the complex conjugate roots is 10... how does that help me find the real root? Am I being dense?
if i have 2 roots and i know their product, and also know the product of all 3 roots then i can work out the third root.
9. AAAAAAAAAAAAAAAAAAAAAAAAAH I've got it! Thank you all very much!!!!! I was being completely dense
10. (Original post by mathz)
if i have 2 roots and i know their product, and also know the product of all 3 roots then i can work out the third root.
Thanks, that oversight of mine was laughable.
11. (Original post by mathz)
not quite. -d/a is product of roots.
Grrr.

i'm a little rusty here and there. the sum is -b/a yeah?

just remembered...

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