anonymous
  • anonymous
Find a polynomial equation with integer coefficients that has the given numbers as solutions and the indicated degree. -3, 6, 5; degree 3 A) X^3+8x^2+3x+90=0 B) x^3 -8x^2 -3x +90 =0 C) x^3 -8x^2 -3x - 90 =0 D) x^3 - 8x^2 +3x - 90=0 E) x^3 +8x^2 -3x+90=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
well, it looks like theve given you all the number values to play with, so its just a matter of determing signs
amistre64
  • amistre64
the simplest way to do this would be to use synthetic division to see what zeros out; or simply apply the the values to see what fits
amistre64
  • amistre64
descartes sign rules says that all +s means no postive roots exists all negative signs means the no negative roots exist, since we have a collection of pos and neg roots we can cancel out those with all like signs

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amistre64
  • amistre64
one negative root would imply we have one sign change in the negative version
anonymous
  • anonymous
so what's the answer...because I've never been taught long division.
amistre64
  • amistre64
i think we can narrow it down to b or e
amistre64
  • amistre64
of course you could just rebuild the polynomial from the roots given to get an exacting result
anonymous
  • anonymous
how do I do that??
amistre64
  • amistre64
you go in the opposite direction of finding roots. The idea is, if you can go in one direction, you should be able to turn around and go back again
anonymous
  • anonymous
you should do this one for me...I'm sorry, Im not getting this
amistre64
  • amistre64
umm, nah. im not really up for that option.
amistre64
  • amistre64
if your not getting it, then you should take the time to review your material on this subject to get better acquianted with it
anonymous
  • anonymous
the simplest polynomial with roots of -3, 6, 5 with a degree of 3 is f(x)=(x+3)(x-6)(x-5) if you expand this, you should get what you're looking for
anonymous
  • anonymous
B) \[x^3-8 x^2-3 x+90=(x-6) (x-5) (x+3) \]

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