Use Descartes' rule of signs to describe possible combinations of roots for the polynomial h(x)=2x^3+5x^2-31x-15
http://www.purplemath.com/modules/drofsign.htm see if this helps and let me know
would there be 2 combinations of roots?
one positive root...
2 or 0 negative roots
h(x)= + 2x^3 + 5x^2 - 31x - 15 no yes no signs changed only once, so 1 or 0 real positive roots now checking f( -x ) h( -x)= - 2x^3 - 5x^2 + 31x - 15 no yes yes it changed signs twice, so 2 or 0 negative real roots
what do you mean by 2 or 0 negative roots?
hmm one sec... shoot... I missed the \(x^2\)
not 1 or 0 positive root, 1 positive root
now checking f( -x ) h( -x)= - 2x^3 + 5x^2 + 31x - 15 yes no yes it changed signs twice, so 2 or 0 negative real roots
I'm really confused now! For the function h(x)=2x^3+5x^2-31x-15, it asked for the possible combinations of roots, so the possible combinations are only 1 positive root, and 2 or 0 negative roots?
complex roots always come in conjugate pairs whenever a polynomial has real coefficients so you have to account for that possibility. if there are 2 negative roots, then they're real. otherwise, the polynomial may have to complex roots and in that case they would not cross the x axis anywhere (in the real cartesian coordinates)
oh okay @pgpilot326! that makes sense :)
"two", not "to"
always count down the number of changes from the max, by two to account for possible conjugate pairs. so if you came up with 3 changes, then the possibilities are 3 or 1 (counting down by 2).
okay :) so again, just to confirm the answer i put down 1 positive root and 2 or 0 negative roots for the possible roots?
you got it girl!
thank you so much! you're awesome! :)
\[\tiny\text{Lana...}\] \[\small\text{Lana...}\] \[\text{Lana...}\] \[\large\text{Lana...}\] \[\huge\text{Lana!!!}\] "What!!!" \[\small\text{Danger Zone!}\]
You are!
Haha! Thank you :)
I take it you've seen Archer?
I can't remember when, but yeah! :)
cool. keep at it!
thanks i will, you too! :)
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