Can anyone do this trial question? (1 Viewer)

lolmb · Nov 13, 2014

At least one zero of P(x)=3x^3 + px^2 + 15x + 10 is purely imaginary and p is real. Find p and hence write P(x) as a product of linear factors.

Thank You!

seventhroot · Nov 13, 2014

Use the sum and product of roots

HeroicPandas · Nov 13, 2014

At least 1 root is imaginary, so by the conjugate root theorem, it's conjugate is also a root

Let z be a purely imaginary root

Roots = ib, ib(conj), something

sum of roots = -p/3 = real

so, ib + ib(conj)+ something = real

ib - ib+ something = real

something = real

So roots are: imaginary, conjugate of imaginary and real

So roots are: ib, -ib, c

integral95 · Nov 13, 2014

lolmb said:
At least one zero of P(x)=3x^3 + px^2 + 15x + 10 is purely imaginary and p is real. Find p and hence write P(x) as a product of linear factors.

Thank You!

lel

the complex roots are purely imaginary, so they're simply in the form bi ,-bi (b is real)

$$let the roots be$ \ \alpha \ bi, \ -bi \\ $using Product of roots you get$ \ \alpha(\beta)^2 = \frac{-10}{3} \\ $us sum of roots 2 at a time to get$ \ b^2 = 5 \\ \\ $sub that in previous equation to get$ \ \ \alpha(5) = \frac{-10}{3} \Rightarrow \alpha = \frac{-2}{3} \\ \\ $sum of roots gives$ \ \ \alpha = -\frac{p}{3} = -\frac{2}{3} \\ \\ $So \ p = 2$ \\ \\ \\ \\ $roots are -2/3, 5i and -5i so$ \ \ P(x) = (3x+2)(x^2+5) = (3x+2)(x+5i)(x-5i)$

HeroicPandas · Nov 13, 2014

oh yeh woops lol

Amundies · Nov 13, 2014

integral95, you missed an "i" in the final bracket

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Can anyone do this trial question? (1 Viewer)

lolmb

New Member

seventhroot

gg no re

HeroicPandas

Heroic!

integral95

Well-Known Member

HeroicPandas

Heroic!

Amundies

Commander-in-Chief

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