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

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

Users Who Are Viewing This Thread (Users: 0, Guests: 1)