Question

PLEASE HELP

How can i find the number of complex roots of f(x)=3x^3 -2x^2+5x+3 by using the Fundamental Theorem of Algebra and the Descrates Rule of Signs

Answer 1

@dalia_lam can you help?

Answer 2

@aidahjaan1

Answer 3

I'm sorry, I haven't learnt it yet :(

Answer 4

@FutureMathProfessor can you help please :(?

Answer 5

its ok @dalia_lam thanks anyways

Answer 6

@ash2326

Answer 7

In the original function there are 2 sign changes so there are either 2 or 0 positive real roots.
Plug in -x for x: f(x)=3x^3 -2x^2+5x+3
f(-x) = 3(-x)^3 - 2(-x)^2 + 5(-x) +3
= -3x^3 - 2x^2 -5x + 3
There is only one sign change here. So there is one negative real root.
Remember the function is a cubic (degree 3) so there are a total of three complex roots. Complex roots always come in pairs called complex conjugates like a+bi and a-bi. So if there are 3 total roots and 1 negative real root and 0 positive real roots there must be 2 complex roots. If there's 1 negative real root and 2 positive real roots then there are no complex roots.
In this case if you graph the function you'll see there is only one real root - a negative one. There are zero positive real roots. So there must be 2 complex roots.

Answer 8

this helped me so much thank you!!! <3 such a great help @Titanic12