## kirk.freedman 2 years ago Find all the zeroes of each equation. x^3 – x^2 + 5x – 5 = 0

1. kirk.freedman

the sum of all coefficients be 0 (1-1+5-5=0) it means x=1 is the root or (x-1) is one of the factor that polynomial find the other factor by using synthetic division, then for zeroes take all factors be 0 and solve for x

3. kirk.freedman

So I got x^2+5 When I used synthetic division. What now?

4. sirm3d

equate it to zero and use the quadratic formula.

5. kirk.freedman

Oh wait so if I have x^2+5=0 then x=i sqrt 5? That's one of my answers?

6. kirk.freedman

These Are the possible answers btw |dw:1354664328662:dw|

just to check the value of D(discrimininant) with D=b^2-4ac D=0^2-4*1*5=-20 it means no solution for x real, but if u want representing x to in complex number use the quadratic formula : x=[-b +- sqrt(b^2-4ac)]/(2a)

8. kirk.freedman

What is a,b, and c for the quadratic formula?

9. kirk.freedman

is it 1,1, and 5?

actually the general form a quadratic equation is ax^2 + bx + c = 0 so given x^2 + 5 = 0 it means a=1, b=0, c=5

11. kirk.freedman

But the original formula is x^3 – x^2 + 5x – 5 = 0 wouldn't b be 5 since it's 5x?? I'm confused there.

12. kirk.freedman

Where do you get the x^2+5=0 formula?

13. kirk.freedman

Oh nvm, I got it. Thanks for the help!

14. sirm3d