## toadytica305 Group Title List all the real and imaginary zeros of f(x) = x^3+x^2+ 13x - 15... 1 is a zero of the function 2 years ago 2 years ago

1. satellite73 Group Title

did you factor? you already knew 1 was a zero, from the last question

2. Callisto Group Title

I think the question asks to solve f(x) =0 Since the question gives you 1 is a zero of the function. (x-1) is a factor. Now, do the division to get the other factor first, can you?

3. satellite73 Group Title

once you factor, you find the other zeros using the quadratic formula

You said (x−1)(x^2+2x+15) This gives zeros of 1, 3, 5, 15?

5. Callisto Group Title

Nope!~ f(x) =0 x^3+x^2+ 13x - 15 =0 (x−1)(x^2+2x+15) =0 So, x-1 =0 -(1) or (x^2+2x+15) =0 -(2) You can solve (1) easily. And solve (2) using quadratic formula!~

6. Callisto Group Title

For a quadratic equation ax^2 + bx + c=0 $x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$^ Quadratic formula

i get $(-4\sqrt{15}) /2$

8. Callisto Group Title

It doesn't seems right... $x=\frac{-2 \pm \sqrt{2^2 - 4(1)(15)}}{2(1)} =?$

-2i√14?

10. Callisto Group Title

Nope.. $x=\frac{-2\pm \sqrt {-56}}{2} = \frac{-2\pm -2\sqrt {-14}}{2} = -1 \pm -14i$

11. satellite73 Group Title

$\sqrt{14}i$

12. satellite73 Group Title

a typo

13. Callisto Group Title

Sorry :S $-1\pm -\sqrt{14}i$

Im still not sure how I get the real and imaginary zeros off that...

15. Callisto Group Title

the real one is 1 the imaginary ones are $-1\pm -\sqrt{14}i$ Actually, the question just asks to you to solve f(x) =0 ... I think..