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## 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

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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

4. toadytica305 Group Title

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

7. toadytica305 Group Title

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

8. Callisto Group Title

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

9. toadytica305 Group Title

-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 $-1\pm -\sqrt{14}i$

14. toadytica305 Group Title

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..

16. toadytica305 Group Title

Thankssss!

17. Callisto Group Title

Welcome!!~~~