Question

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

Answer 1

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

Answer 2

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?

Answer 3

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

Answer 4

You said

(x−1)(x^2+2x+15)

This gives zeros of 1, 3, 5, 15?

Answer 5

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!~

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

-2i√14?

Answer 10

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

Answer 11

\[\sqrt{14}i\]

Answer 12

a typo

Answer 13

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

Answer 14

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

Answer 15

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..

Answer 16

Thankssss!

Answer 17

Welcome!!~~~