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

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?

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

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

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

-2i√14?

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

\[\sqrt{14}i\]

a typo

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

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

Thankssss!

Welcome!!~~~