Find all the real zeros of the polynomial
P(x)=x^3 - 3 x^2 - 11 x + 5.
there should be three. I did synthetic division and the positive 5 gave me a zero, I was stuck on how to find the other 2 zeros

Question

Find all the real zeros of the polynomial
P(x)=x^3 - 3 x^2 - 11 x + 5.
there should be three. I did synthetic division and the positive 5 gave me a zero, I was stuck on how to find the other 2 zeros

anonymous · Answer

divide $P(x)$ with $(x-5)$ because that was one of the solution, then solve for the polynomial equation.

anonymous · Answer

don't understand, I when i did do that my zero for that was 5. so then i ended up with x^2+2x-1   and I don't know how to move on from there

anonymous · Answer

then you can solve for the zeros in $x^{2}+2x-1=0$

anonymous · Answer

when i was factoring i couldn't come up with the correct answer: (x +2)(x-1)  it doesn't give me x^2+2x-1  no matter which way i flipped flopped it

anonymous · Answer

omg. i was trying to find x^2-2x-1...no wonder i couldnt find the answer

anonymous · Answer

Hint: use the quadratic formula$$\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$$

anonymous · Answer

do you get it yet?

anonymous · Answer

yeah so far i got -2 +/-sqrt8 all divided by 2...now i can take the 2 and reduce everything by it (right??), ending up with -1+/-sqrt4 which would be -1+/-2 which would be 1+2=3 and 1-2=1, right? or no?

anonymous · Answer

no you cannot divide sqrt8 by 2 to get sqrt4 because sqrt8 actually is not twice sqrt4. sqrt8=sqrt4*sqrt2=2sqrt2, so sqrt8/2=sqrt2, not 2

anonymous · Answer

yeah, I wasn't sure about that.

anonymous · Answer

so its +/-sqrt2

anonymous · Answer

so the zeros are $5$ and $1 \pm \sqrt{2}$

anonymous · Answer

thank you so much for your help.

anonymous · Answer

no problem!