Find all the real zeros of the function:
f(x)=x^4-5x^3+7x^2+3x-10

Question

Find all the real zeros of the function:
f(x)=x^4-5x^3+7x^2+3x-10

tkhunny · Answer

Have you considered the factors of 10?

anonymous · Answer

Yes +-1,+-2,+-5,+-10 but I can't get from there I'm using synthetic division

tkhunny · Answer

Thaat's okay.  That will find only the RATIONAL zeros.  If they are not Rational, that is the right result.

tkhunny · Answer

On the other hand, you should try x = 2 and x = -1 again.

anonymous · Answer

But what I end up with is (x^3-6x^2) +(13x-10) and that doesn't factor out correctly

tkhunny · Answer

I don't know what that is.  How did you come to that condition?

anonymous · Answer

I did synthetic division using -1

tkhunny · Answer

Please demonstrate.  You should not end up with that.

tkhunny · Answer

|dw:1387408015594:dw|

Well, there's the numbers for x = 2.  I wish it had looked better.

tkhunny · Answer

I see.  1, -6, 13, -10 is good.

Now do x = 2.

anonymous · Answer

Ok I tried 2 and I got that as we'll but then after you have to turn it into a poynomial equatiom

anonymous · Answer

*well

tkhunny · Answer

Did you get $x^{2} - 4x + 5 = 0$?

anonymous · Answer

No I got x^3 -3x^2 +x+5

tkhunny · Answer

We are going around in circles.

x^4-5x^3+7x^2+3x-10 = (x-2)(x^3 - 3x^2 + x + 5)
x^4-5x^3+7x^2+3x-10 = (x+1) (x^3 - 6x^2 + 13x-10)

Pick one and do the other to it.

x^4-5x^3+7x^2+3x-10 = (x+1)(x-2)(x^2 − 4x + 5)

anonymous · Answer

Ok thanks I guess that was easier