Question

What are the possible rational zeros of f(x) = x4 + 2x3 - 3x2 - 4x - 12?

Can some one show me how to do this?

Answer 1

First you can try x= -1,0,1,2
I reckon that x = 2 will give you f(x) = 0
After that you can do a long division on the original quartic using x-2 as the divisor

Answer 2

I would recommend using the Quadratic Formula for a big polynomial function like that one

Answer 3

use long division

Answer 4

(x-2)(x^3+4x^2+5x+6)=0

Answer 5

do the same thing again

Answer 6

(x-2)(x+3)(x^2+x+2)

Answer 7

the last one use :
\[\frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }\]

Answer 8

ax^2 +bx +c
so a=1, b=1, c=2
\[\frac{ 1\pm \sqrt{1-4*1*2} }{ 2*1 }=\frac{ 1\pm \sqrt{1-8} }{ 2 }=\frac{ 1 }{ 2 }\pm \frac{ \sqrt{-7} }{2 }\]

Answer 9

these roots are not real

Answer 10

rational zeros

Answer 11

Ccs. Quadratic formula doesn't work for a quartic

Answer 12

Well done AJ. That's two real roots