# How do you find all the zeros of #P(x) = x^4 + x^3 - 2x^2 + 4x – 24#?

##### 1 Answer

Zeros:

#### Explanation:

#P(x) = x^4+x^3-2x^2+4x-24#

By the rational root theorem, any *rational* zeros of

That means that the only possible *rational* zeros are:

#+-1# ,#+-2# ,#+-3# ,#+-4# ,#+-6# ,#+-8# ,#+-12# ,#+-24#

Trying each in turn, we find:

#P(1) = 1+1-2+4-24 = -20#

#P(-1) = 1-1-2-4-24 = -30#

#P(2) = 16+8-8+8-24 = 0#

So

#x^4+x^3-2x^2+4x-24 = (x-2)(x^3+3x^2+4x+12)#

Note that the ratio between the first and second terms in the remaining cubic is the same as that between its third and fourth terms, so this cubic will factor by grouping:

#x^3+3x^2+4x+12#

#=(x^3+3x^2)+(4x+12)#

#=x^2(x+3)+4(x+3)#

#=(x^2+4)(x+3)#

Hence zero

The remaining quadratic factor is always positive for Real values of

#(2i)^2 = 4i^2 = -4#