# What are all the possible rational zeros for #f(x)=4x^5-2x^4+30x^3-15x^2+50x-25#?

##### 1 Answer

#### Answer:

See explanation...

#### Explanation:

#f(x) = 4x^5-2x^4+30x^3-15x^2+50x-25#

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

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

#+-1/4, +-1/2, +-1, +-5/4, +-5/2, +-5, +-25/4, +-25/2, +-25#

In fact, note that the signs of the coefficients alternate, with no powers of

So the only possible rational zeros are:

#1/4, 1/2, 1, 5/4, 5/2, 5, 25/4, 25/2, 25#

Trying each in turn, we find:

#f(1/2) = 4(1/32)-2(1/16)+30(1/8)-15(1/4)+50(1/2)-25#

#color(white)(f(1/2))=(1-1+30-30+200-200)/8 = 0#

So

#4x^5-2x^4+30x^3-15x^2+50x-25 = (2x-1)(2x^4+15x^2+25)#

We can factor the remaining quartic as a quadratic in

#2x^4+15x^2+25 = 2x^4+10x^2+5x^2+25#

#color(white)(2x^4+15x^2+25) = 2x^2(x^2+5)+5(x^2+5)#

#color(white)(2x^4+15x^2+25) = (2x^2+5)(x^2+5)#

Hence zeros:

#x = +-sqrt(10)/2i" "# and#" "x = +-sqrt(5)i#