# What are all the possible rational zeros for #f(x)=2x^4-7x^3-2x^2-7x-4# and how do you find all zeros?

##### 1 Answer

#### Explanation:

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

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

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

Before we embark on trying these, note that the coefficients

#f(i) = 2i^4-7i^3-2i^2-7x-4 = 2+7i+2-7i-4 = 0#

So

Since the coefficients of

Hence

#2x^4-7x^3-2x^2-7x-4 = (x^2+1)(2x^2-7x-4)#

We can factor the remaining quadratic using an AC method:

Find a pair of factors of

The pair

Use this pair to split the middle term and factor by grouping:

#2x^2-7x-4 = (2x^2-8x)+(x-4)#

#color(white)(2x^2-7x-4) = 2x(x-4)+1(x-4)#

#color(white)(2x^2-7x-4) = (2x+1)(x-4)#

Hence the other two zeros are:

#x=-1/2# and#x = 4#