# What are all the possible rational zeros for #f(x)=5x^4+32x^2-21#?

##### 1 Answer

The "possible" rational zeros are:

#+-1/5, +-3/5, +-1, +-7/5, +-3, +-21/5, +-7, +-21#

The actual zeros are:

#+-sqrt(15)/5" "# and#" "+-sqrt(7)i#

#### Explanation:

Given:

#f(x) = 5x^4+32x^2-21#

If we apply the rational root theorem directly, any *rational* zeros of

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

#+-1/5, +-3/5, +-1, +-7/5, +-3, +-21/5, +-7, +-21#

Note also that when

#f(x) >= 5+32-21 = 16#

So the only possible rational zeros are:

#+-1/5, +-3/5#

and since all the terms in

So we only need to check

#f(1/5) = 5(1/625)+32(1/25)-21 = (1+160-2625)/125 = -2464/125#

#f(3/5) = 5(81/625)+32(9/25)-21 = (81+1440-2625)/125 = -1104/125#

So

We can factor

Look for a pair of factors of

The pair

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

#5x^4+32x^2-21 = (5x^4+35x^2)-(3x^2+21)#

#color(white)(5x^4+32x^2-21) = 5x^2(x^2+7)-3(x^2+7)#

#color(white)(5x^4+32x^2-21) = (5x^2-3)(x^2+7)#

Hence we can see that the zeros of

#x = +-sqrt(3/5) = +-sqrt(15)/5" "# and#" "x = +-sqrt(7)i#