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#