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

1 Answer
Jan 8, 2017

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 #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-21# and #q# a divisor of the coefficient #5# of the leading term.

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 #abs(x) >= 1#, then:

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

So the only possible rational zeros are:

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

and since all the terms in #x# are of even degree, if #x_1# is a zero then so is #-x_1#.

So we only need to check #x=1/5# and #x=3/5#...

#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 #f(x)# has no rational zeros.

#color(white)()#
We can factor #f(x)# as a quadratic in #x^2# using an AC method:

Look for a pair of factors of #AC=5*21 = 105# with difference #B=32#

The pair #35, 3# works.

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 #f(x)# are actually:

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