What are all the possible rational zeros for #y=x^4-2x^3-21x^2+22x+40# and how do you find all zeros?

2 Answers
Nov 1, 2016

Answer:

All the zeros are for #x=-1, 2, -4, 5#

Explanation:

Let #f(x)=x^4-2x^3-21x^2+22x+40#
By trial and error
#f(-1)=1+2-21-22+40=0#
And #f(2)=16-16-84+44+40=0#
Therefore, #(x+1)(x-2)# is a factor
We have tto make a long division
#x^4-2x^3-21x^2+22x+40##color(white)(aaaa)##∣##x^2-x-2#
#x^4-x^3-2x^2##color(white)(aaaaaaaaaaaaaaaa)##∣##x^2-x-20#
#color(white)(a)##0-x^3-19x^2+22x#
#color(white)(aaa)##-x^3+x^2+2x#
#color(white)(aaaa)##0-20x^2+20x+40#
#color(white)(aaaaaa)##-20x^2+20x+40#
So #x^2-x-20=(x+4)(x-5)# these are factors
all the zeros are for #x=-1, 2, -4, 5#

Nov 1, 2016

Answer:

The "possible" rational zeros are:

#+-1, +-2, +-4, +-5, +-8, +-10, +-20, +-40#

The actual zeros are: #-1#, #2#, #5# and #-4#.

Explanation:

#color(white)()#
#f(x) = x^4-2x^3-21x^2+22x+40#

By the rational roots theorem, 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 #40# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-2, +-4, +-5, +-8, +-10, +-20, +-40#

Start by trying each in turn:

#f(1) = 1-2-21+22+40 = 40#

#f(-1) = 1+2-21-22+40 = 0#

So #color(blue)(x=-1)# is a zero and #(x+1)# a factor:

#x^4-2x^3-21x^2+22x+40 = (x+1)(x^3-3x^2-18x+40)#

Let #g(x) = x^3-3x^2-18x+40#

#g(-1) = -1-3+18+40 = 54#

#g(2) = (2)^3-3(2)^2-18(2)+40 = 8-12-36+40 = 0#

So #color(blue)(x=2)# is a zero and #(x-2)# a factor:

#x^3-3x^2-18x+40 = (x-2)(x^2-x-20)#

To factor the remaining quadratic note, find a pair of factors of #20# which differ by #1#. The pair #5, 4# works, so we find:

#x^2-x-20 = (x-5)(x+4)#

Hence the other two zeros are #color(blue)(x = 5)# and #color(blue)(x = -4)#.