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

Question

Theodore Amidon · Accepted Answer

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

Ariana Alvarez · Answer

The "possible" rational zeros are:

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

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

#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)#.

Liam Brennan · Answer

undefined To find all possible rational zeros for the polynomial $y = x^4 - 2x^3 - 21x^2 + 22x + 40$, you can use the Rational Root Theorem. The Rational Root Theorem states that any rational zero of a polynomial must be of the form $\frac{p}{q}$, where $p$ is a factor of the constant term (40 in this case) and $q$ is a factor of the leading coefficient (1 in this case).

So, the possible rational zeros are all the combinations of factors of 40 divided by factors of 1. Factors of 40 are ±1, ±2, ±4, ±5, ±8, ±10, ±20, ±40, and factors of 1 are ±1.

Combining them, we get the list of possible rational zeros:

$\pm 1, \pm 2, \pm 4, \pm 5, \pm 8, \pm 10, \pm 20, \pm 40$

To find all zeros, you can use methods like synthetic division or polynomial long division along with factoring or numerical methods like Newton's method or the quadratic formula to find any remaining zeros.