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
All the zeros are for
Explanation:
Let
By trial and error
And
Therefore,
We have tto make a long division
So
all the zeros are for
The "possible" rational zeros are:
#+-1, +-2, +-4, +-5, +-8, +-10, +-20, +-40#
The actual zeros are:
Explanation:
By the rational roots theorem, any rational zeros of
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
#x^4-2x^3-21x^2+22x+40 = (x+1)(x^3-3x^2-18x+40)#
Let
#g(-1) = -1-3+18+40 = 54#
#g(2) = (2)^3-3(2)^2-18(2)+40 = 8-12-36+40 = 0#
So
#x^3-3x^2-18x+40 = (x-2)(x^2-x-20)#
To factor the remaining quadratic note, find a pair of factors of
#x^2-x-20 = (x-5)(x+4)#
Hence the other two zeros are