What are all the possible rational zeros for #f(x)=x^3-6x^2+7x+4# and how do you find all zeros?
1 Answer
The "possible" rational zeros are:
The actual zeros are:
Explanation:
#f(x) = x^3-6x^2+7x+4#
By the rational roots theorem, any rational zeros of
That means that the only possible rational zeros are:
#+-1, +-2, +-4#
Trying each in turn, we eventually find:
#f(4) = 64-96+28+4 = 0#
So
#x^3-6x^2+7x+4 = (x-4)(x^2-2x-1)#
The remaining quadratic can be factored by completing the square:
#x^2-2x-1 = x^2-2x+1-2#
#color(white)(x^2-2x-1) = (x+1)^2-sqrt(2)^2#
#color(white)(x^2-2x-1) = ((x+1)-sqrt(2))((x+1)+sqrt(2))#
#color(white)(x^2-2x-1) = (x+1-sqrt(2))(x+1+sqrt(2))#
Hence the other two zeros are:
#x = -1+-sqrt(2)#