What are all the possible rational zeros for #f(x)=x^3+5x^2-11x-6# and how do you find all zeros?
1 Answer
Explanation:
#f(x) = x^3+5x^2-11x-6#
By the rational roots theorem, any rational zeros of
Thats means that the only possible rational zeros are:
#+-1, +-2, +-3, +-6#
In addition, note that the signs of the coefficients of
#f(1) = 1+5-11+6 = 1#
#f(2) = 8+5(4)-11(2)-6 = 8+20-22-6 = 0#
So
#x^3+5x^2-11x-6 = (x-2)(x^2+7x+3)#
The remaining two zeros are roots of the quadratic:
#x = (-b+-sqrt(b^2-4ac))/(2a)#
#color(white)(x) = (-7+-sqrt(7^2-4(1)(3)))/(2*1)#
#color(white)(x) = (-7+-sqrt(49-12))/2#
#color(white)(x) = -7/2+-sqrt(37)/2#