# 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

#### 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#