# How do you find all zeros with multiplicities of #f(x)=36x^4-12x^3-11x^2+2x+1#?

##### 1 Answer

#### Explanation:

Given:

#f(x) = 36x^4-12x^3-11x^2+2x+1#

By the rational roots theorem, any rational zeros of

That means that the only possible rational zeros are:

#+-1/36, +-1/18, +-1/12, +-1/9, +-1/6, +-1/4, +-1/3, +-1/2, +-1#

Note that

#f(1/2) = 36(color(blue)(1/2))^4-12(color(blue)(1/2))^3-11(color(blue)(1/2))^2+2(color(blue)(1/2))+1#

#color(white)(f(1/2)) = 36/16-12/8-11/4+2/2+1#

#color(white)(f(1/2)) = (9-6-11+4+4)/4#

#color(white)(f(1/2)) = 0#

So

#36x^4-12x^3-11x^2+2x+1 = (2x-1)(18x^3+3x^2-4x-1)#

Trying

#18(color(blue)(1/2))^3+3(color(blue)(1/2))^2-4(color(blue)(1/2))-1 = 18/8+3/4-4/2-1#

#color(white)(18(1/2)^3+3(1/2)^2-4(1/2)-1) = (9+3-8-4)/4#

#color(white)(18(1/2)^3+3(1/2)^2-4(1/2)-1) = 0#

So

#18x^3+3x^2-4x-1 = (2x-1)(9x^2+6x+1)#

#color(white)(18x^3+3x^2-4x-1) = (2x-1)(3x+1)^2#

So the remaining zero is

How did I get from

Note that

graph{36x^4-12x^3-11x^2+2x+1 [-1.2, 1.2, -0.74, 1.76]}