Question

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

Answer 1

`f(x)` has zeros `1/2` with multiplicity `2` and `-1/3` with multiplicity `2`

Explanation:

Given:

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

By the rational roots theorem, any rational zeros of `f(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `1` and `q` a divisor of the coefficient `36` of the leading term.

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 `36 > 26 = 12+11+2+1`, so `+-1` are not zeros.

`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 `x=1/2` is a zero and `(2x-1)` a factor:

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

Trying `x=1/2` with the remaining cubic factor, we find:

`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 `x=1/2` is a zero again and `(2x-1)` a factor again:

`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 `x=-1/3` with multiplicity `2`.

How did I get from `9x^2+6x+1` to `(3x+1)^2` ?

Note that `961 = 31^2` and the multiplication `31*31 = 961` involves no carrying of digits. So it's just like `9x^2+6x+1 = (3x+1)^2` with `x=10`.

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