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

Mar 5, 2017

$f \left(x\right)$ has zeros $\frac{1}{2}$ with multiplicity $2$ and $- \frac{1}{3}$ with multiplicity $2$

Explanation:

Given:

$f \left(x\right) = 36 {x}^{4} - 12 {x}^{3} - 11 {x}^{2} + 2 x + 1$

By the rational roots theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ 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:

$\pm \frac{1}{36} , \pm \frac{1}{18} , \pm \frac{1}{12} , \pm \frac{1}{9} , \pm \frac{1}{6} , \pm \frac{1}{4} , \pm \frac{1}{3} , \pm \frac{1}{2} , \pm 1$

Note that $36 > 26 = 12 + 11 + 2 + 1$, so $\pm 1$ are not zeros.

$f \left(\frac{1}{2}\right) = 36 {\left(\textcolor{b l u e}{\frac{1}{2}}\right)}^{4} - 12 {\left(\textcolor{b l u e}{\frac{1}{2}}\right)}^{3} - 11 {\left(\textcolor{b l u e}{\frac{1}{2}}\right)}^{2} + 2 \left(\textcolor{b l u e}{\frac{1}{2}}\right) + 1$

$\textcolor{w h i t e}{f \left(\frac{1}{2}\right)} = \frac{36}{16} - \frac{12}{8} - \frac{11}{4} + \frac{2}{2} + 1$

$\textcolor{w h i t e}{f \left(\frac{1}{2}\right)} = \frac{9 - 6 - 11 + 4 + 4}{4}$

$\textcolor{w h i t e}{f \left(\frac{1}{2}\right)} = 0$

So $x = \frac{1}{2}$ is a zero and $\left(2 x - 1\right)$ a factor:

$36 {x}^{4} - 12 {x}^{3} - 11 {x}^{2} + 2 x + 1 = \left(2 x - 1\right) \left(18 {x}^{3} + 3 {x}^{2} - 4 x - 1\right)$

Trying $x = \frac{1}{2}$ with the remaining cubic factor, we find:

$18 {\left(\textcolor{b l u e}{\frac{1}{2}}\right)}^{3} + 3 {\left(\textcolor{b l u e}{\frac{1}{2}}\right)}^{2} - 4 \left(\textcolor{b l u e}{\frac{1}{2}}\right) - 1 = \frac{18}{8} + \frac{3}{4} - \frac{4}{2} - 1$

$\textcolor{w h i t e}{18 {\left(\frac{1}{2}\right)}^{3} + 3 {\left(\frac{1}{2}\right)}^{2} - 4 \left(\frac{1}{2}\right) - 1} = \frac{9 + 3 - 8 - 4}{4}$

$\textcolor{w h i t e}{18 {\left(\frac{1}{2}\right)}^{3} + 3 {\left(\frac{1}{2}\right)}^{2} - 4 \left(\frac{1}{2}\right) - 1} = 0$

So $x = \frac{1}{2}$ is a zero again and $\left(2 x - 1\right)$ a factor again:

$18 {x}^{3} + 3 {x}^{2} - 4 x - 1 = \left(2 x - 1\right) \left(9 {x}^{2} + 6 x + 1\right)$

$\textcolor{w h i t e}{18 {x}^{3} + 3 {x}^{2} - 4 x - 1} = \left(2 x - 1\right) {\left(3 x + 1\right)}^{2}$

So the remaining zero is $x = - \frac{1}{3}$ with multiplicity $2$.

How did I get from $9 {x}^{2} + 6 x + 1$ to ${\left(3 x + 1\right)}^{2}$ ?

Note that $961 = {31}^{2}$ and the multiplication $31 \cdot 31 = 961$ involves no carrying of digits. So it's just like $9 {x}^{2} + 6 x + 1 = {\left(3 x + 1\right)}^{2}$ with $x = 10$.

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