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

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

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Answer 1 · 2017-03-05T14:57:12

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

Explanation:

Given:

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

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

That means that the only possible rational zeros are:

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

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

$f (\frac{1}{2}) = 36 {(\frac{1}{2})}^{4} - 12 {(\frac{1}{2})}^{3} - 11 {(\frac{1}{2})}^{2} + 2 (\frac{1}{2}) + 1$ $f(1/2) = 36(color(blue)(1/2))^4-12(color(blue)(1/2))^3-11(color(blue)(1/2))^2+2(color(blue)(1/2))+1$

$f (\frac{1}{2}) = \frac{36}{16} - \frac{12}{8} - \frac{11}{4} + \frac{2}{2} + 1$ $color(white)(f(1/2)) = 36/16-12/8-11/4+2/2+1$

$f (\frac{1}{2}) = \frac{9 - 6 - 11 + 4 + 4}{4}$ $color(white)(f(1/2)) = (9-6-11+4+4)/4$

$f (\frac{1}{2}) = 0$ $color(white)(f(1/2)) = 0$

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

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

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

$18 {(\frac{1}{2})}^{3} + 3 {(\frac{1}{2})}^{2} - 4 (\frac{1}{2}) - 1 = \frac{18}{8} + \frac{3}{4} - \frac{4}{2} - 1$ $18(color(blue)(1/2))^3+3(color(blue)(1/2))^2-4(color(blue)(1/2))-1 = 18/8+3/4-4/2-1$

$18 {(\frac{1}{2})}^{3} + 3 {(\frac{1}{2})}^{2} - 4 (\frac{1}{2}) - 1 = \frac{9 + 3 - 8 - 4}{4}$ $color(white)(18(1/2)^3+3(1/2)^2-4(1/2)-1) = (9+3-8-4)/4$

$18 {(\frac{1}{2})}^{3} + 3 {(\frac{1}{2})}^{2} - 4 (\frac{1}{2}) - 1 = 0$ $color(white)(18(1/2)^3+3(1/2)^2-4(1/2)-1) = 0$

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

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

$18 x^{3} + 3 x^{2} - 4 x - 1 = (2 x - 1) {(3 x + 1)}^{2}$ $color(white)(18x^3+3x^2-4x-1) = (2x-1)(3x+1)^2$

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

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

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

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

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

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How do you find all zeros with multiplicities of f(x)=36x4−12x3−11x2+2x+1f(x)=36x^4-12x^3-11x^2+2x+1?

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