How do you find all the zeros of $f (x) = x^{4} - 3.3 x^{3} + 2.3 x^{2} + 0.6 x$ $f(x) = x^4 - 3.3x^3 + 2.3x^2 + 0.6x$ ?

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Answer 1 · 2016-05-05T16:13:17

Zeros of $f (x)$ $f(x)$ are ${0, 2, \frac{3}{2}, - \frac{1}{5}}$ ${0,2,3/2,-1/5}$

$f (x) = x^{4} - 3.3 x^{3} + 2.3 x^{2} + 0.6 x$ $f(x)=x^4-3.3x^3+2.3x^2+0.6x$

= $\frac{x}{10} (10 x^{3} - 33 x^{2} + 23 x + 6)$ $x/10(10x^3-33x^2+23x+6)$

Hence one of the zeros is $0$ $0$ . Another zero could be a factor of $6$ $6$

As putting $x = 2$ $x=2$ in $(10 x^{3} - 33 x^{2} + 23 x + 6)$ $(10x^3-33x^2+23x+6)$ , we get

$10 \cdot 2^{3} - 33 \cdot 2^{2} + 23 \cdot 2 + 6 = 80 - 132 + 46 + 6 = 0$ $10*2^3-33*2^2+23*2+6=80-132+46+6=0$

Hence, $2$ $2$ is another zero and dividing $10 x^{3} - 33 x^{2} + 23 x + 6$ $10x^3-33x^2+23x+6$ by $(x - 2)$ $(x-2)$ , we get

$10 x^{3} - 33 x^{2} + 23 x + 6 = (x - 2) (10 x^{2} - 13 x - 3)$ $10x^3-33x^2+23x+6=(x-2)(10x^2-13x-3)$

or $(x - 2) (10 x^{2} - 15 x + 2 x - 3)$ $(x-2)(10x^2-15x+2x-3)$

or $(x - 2) (5 x (2 x - 3) + 1 \cdot (2 x - 3))$ $(x-2)(5x(2x-3)+1*(2x-3))$

= $(x - 2) (5 x + 1) (2 x - 3)$ $(x-2)(5x+1)(2x-3)$

Hence other zeros are given by $2 x - 3 = 0$ $2x-3=0$ and $5 x + 1 = 0$ $5x+1=0$ i.e. are $\frac{3}{2}$ $3/2$ and $- \frac{1}{5}$ $-1/5$

Hence zeros of $f (x)$ $f(x)$ are ${0, 2, \frac{3}{2}, - \frac{1}{5}}$ ${0,2,3/2,-1/5}$

How do you find all the zeros of f(x) = x^4 - 3.3x^3 + 2.3x^2 + 0.6xf(x)=x4−3.3x3+2.3x2+0.6xf(x) = x^4 - 3.3x^3 + 2.3x^2 + 0.6x?