How do you find all the real and complex roots of $F (x) = - 9 x^{4} + 12 x^{3} + 40 x^{2} - 60 x + 25$ $F(x) = -9x^4 + 12x^3+40x^2-60x+25$ ?

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How do you find all the real and complex roots of $F (x) = - 9 x^{4} + 12 x^{3} + 40 x^{2} - 60 x + 25$ $F(x) = -9x^4 + 12x^3+40x^2-60x+25$ ?

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Answer 1 · 2016-05-26T20:58:35

$x_{1} = \frac{2}{3} + \frac{1}{3} i$ $x_1 = 2/3+1/3i$

$x_{2} = \frac{2}{3} - \frac{1}{3} i$ $x_2 = 2/3-1/3i$

$x_{3} = \sqrt{5}$ $x_3 = sqrt(5)$

$x_{4} = - \sqrt{5}$ $x_4 = -sqrt(5)$

Explanation:

Given:

$f (x) = - 9 x^{4} + 12 x^{3} + 40 x^{2} - 60 x + 25$ $f(x) = -9x^4+12x^3+40x^2-60x+25$

By the rational root theorem, any rational zeros of $f (x)$ $f(x)$ will be expressible in the form $\frac{p}{q}$ $p/q$ with integers $p, q$ $p, q$ , with $p$ $p$ a divisor of the constant term $25$ $25$ and $q$ $q$ a divisor of the coefficient $- 9$ $-9$ of the leading term.

That means that the only possible rational zeros of $f (x)$ $f(x)$ are:

$\pm \frac{1}{9}, \pm \frac{1}{3}, \pm \frac{5}{9}, \pm 1, \pm \frac{5}{3}, \pm \frac{25}{9}, \pm 5, \pm \frac{25}{3}, \pm 25$ $+-1/9, +-1/3, +-5/9, +-1, +-5/3, +-25/9, +-5, +-25/3, +-25$

None of these work, so $f (x)$ $f(x)$ has no rational zeros.

Take a deep breath and start trying a full blown quartic solution:

Multiply through by $- 9$ $-9$ to make the sums easier:

$- 9 f (x) = 81 x^{4} - 108 x^{3} + 360 x^{2} - 540 x + 225$ $-9f(x) = 81x^4-108x^3+360x^2-540x+225$

$= {(3 x - 1)}^{4} - 46 {(3 x - 1)}^{2} + 92 (3 x - 1) - 88$ $=(3x-1)^4-46(3x-1)^2+92(3x-1)-88$

Let $t = 3 x - 1$ $t=3x-1$

We want to solve:

$0 = t^{4} - 46 t^{2} + 92 t - 88$ $0 = t^4-46t^2+92t-88$

$= (t^{2} - a t + b) (t^{2} + a t + c)$ $=(t^2-at+b)(t^2+at+c)$

$= t^{4} + (b + c - a^{2}) t^{2} + a (b - c) t + b c$ $=t^4+(b+c-a^2)t^2+a(b-c)t+bc$

Equating coefficients and rearranging a little:

${ (b+c = a^2-46), (b-c = 92/a), (bc = -88) :}$

Then:

$(a^2-46)^2 = (b+c)^2 = (b-c)^2+4bc = (92/a)^2-352$

Expanding both ends:

$a^4-92a^2+2116 = 8464/a^2-352$

Multiplying through by $a^2$ and rearranging a little, this becomes:

$(a^2)^3-92(a^2)^2+2468(a^2)-8464 = 0$

Let $y=a^2$

Let $g(y) = y^3-92y^2+2468y-8464$

By the rational root theorem, any rational zeros of $g(y)$ are factors of $8464$ .

Trying the first few we find:

$g(4) = 64-1472+9872-8464=0$

So one solution is $y = 4$ , that is $a^2=4$ . So we can put $a=2$ .

${ (b+c = a^2-46 = 2^2-46 = -42), (b-c = 92/a = 92/2 = 46), (bc = -88) :}$

Hence:

$b=2$ and $c=-44$

So we have:

$t^4-46t^2+92t-88 = (t^2-2t+2)(t^2+2t-44)$

which has zeros:

$t_1 = 1+i$

$t_2 = 1-i$

$t_3 = -1+3sqrt(5)$

$t_4 = -1-3sqrt(5)$

Then $x = (t+1)/3$ , hence zeros of the original quartic:

$x_1 = 2/3+1/3i$

$x_2 = 2/3-1/3i$

$x_3 = sqrt(5)$

$x_4 = -sqrt(5)$

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How do you find all the real and complex roots of $F (x) = - 9 x^{4} + 12 x^{3} + 40 x^{2} - 60 x + 25$ $F(x) = -9x^4 + 12x^3+40x^2-60x+25$ ?

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How do you find all the real and complex roots of F(x) = -9x^4 + 12x^3+40x^2-60x+25F(x)=−9x4+12x3+40x2−60x+25F(x) = -9x^4 + 12x^3+40x^2-60x+25?

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How do you find all the real and complex roots of $F (x) = - 9 x^{4} + 12 x^{3} + 40 x^{2} - 60 x + 25$ $F(x) = -9x^4 + 12x^3+40x^2-60x+25$ ?