How do you find all the zeros of $x^{5} - 2 x^{4} - 4 x^{3} - 10 x^{2} - 21 x - 12$ $x^5-2x^4-4x^3-10x^2-21x-12$ ?

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Answer 1 · 2016-05-24T17:17:01

${(x + 1)}^{2} (x - 4) (x + i \sqrt{3}) (x + i \sqrt{3})$ $(x+1)^2(x-4)(x+i sqrt(3))(x+i sqrt(3))$

The polynomial has constant coefficient $- 12$ $-12$ which is the product of its roots.

Trying the sequence of $- 12$ $-12$ factors given by

${\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12}$ ${pm 1,pm2, pm 3, pm 4, pm 6, pm 12}$

we get

$p (- 1) = p (4) = 0$ $p(-1) = p(4) = 0$

then at last we have found two roots ( keep in mind that some of this roots can be multiple). So we can write

$p (x) = x^{5} - 2 x^{4} - 4 x^{3} - 10 x^{2} - 21 x - 12$ $p(x) = x^5 - 2 x^4 - 4 x^3 - 10 x^2 - 21 x - 12$

and also

$p (x) - (x + 1) (x - 4) (x^{3} + a x^{2} + b x + c) = 0$ $p(x) - (x+1)(x-4)(x^3+a x^2+b x+c) = 0$ for all $x$ $x$

Taking the coefficients equal to zero we have

$((12 - 4 c=0), (21 - 4 b - 3 c=0),(10 - 4 a - 3 b + c=0), (-3 a + b=0), (-1 + a=0))$ .

Solving for $a,b,c$ we get

$(a = 1, b = 3, c =3)$

Focusing now in the roots of

$x^3+x^2+3x+3=0$

and having in mind the ideas about the composition of the constant term we find that $x = -1$ is a root so we can write

$x^3+x^2+3x+3 = (x+1)(x^2+d x + e)$

proceeding as before we find $d = 0, e = 3$ . Putting all together we have

$p(x)=(x+1)^2(x-4)(x+i sqrt(3))(x+i sqrt(3))$

How do you find all the zeros of x^5-2x^4-4x^3-10x^2-21x-12x5−2x4−4x3−10x2−21x−12x^5-2x^4-4x^3-10x^2-21x-12?