# Solving a Quintic polynomial

The polynomial

$F(x) = x^5-9x^4+24x^3-24x^2+23x-15$

has roots $x=1$ and $x=j$. Calculate all the roots of the polynomial.

I was told I had to use radicals or similar to solve this but after reading up on it I'm still confused about how to solve it.

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If you don't want to factor, then you can assume $F(x) = (x-1)(x-j)(x+j)(x-a)(x-b).$ Expand $F(x),$ equate with $x^5-9x^4+24x^3-24x^2+23x-15,$ and solve for $ab.$ Sanity check: $a^2+b^2=34.$ – user2468 Jan 8 '12 at 18:53

if $F(a)=0$, then $(x-a)|F(x)$. also, if a polynomial with real cofficients has a complex root, then the complex conjugate is also a root. so $F$ is divisible by $(x-1)$ and $x^2+1=(x-i)(x+i)$ (according to you). after dividing by these, you will have a polynomial of degree 2, which you can easily factor: $$F/(x^2+1)=x^3-9x^2+23x-15=G$$ $$G/(x-1)=x^2-8x+15=(x-5)(x-3)$$ so $$F(x)=(x-1)(x-3)(x-5)(x^2+1)$$

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Knowing that one root is $x = 1$ means $F(x)$ has a factor $x - 1$. So you can either obtain the complementary factor by long division, or note that:

\begin{align}F(x) &= x^4(x - 1) + 24x^2(x - 1) - 8x^4 + 23x - 15 \\ &= x^4(x - 1) + 24x^2(x - 1) - 8x(x^3 - 1) + 15(x - 1)\end{align}

so using $x^3 - 1 = (x - 1)(x^2 + x + 1)$, the complementary factor can be read off as:

$x^4 - 8x^3 + 16x^2 - 8x + 15$

Now if a polynomial with real coefficients has a complex root then it must have the conjugate of that value as another root, and thus having a root $j$ (which I presume denotes $-1^\frac{1}{2}$) means it must have a root $-j$, and thus a factor $(x - j)(x + j)$ which is $x^2 + 1$.

Dividing this out leaves a quadratic factor $x^2 - 8 x + 15$, which by inspection (or solving in the usual way) factors as $(x - 3)(x - 5)$.

edit: DOH! I didn't see yoyo's reply until I had posted mine! May as well leave it now.

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Thanks for editing Martin - I must remember that align trick. Also, if doing this systematically via long division (rather than my slightly hacky shortcut) then starting with $x^2 + 1$, as yoyo did, is better because it gets the degree down faster. – John R Ramsden Jan 8 '12 at 20:58

As you know that x=1 is a solution, factor (x-1) out of the equation then solve the resulting quartic equation with the formula. You can see the formula at http://upload.wikimedia.org/wikipedia/commons/9/99/Quartic_Formula.svg.

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