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.
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.
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 $$ F(x)=(x-1)(x-3)(x-5)(x^2+1) $$
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.
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.