Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

2 Answers

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

share|improve this answer
add comment

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.