# Roots of Polynomial

• May 20th 2010, 01:44 AM
Lukybear
Roots of Polynomial
Let x=a be a root fo the quartic polynomial P(x) = x^4 + Ax^3 + Bx^2 + Ax + 1 where (2+B)^2=/ 4A^2

Show that x=1/a is a root
• May 20th 2010, 03:26 AM
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Quote:

Originally Posted by Lukybear
Let x=a be a root fo the quartic polynomial P(x) = x^4 + Ax^3 + Bx^2 + Ax + 1 where (2+B)^2=/ 4A^2

Show that x=1/a is a root

It would be helpful if you clarified what should be in place of the typo I marked in red above.
• May 20th 2010, 03:12 PM
Lukybear
Does not equal to is =/=

sorry.
• May 20th 2010, 04:27 PM
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So we have $P(a) = a^4 + Aa^3 + Ba^2 + Aa + 1 = 0$.

In order for $\frac{1}{a}$ to be defined, we cannot have $a=0$. Setting $a = 0$ we see that the above equation reduces to $1 = 0$, so we can safely say that $a \neq 0$.

Notice that $P\left(\frac{1}{a}\right) = \left(\frac{1}{a}\right)^4 + A\left(\frac{1}{a}\right)^3 + B\left(\frac{1}{a}\right)^2 + A\left(\frac{1}{a}\right) + 1 = \frac{1}{a^4}+\frac{A}{a^3}+\frac{B}{a^2}+\frac{A} {a}+1$.

Multiply both sides by $a^4$ to remove fractions.

$a^4\cdot P\left(\frac{1}{a}\right)=a^4 + Aa^3 + Ba^2 + Aa + 1$

Look familiar?

I'm still shaky on the restriction $(2+B)^2 \neq 4A^2$, which seems unnecessary. I'll look into it but maybe someone with keener insight can elucidate. Or perhaps it's just unnecessary.
• May 20th 2010, 11:54 PM
Lukybear
Ive pretty much got that. But the only thing i dont understand is that if we multiply a^4 to the polynomial, it changes dosent it. Since a^4 is not a roots of unity. Or are we to argue that since a^4 P(1/a) = 0 then p(1/a) = 0 via division?

The restriction of (2+B)^2 =/=4A^2 is i think unncessary. Think that is another part of the question.
• May 21st 2010, 06:25 AM
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Quote:

Originally Posted by Lukybear
Or are we to argue that since a^4 P(1/a) = 0 then p(1/a) = 0 via division?

Yes. To put it another way, whenever we have $xy = 0$ then either $x = 0$ or $y = 0$, or both.