# Polynomials question (1 Viewer)

#### leehuan

##### Well-Known Member
I only need the second part of part (iii). Everything else I could get out but I get something wrong for the proof.

$\bg_white P(x)=x^4+Ax^3+Bx^2+Ax+1\\ x=\alpha\text{ is a zero.}$

$\bg_white \text{(i) Given that }(2+B)^2\neq 4A^2\text{, show that }\alpha \neq 0, \pm 1$

$\bg_white \text{(ii) Show that }x=\frac{1}{\alpha}\text{ is a root.}$

$\bg_white \text{(iii) Deduce that if }\alpha\text{ is a multiple root, it has multiplicity 2 }\textbf{and this: }4B=8+A^2$

I had the four roots being alpha, alpha, 1/alpha, 1/alpha

But then I get 4B = A^2, without the eight

#### InteGrand

##### Well-Known Member
I only need the second part of part (iii). Everything else I could get out but I get something wrong for the proof.

$\bg_white P(x)=x^4+Ax^3+Bx^2+Ax+1\\ x=\alpha\text{ is a zero.}$

$\bg_white \text{(i) Given that }(2+B)^2\neq 4A^2\text{, show that }\alpha \neq 0, \pm 1$

$\bg_white \text{(ii) Show that }x=\frac{1}{\alpha}\text{ is a root.}$

$\bg_white \text{(iii) Deduce that if }\alpha\text{ is a multiple root, it has multiplicity 2 }\textbf{and this: }4B=8+A^2$

I had the four roots being alpha, alpha, 1/alpha, 1/alpha

But then I get 4B = A^2, without the eight
$\bg_white \noindent Let the roots be \alpha, \alpha, \alpha ^{-1}, \alpha ^{-1}. From sum of roots: \alpha + \alpha + \alpha ^{-1} + \alpha ^{-1} = - A \Rightarrow -A = 2 \left(\alpha + \alpha ^{-1}\right) \quad (1).$

$\bg_white \noindent From sum of pairs of roots, \alpha^2 + 1 + 1 + 1 + 1 + \alpha^{-2} = B \Rightarrow B = \alpha^2+ \alpha ^{-2} + 4 \quad (2).$

$\bg_white \noindent Using the identity \alpha^2 + \alpha ^{-2} = \left(\alpha + \alpha ^{-1}\right)^2 -2, and in view of (1), we have B = \left(-\frac{A}{2}\right)^2 -2 + 4 = \frac{A^2}{4} + 2. Multiplying through by 4, we have, 4B = A^2 + 8, as required.$

#### leehuan

##### Well-Known Member
Thanks lol I messed up my sum of two roots at a time