Polynomial Questions (1 Viewer)

sunmoonlight · Oct 12, 2023

How to do Q22 and 26?

Unovan · Oct 12, 2023

a^n+1 + b^n + 1 = 0 (by factor theorem)

dont have time to look at it deeply cos im on the way to english lmao but i think factor theorem will unlock 22

Drongoski · Oct 12, 2023

Q22

$P(x) = ax^{n+1} + bx^n + 1\\ \\ P'(x) = (n+1)ax^n + nbx^{n-1}\\ \\ \therefore P(1) = a + b + 1 = 0 \implies a+ b = -1 \\ \\ P'(1) = (n+1)a + nb = 0 \implies n(a+b) + a = 0 \\ \\ \implies -1 \times n + a = 0 \implies a = n \\ \\ \therefore a+b+1 = 0 \implies n+b+1 = 0 \implies b = -(n+1)$

Luukas.2 · Oct 12, 2023

sunmoonlight said:
View attachment 40423

Start question 26 by letting the three roots be $\alpha,\ \alpha, \text{ and } \beta \text{ where } \beta \neq \alpha$ .

Root theory (sum of roots, product of roots, etc) can then be used to get $a = \alpha$ for part (a) and then use the three equations to eliminate $\alpha$ and $\beta$ to get the relationship between the coefficients.

Alternatively, from the factor and multiple root theorems, you know that $P(\alpha) = P'(\alpha) = 0$ , which is another approach to part (a). Finding $\beta$ and substituting $P(\beta) = 0$ can then yield the relationships between coefficients.

Using root theory:

$\begin{align*} \text{Sum of roots:} \qquad \alpha + \alpha + \beta &= -\frac{3p}{1} \qquad \implies \qquad \beta = -3p - 2\alpha \qquad \qquad \text{. . . (1)} \\ \text{Sum of roots in pairs:} \qquad \alpha \times \alpha + \alpha \times \beta + \beta \times \alpha &= \frac{3q}{1} \qquad \implies \qquad \alpha^2 + 2\alpha\beta = 3q \qquad \qquad \text{. . . (2)} \\ \text{Product of roots:} \qquad \alpha \times \alpha \times \beta &= -\frac{r}{1} \qquad \implies \qquad \alpha^2 = -\frac{r}{\beta} \qquad \qquad \text{. . . (3)} \\ \\ \text{Put (1) into (2):} \qquad \alpha^2 + 2\alpha(-3p - 2\alpha) &= 3q \\ \alpha^2 + 2p\alpha &= -q \\ (\alpha + p)^2 &= p^2 - q \\ \alpha &= -p \pm \sqrt{p^2 - q} \end{align*}$

The same equation results from the multiple root theorem:

$\begin{align*} P(x) &=x^3 + 3px^2 + 3qx + r \\ P'(x) &= 3x^2 + 6px + 3q \\ \text{Noting that $P'(\alpha) = 0$:} \qquad \qquad 0 &= 3\left(\alpha^2 + 2p\alpha + q\right) \\ \alpha^2 + 2p\alpha + p^2 &= -q + p^2 \\ \left(\alpha + p\right)^2 &= p^2 - q \end{align*}$

You could put each possible solution into $P(x)$ to establish which is the double root.

The coefficient result will arise from putting $\alpha$ and $\beta$ into one of the root theory equations and rearranging.

sunmoonlight · Oct 12, 2023

For Q26, after you find out what $a$ is, how do you prove it? (I still don't get that part).

Luukas.2 · Oct 12, 2023

sunmoonlight said:
For Q26, after you find out what $a$ is, how do you prove it? (I still don't get that part).

Prove that the value is

$a=\frac{pq-r}{2\left(q - p^2\right)$

means that you need to find the value and, in so doing, establish that it can be no other value.

So, if you used the approach I noted above with $\alpha = -p \pm \sqrt{p^2 - q}$ , you would need to establish which of these two possible values is the double root.

sunmoonlight · Oct 12, 2023

How do you find the value of a? like r is unknown

Luukas.2 · Oct 13, 2023

sunmoonlight said:
For Q26, after you find out what $a$ is, how do you prove it? (I still don't get that part).

It would also be sufficient to take the given value for $a$ and show that $P(a) = P'(a) = 0$ .

sunmoonlight · Oct 15, 2023

How to do (i)?

tywebb · Oct 15, 2023

sunmoonlight said:
View attachment 40570
How to do (i)?

$\text{``You may use the fact that }4\cos^3\theta+3\cos\theta=\cos3\theta.\text{ (Do NOT prove this)"$

Well you better not prove it because it's wrong!

The third degree Chebyshev polynomial of the first kind T₃(x)=4x³-3x not 4x³+3x so whoever wrote the paper clearly don't know their Chebyshev polynomials

Changing 4cos³θ+3cosθ=cos3θ to 4cos³θ-3cosθ=cos3θ results in the easy solution

$\text{Suppose }x=r\cos\theta.\text{ Then }$
$\begin{aligned}x^3-px+q&=\textstyle r^3\cos^3\theta-pr\cos\theta-\frac{r^3}{4}\cos3\theta\\&=\textstyle\frac{4p}{3}\sqrt\frac{4p}{3}\cos^3\theta-p\sqrt\frac{4p}{3}\cos\theta-\frac{1}{4}\cdot\frac{4p}{3}\sqrt\frac{4p}{3}(4\cos^3\theta-3\cos\theta)\\&=0\end{aligned}$
$\text{Hence }r\cos\theta\text{ is a root of }x^3-px+q=0$

Polynomial Questions (1 Viewer)

sunmoonlight

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