Further Algebra

Further algebra builds on the polynomial and algebraic techniques from A Level mathematics, Extending to partial fractions with irreducible quadratics, the relationships between roots and Coefficients of polynomial equations, and systematic summation of series using the method of Differences.

To divide $P(x)$ by $(ax + b)$Perform polynomial long division (or synthetic division) to obtain:

$P(x) = (ax + b)Q(x) + R$

Where $Q(x)$ is the quotient and $R$ is a constant remainder.

Let $P(x)$ be divided by $(x - c)$:

$P(x) = (x - c)Q(x) + R$

For some polynomial $Q(x)$ and constant $R$. Setting $x = c$:

$P(c) = (c - c)Q(c) + R = 0 + R = R$

$\boxed{P(c) = R}$

$\square$

Definition. If $P(c) = 0$Then $(x - c)$ is a factor of $P(x)$. This is the factor theorem.

This follows directly from the remainder theorem: if the remainder is zero, the divisor is a factor.

When a polynomial has unknown coefficients, use the factor theorem by substituting known roots, or Use the remainder theorem by evaluating at specified points.

In A Level, partial fractions involved only linear factors. In further mathematics, denominators may Contain irreducible quadratic factors, requiring a different decomposition.

Definition. A quadratic $x^2 + cx + d$ is irreducible if it has no real roots, i.e. $\Delta = c^2 - 4d < 0$.

$\boxed{\frac{px + q}{(ax + b)(x^2 + cx + d)} = \frac{A}{ax + b} + \frac{Bx + C}{x^2 + cx + d}}$

The numerator of the irreducible quadratic factor is always linear ($Bx + C$), not just a constant.

$\boxed{\frac{px^2 + qx + r}{(x^2 + a)^2} = \frac{Ax + B}{x^2 + a} + \frac{Cx + D}{(x^2 + a)^2}}$

When the irreducible quadratic is repeated, the numerators follow the same pattern as repeated Linear factors.

$\boxed{\frac{px^2 + qx + r}{(x^2 + cx + d)(x^2 + ex + f)} = \frac{Ax + B}{x^2 + cx + d} + \frac{Cx + D}{x^2 + ex + f}}$

Each distinct irreducible quadratic factor contributes a linear numerator.

Info: info OCR cover Types 1 and 2. CIE covers Type 1 extensively in P3.

If $P(x) = ax^3 + bx^2 + cx + d = a(x - \alpha)(x - \beta)(x - \gamma)$ where $\alpha, \beta, \gamma$ Are the roots, then:

$\boxed{\alpha + \beta + \gamma = -\frac{b}{a}}$

$\boxed{\alpha\beta + \alpha\gamma + \beta\gamma = \frac{c}{a}}$

$\boxed{\alpha\beta\gamma = -\frac{d}{a}}$

Let $P(x) = ax^3 + bx^2 + cx + d = a(x - \alpha)(x - \beta)(x - \gamma)$.

Expanding the RHS:

$a[(x - \alpha)(x - \beta)(x - \gamma)] = a[x^3 - (\alpha + \beta + \gamma)x^2 + (\alpha\beta + \alpha\gamma + \beta\gamma)x - \alpha\beta\gamma]$

$= ax^3 - a(\alpha + \beta + \gamma)x^2 + a(\alpha\beta + \alpha\gamma + \beta\gamma)x - a\alpha\beta\gamma$

Comparing coefficients with $ax^3 + bx^2 + cx + d$:

• $x^2$: $-a(\alpha + \beta + \gamma) = b \implies \alpha + \beta + \gamma = -\dfrac{b}{a}$
• $x^1$: $a(\alpha\beta + \alpha\gamma + \beta\gamma) = c \implies \alpha\beta + \alpha\gamma + \beta\gamma = \dfrac{c}{a}$
• $x^0$: $-a\alpha\beta\gamma = d \implies \alpha\beta\gamma = -\dfrac{d}{a}$

$\square$

For $P(x) = ax^4 + bx^3 + cx^2 + dx + e = a(x - \alpha)(x - \beta)(x - \gamma)(x - \delta)$:

$\boxed{\sum\alpha = \alpha + \beta + \gamma + \delta = -\frac{b}{a}}$

$\boxed{\sum\alpha\beta = \alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta = \frac{c}{a}}$

$\boxed{\sum\alpha\beta\gamma = -\frac{d}{a}}$

$\boxed{\alpha\beta\gamma\delta = \frac{e}{a}}$

Using the elementary symmetric sums, we can express other symmetric functions:

• $\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \alpha\gamma + \beta\gamma)$
• $\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma} = \dfrac{\alpha\beta + \alpha\gamma + \beta\gamma}{\alpha\beta\gamma}$
• $\alpha^2\beta + \alpha^2\gamma + \beta^2\alpha + \beta^2\gamma + \gamma^2\alpha + \gamma^2\beta = (\alpha + \beta + \gamma)(\alpha\beta + \alpha\gamma + \beta\gamma) - 3\alpha\beta\gamma$

The following summation formulae are essential:

$\boxed{\sum_{r=1}^{n} r = \frac{n(n+1)}{2}}$

$\boxed{\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}}$

$\boxed{\sum_{r=1}^{n} r^3 = \left[\frac{n(n+1)}{2}\right]^2}$

To find $\displaystyle\sum_{r=1}^{n} f(r)$ where $f(r)$ can be written as $g(r) - g(r+1)$:

$\sum_{r=1}^{n} f(r) = \sum_{r=1}^{n} [g(r) - g(r+1)] = g(1) - g(n+1)$

This is a telescoping sum — all intermediate terms cancel.

Note that $r^3 - (r-1)^3 = 3r^2 - 3r + 1$So $r^2 = \dfrac{r^3 - (r-1)^3 + 3r - 1}{3}$.

Summing from $r = 1$ to $n$:

$\sum_{r=1}^{n} r^2 = \frac{1}{3}\sum_{r=1}^{n}[r^3 - (r-1)^3] + \sum_{r=1}^{n} r - \frac{n}{3}$

The first sum telescopes: $\sum_{r=1}^{n}[r^3 - (r-1)^3] = n^3 - 0 = n^3$.

$\sum_{r=1}^{n} r^2 = \frac{n^3}{3} + \frac{n(n+1)}{2} - \frac{n}{3} = \frac{2n^3 + 3n^2 + 3n + 2n^2 + 2n - 2n}{6} \cdot \frac{1}{1}$

More carefully:

$\sum_{r=1}^{n} r^2 = \frac{n^3}{3} + \frac{n(n+1)}{2} - \frac{n}{3} = \frac{2n^3 + 3n(n+1) - 2n}{6} = \frac{2n^3 + 3n^2 + 3n - 2n}{6}$

$= \frac{2n^3 + 3n^2 + n}{6} = \frac{n(2n^2 + 3n + 1)}{6} = \frac{n(n+1)(2n+1)}{6} \quad \square$

$\boxed{\sum_{r=1}^{n} r(r+1) = \frac{n(n+1)(n+2)}{3}}$

$\boxed{\sum_{r=1}^{n} r(r+1)(r+2) = \frac{n(n+1)(n+2)(n+3)}{4}}$

$\displaystyle\sum_{r=1}^{n} \binom{r+k}{k+1} = \binom{n+k+1}{k+2}$.

When the general term of a series can be decomposed into partial fractions, the method of Differences often applies.

To find $\displaystyle\sum_{r=1}^{n} r \cdot a_r$ where $a_r = f(r) - f(r-1)$:

$\sum_{r=1}^{n} r \cdot a_r = \sum_{r=1}^{n} r[f(r) - f(r-1)] = nf(n) - \sum_{r=0}^{n-1} f(r)$

This is known as the summation by parts technique.

Base case ($n = 1$): $\displaystyle\sum_{r=1}^{1} r = 1 = \frac{1 \times 2}{2}$. ✓

Inductive step. Assume $\displaystyle\sum_{r=1}^{k} r = \frac{k(k+1)}{2}$. Then:

$\sum_{r=1}^{k+1} r = \frac{k(k+1)}{2} + (k+1) = \frac{k(k+1) + 2(k+1)}{2} = \frac{(k+1)(k+2)}{2}$

✓ $\square$

Base case ($n = 1$): $1^3 = 1 = \left[\dfrac{1 \times 2}{2}\right]^2 = 1$. ✓

Inductive step. Assume $\displaystyle\sum_{r=1}^{k} r^3 = \left[\frac{k(k+1)}{2}\right]^2$. Then:

$\sum_{r=1}^{k+1} r^3 = \left[\frac{k(k+1)}{2}\right]^2 + (k+1)^3 = \frac{k^2(k+1)^2}{4} + \frac{4(k+1)^3}{4}$

$= \frac{(k+1)^2[k^2 + 4(k+1)]}{4} = \frac{(k+1)^2(k+2)^2}{4} = \left[\frac{(k+1)(k+2)}{2}\right]^2$

✓ $\square$

$\boxed{P(c) = R \quad \mathrm{(Remainder Theorem)}}$

$\boxed{\frac{px + q}{(ax + b)(x^2 + cx + d)} = \frac{A}{ax + b} + \frac{Bx + C}{x^2 + cx + d}}$

$\boxed{\alpha + \beta + \gamma = -\frac{b}{a}, \quad \alpha\beta + \alpha\gamma + \beta\gamma = \frac{c}{a}, \quad \alpha\beta\gamma = -\frac{d}{a}}$

$\boxed{\sum_{r=1}^{n} r = \frac{n(n+1)}{2}, \quad \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}, \quad \sum_{r=1}^{n} r^3 = \left[\frac{n(n+1)}{2}\right]^2}$

$\boxed{\sum_{r=1}^{n} [g(r) - g(r+1)] = g(1) - g(n+1)}$

Problem 1. Express $\dfrac{2x^2 + 3x + 4}{(x + 2)(x^2 + 2x + 5)}$ in partial fractions.

Problem 2. The equation $x^3 - 4x^2 + x + 6 = 0$ has roots $\alpha, \beta, \gamma$. Find the Value of $\dfrac{1}{\alpha\beta} + \dfrac{1}{\alpha\gamma} + \dfrac{1}{\beta\gamma}$.

Problem 3. Express $\dfrac{3x + 1}{(x^2 + 1)(x^2 + 4)}$ in partial fractions.

Problem 4. Find $\displaystyle\sum_{r=1}^{n} \frac{2}{r(r+1)(r+2)}$.

Problem 5. The equation $3x^3 + px^2 + qx + 12 = 0$ has roots $\alpha, \beta, \gamma$ such that $\alpha + \beta + \gamma = 4$ and $\alpha\beta\gamma = -4$. Find $p$, $q$And $\alpha\beta + \alpha\gamma + \beta\gamma$.

Problem 6. Find $\displaystyle\sum_{r=1}^{n} \frac{1}{r(r+3)}$.

Problem 7. The polynomial $P(x) = x^4 + ax^3 + bx^2 + cx + d$ has roots $\alpha, \beta, \gamma, \delta$. Given that $\alpha + \beta = 3$, $\gamma + \delta = -5$, and $\alpha\beta = 2$Find $a$ and $b$.

Problem 8. Prove by induction that $\displaystyle\sum_{r=1}^{n} r(r+1) = \frac{n(n+1)(n+2)}{3}$ For all $n \in \mathbb{Z}^+$.

Problem 9. Express $\dfrac{x^2 + 3x + 2}{(x^2 + 2x + 3)^2}$ in partial fractions.

Problem 10. The cubic equation $x^3 + px^2 + qx + r = 0$ has roots $\alpha, \beta, \gamma$ where $\beta = 2\alpha$ and $\gamma = 3\alpha$. Express $p$, $q$And $r$ in terms of $\alpha$And hence Find the roots when $p = -6$.

Problem. Find the coefficient of $x^4$ in the expansion of $(1 - 2x)^{-1/2}$ up to and including The term in $x^4$.

Solution. Using the general binomial expansion for $|x| < \dfrac{1}{2}$:

$(1+y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \frac{n(n-1)(n-2)}{3!}y^3 + \frac{n(n-1)(n-2)(n-3)}{4!}y^4 + \cdots$

With $n = -\dfrac{1}{2}$ and $y = -2x$:

$(1-2x)^{-1/2} = 1 + \left(-\frac{1}{2}\right)(-2x) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}(-2x)^2 + \cdots$

$= 1 + x + \frac{3}{8}(4x^2) + \cdots = 1 + x + \frac{3}{2}x^2 + \cdots$

The $x^4$ coefficient: $\dfrac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)\left(-\frac{7}{2}\right)}{24}(16) = \dfrac{105}{16} \cdot \dfrac{16}{24} = \dfrac{105}{24} = \boxed{\dfrac{35}{8}}$.

Problem. The cubic equation $x^3 - 3x^2 + 4 = 0$ has roots $\alpha, \beta, \gamma$. Find the Value of $\alpha^2 + \beta^2 + \gamma^2$.

Solution. By Vieta’s formulae: $\alpha + \beta + \gamma = 3$ and $\alpha\beta + \beta\gamma + \gamma\alpha = 0$.

$\alpha^2 + \beta^2 + \gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = 9 - 0 = \boxed{9}$

Problem. Find $\displaystyle\sum_{r=1}^{n} \frac{1}{r(r+1)}$ and deduce $\displaystyle\sum_{r=1}^{\infty} \frac{1}{r(r+1)}$.

Solution. $\dfrac{1}{r(r+1)} = \dfrac{1}{r} - \dfrac{1}{r+1}$.

$\sum_{r=1}^{n} \frac{1}{r(r+1)} = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right) = 1 - \frac{1}{n+1} = \frac{n}{n+1}$