Algebraic Methods

Polynomial Division

Method 1

Find $(4x^4 - 17x^2 + 4) \div (2x + 1)$

$$\begin{array}{r} 2x^3 - x^2 - 8x + 4 \ \\ 2x + 1 \mathbin{\overline{\smash{\big)}\, 4x^4 + 0x^3 - 17x^2 + 0x + 4}} \\ \underline{4x^4 + 2x^3 \phantom{- 17x^2 + 0x + 4}} \\ -2x^3 - 17x^2 \phantom{+ 0x + 4} \\ \underline{-2x^3 - x^2 \phantom{+ 0x + 4}} \\ -16x^2 + 0x \phantom{+ 4} \\ \underline{-16x^2 - 8x \phantom{+ 4}} \\ 8x + 4 \\ \underline{8x + 4} \\ 0 \end{array}$$

So $4x^4 - 17x^2 + 4 = (2x + 1)(2x^3 - x^2 - 8x + 4)$

Method 2

Only works if divisor is a factor

Divide $x^3 + 2x^2 - 17x + 6$ by $(x-3)$

$$\begin{align} x^3 + 2x^2 - 17x + 6 &= (x-3)(ax^2+bc+c) \\ \\ a&=1 \\ \\ -3c&=6 \\ c&=-2 \\ \\ 2&=b-3a \\ 2&=b-3 \\ 5&=b \end{align}$$

Homework

Factor Theorem

The factor theorem states that if $(x-p)$ is a factor, $f(p) = 0$, and vice versa

Mathematical Proof

Methods of Proof

• Exhaustion
  • Break down into smaller cases and prove individually to prove a statement
• Counter-example
  • Use one example to disprove a statement

Example 12

Prove that all square numbers are either a multiple of 4 or 1 more than a multiple of 4

$$\begin{align} \text{Odd case} \\ (2n+1)^2 &= 4n^2 + 4n + 1 \\ &= 4(n^2+n) + 1 \\ \implies \text{One more than a multiple of 4} \\ \\ \text{Even case} \\ (2n)^2 = 4n^2 \\ \implies \text{Multiple of 4} \\ \\ \text{Q.E.D.} \end{align}$$

Example 13

Prove that the following statement is not true:

The sum of two consecutive prime numbers is always even

$$\begin{gather} 2 + 3 = 5\text{ which is not even} \\ \text{Q.E.D.} \end{gather}$$

Example 14

Prove that for all positive values of $x$ and $y$:

$$\frac{x}{y} + \frac{y}{x} \geq 2$$ $$\begin{align} \frac{x}{y} + \frac{y}{x} &= 2 \\ \frac{x^2 + y^2}{xy} &= 2xy\\ x^2 - 2xy + y^2 &= 0 \\ (x-y)^2 &= 0 \\ \\ \text{consider }(x-y)^2 \\ (x-y)^2 &\geq 0 \\ x^2 - 2xy + y^2 &\geq 0 \\ \frac{x^2 - 2xy + y^2}{xy} &\geq 0\text{ where } xy \neq 0\text{ and $xy$ cannot be negative} \\ \frac{x}{y} - 2 + \frac{y}{x} &\geq 0 \\ \left( \frac{x}{y} + \frac{y}{x} \right) - 2 &\geq 0 \\ \therefore \frac{x}{y} + \frac{y}{x} &\geq 2 \end{align}$$

Use a counter-example to show that this is not true when $x$ and $y$ are not both positive

$$\begin{align} \text{Let }x=-1,\ y=2 \\ \frac{-1}{2} + \frac{2}{-1} &= -\frac{1}{2} + -\frac{2}{1} \\ &= -\frac{1}{2} - 2 \\ &= -\frac{5}{2} \\ \text{Q.E.D.} \end{align}$$