Algebraic Manipulation and Polynomial Theorems

Expanding Brackets and Collecting Like Terms

In the ESAT, you are expected to be able to multiply out brackets and collect like terms efficiently. Collecting like terms refers to the process of grouping all constant terms together, then grouping all $x$ terms, then all $x^2$ terms, and continuing this for every power of $x$ present in the expression. This organisation ensures that the final polynomial is in its simplest standard form.

Algebraic Factorisation and Long Division

You should be able to factorise simple algebraic expressions, particularly quadratics and expressions with common factors. For higher degree polynomials, such as cubics, factorisation is often achieved using the Factor Theorem. In addition to factorisation, you must be capable of performing algebraic long division by both linear expressions, such as $ax + b$, and quadratic expressions, such as $ax^2 + bx + c$.

Worked Example: Algebraic Long Division

Consider the division of $x^4 + 2x^2 + 3x - 4$ by $x + 3$. To perform this division, it is highly recommended to include a placeholder for any missing powers of $x$. In this case, we include $0x^3$ to keep the columns aligned correctly.

		$x^3$	$-3x^2$	$+11x$	$-30$
$x + 3$	$x^4$	$+0x^3$	$+2x^2$	$+3x$	$-4$
	$x^4$	$+3x^3$	↓	↓	↓
	$0x^4$	$-3x^3$	$+2x^2$	$+3x$	$-4$
		$-3x^3$	$-9x^2$	↓	↓
		$0x^3$	$+11x^2$	$+3x$	$-4$
			$+11x^2$	$+33x$	↓
			$0x^2$	$-30x$	$-4$
				$-30x$	$-90$
				$0x$	$86$

At each stage, division involves only the leading terms, but when we multiply back to find the subtraction value, we use all terms of the divisor. We stop when the remaining value (86) has a lower degree than the divisor ($x + 3$). Thus, 86 is the remainder. We can express the result as:

$x^4 + 2x^2 + 3x - 4 = (x^3 - 3x^2 + 11x - 30)(x + 3) + 86$

This follows the standard division logic where a number like 11 divided by 4 is 2 with a remainder of 3, written as $11 = (2 \times 4) + 3$.

The Factor Theorem

An algebraic expression is a factor of another if it divides into it exactly, meaning there is no remainder. For example, in the expression $(x + 2)(x - 3)$, both $(x + 2)$ and $(x - 3)$ are factors. Using functional notation, $f(x)$ represents an algebraic expression. If $f(x) = x^2 + 3x + 2$, then $f(2) = 2^2 + 3(2) + 2 = 12$.

The Factor Theorem states: If $f(x)$ is a polynomial in $x$, then $f(a) = 0$ if and only if $(x - a)$ is a factor of $f(x)$. This is a bidirectional statement. If we find $f(a) = 0$, then $(x - a)$ must be a factor. Conversely, if we know $(x - a)$ is a factor, then $f(a)$ must equal zero. This theorem links the roots of an equation (where the graph $y = f(x)$ crosses the $x$ axis) to the algebraic factors.

Example 1: Factorising a Quadratic

Factorise $f(x) = x^2 + 3x + 2$. If the factors are $(x - a)$ and $(x - b)$, then $ab = 2$. We test the integer factors of 2: $\pm 1$ and $\pm 2$. Since all terms are positive, we try negative values. $f(-1) = (-1)^2 + 3(-1) + 2 = 1 - 3 + 2 = 0$. By the Factor Theorem, $(x - (-1))$, or $(x + 1)$, is a factor. Through inspection or division, the other factor is $(x + 2)$.

Example 2: Factorising a Cubic

Factorise $f(x) = x^3 + x^2 - 5x + 3$. We test factors of the constant term 3: $1, -1, 3, -3$. $f(1) = 1^3 + 1^2 - 5(1) + 3 = 0$, so $(x - 1)$ is a factor. Dividing $f(x)$ by $(x - 1)$ gives the quadratic $x^2 + 2x - 3$. Factorising this quadratic gives $(x - 1)(x + 3)$. Therefore, $f(x) = (x - 1)^2(x + 3)$.

The Remainder Theorem

The Remainder Theorem states that when a polynomial $f(x)$ is divided by $(x - b)$, the remainder is $f(b)$. This can be shown by writing $f(x) = g(x)(x - b) + R$. Substituting $x = b$ results in $f(b) = g(b)(0) + R$, so $f(b) = R$.

If the divisor is of the form $(px - q)$, the remainder is $f(q/p)$. This is because $px - q = 0$ when $x = q/p$. For more general cases, such as dividing by a quadratic $(x^2 + bx + c)$, the remainder will be of the form $mx + n$. The degree of the remainder is always at least one less than the degree of the divisor.