Algebraic expressions, identities and factoring

Terms and factors

An expression is built from terms joined by $+$ and $-$; each term is a product of factors. In $3x^2 - 5xy + 2$ there are three terms; the first term has the factors $3$, $x$ and $x$. The distinction decides almost everything in this lesson: you may combine like terms (same variables to the same powers), and you may cancel common factors of a numerator and a denominator, but you may never cancel a term.

Expanding

To multiply two brackets, multiply every term of the first by every term of the second and add the results:

$$(2x - 3)(x + 4) = 2x^2 + 8x - 3x - 12 = 2x^2 + 5x - 12$$

Three products come up so often that it pays to know them by sight, in both directions:

Identity	Expanded
$(a + b)^2$	$a^2 + 2ab + b^2$
$(a - b)^2$	$a^2 - 2ab + b^2$
$(a + b)(a - b)$	$a^2 - b^2$

The middle term $2ab$ is the one people lose. $(a + b)^2$ is not $a^2 + b^2$; a quick test with $a = b = 1$ gives $4$ on one side and $2$ on the other.

Factoring is expanding in reverse

Work through these checks in order.

1. Common factor. Take out the largest factor shared by every term: $6x^3 - 9x^2 = 3x^2(2x - 3)$.
2. Difference of squares. Two squares with a minus between them: $x^2 - 49 = (x - 7)(x + 7)$, and $4m^2 - 25 = (2m - 5)(2m + 5)$. A sum of squares such as $x^2 + 49$ does not factor this way.
3. Trinomial $x^2 + bx + c$. Find two numbers that multiply to $c$ and add to $b$. For $x^2 - x - 12$, the numbers are $-4$ and $3$, so it is $(x - 4)(x + 3)$.
4. Trinomial $ax^2 + bx + c$ with $a \neq 1$. Find two numbers that multiply to $ac$ and add to $b$, split the middle term with them, and factor in pairs (grouping).
5. Perfect square trinomial. If the first and last terms are squares and the middle term is twice the product of their roots, it is a square: $x^2 - 10x + 25 = (x - 5)^2$.

After factoring, take out anything that still factors: $2x^3 - 8x = 2x(x^2 - 4) = 2x(x - 2)(x + 2)$. To check a factorisation, expand it back, or substitute a small number such as $x = 3$ into both forms.

Simplifying algebraic fractions

Factor the numerator and the denominator completely, then cancel factors that appear in both. A whole bracket is one factor; a single term inside a bracket is not.

$$\frac{x^2 + 5x}{x^2 - 25} = \frac{x(x + 5)}{(x - 5)(x + 5)} = \frac{x}{x - 5}, \quad x \neq \pm 5$$

The restriction matters. The original fraction is undefined at $x = 5$ and at $x = -5$, and the simplified one still is at $x = 5$; cancelling a factor never makes a forbidden value allowed. When a question adds a condition such as $x \neq 5$, this is the reason.

Identities let you skip the solving

Questions often give two facts and ask for a third expression. If the third expression is linked to the given ones by an identity, use it directly:

• Given $x + y$ and $x - y$, the product gives $x^2 - y^2$.
• Given $a + b$ and $ab$, rearrange $(a + b)^2 = a^2 + 2ab + b^2$ to get $a^2 + b^2 = (a + b)^2 - 2ab$.
• Given $a - b$ and $ab$, similarly $a^2 + b^2 = (a - b)^2 + 2ab$.

When you see $x^2 - y^2$, $a^2 + b^2$ or $(a + b)^2$ in a question, the question is almost always built on one of these identities. Solving for the variables is possible in some cases but slower, and in others the variables are not whole numbers, so the identity is the only sensible route.

Comparisons built on identities

In a quantitative comparison, subtract one quantity from the other and simplify. Comparing $(x + 3)^2$ with $x^2 + 9$, the difference is $6x$, so the answer depends only on the sign of $x$. If the stem says $x$ is negative, Quantity B is greater; if $x$ could be anything, the relationship cannot be determined.