Revision: algebra and functions

How to use this sheet

Mathematics is 24 of the 50 questions on the standard test, and calculators are not allowed, so speed comes from structure: factor before you multiply out, and test the options when solving would take longer. 75 minutes for 50 questions is 90 seconds a question on average; treating that as a target per maths question is our pacing suggestion, not a rule of the test. Read each block below, cover it, and say the rule back.

Factoring identities

Identity	Expanded form	Factored form	Where it saves time
Difference of squares	$a^2 - b^2$	$(a - b)(a + b)$	rationalising roots, mental arithmetic
Square of a binomial	$a^2 \pm 2ab + b^2$	$(a \pm b)^2$	completing the square in circles and parabolas
Sum of cubes	$a^3 + b^3$	$(a + b)(a^2 - ab + b^2)$	cancelling cubic fractions
Difference of cubes	$a^3 - b^3$	$(a - b)(a^2 + ab + b^2)$	simplifying degree-3 polynomials
Square of a trinomial	$a^2 + b^2 + c^2 + 2(ab + bc + ca)$	$(a + b + c)^2$	symmetric three-variable expressions

The difference of squares is also a mental-arithmetic tool: $\frac{2026^2 - 2024^2}{4} = \frac{(2026 - 2024)(2026 + 2024)}{4} = \frac{2 \times 4050}{4} = 2025$, with no large squares computed.

Radicals

• $\sqrt[n]{a^m} = a^{m/n}$ (for $a \ge 0$), so every root is a fractional power and the power rules apply.
• $\sqrt[m]{\sqrt[n]{a}} = \sqrt[mn]{a}$: nested roots multiply their indices.
• $\sqrt{x^2} = \lvert x \rvert$, never simply $x$: an even root is never negative.
• $\sqrt{a^2 + b^2}$ is not $a + b$. Roots do not split over sums ($\sqrt{9 + 16} = 5$, not $7$).

To rationalise a denominator such as $3 - \sqrt{2}$, multiply top and bottom by its conjugate $3 + \sqrt{2}$; the bottom becomes a difference of squares, $9 - 2 = 7$. So $\frac{5}{3 - \sqrt{2}} = \frac{15 + 5\sqrt{2}}{7}$.

Quadratics: the discriminant and Vieta

For $ax^2 + bx + c = 0$ with $a \neq 0$, the discriminant $\Delta = b^2 - 4ac$ counts the real roots: two if $\Delta > 0$, one (repeated) if $\Delta = 0$, none if $\Delta < 0$. When the roots are real, Vieta's formulas give their sum and product without solving:

Expression	In terms of $S = x_1 + x_2 = -\frac{b}{a}$ and $P = x_1 x_2 = \frac{c}{a}$
$x_1^2 + x_2^2$	$S^2 - 2P$
$\frac{1}{x_1} + \frac{1}{x_2}$	$\frac{S}{P}$
$(x_1 - x_2)^2$	$S^2 - 4P$

Sign tables for inequalities

Move everything to one side, factor, and mark where each factor is zero. For a quadratic with $a > 0$ the parabola opens upwards, so it is negative between the roots and positive outside them: $x^2 - 5x + 6 = (x - 2)(x - 3) < 0$ gives $2 < x < 3$.

For a rational inequality such as $\frac{x - 1}{x + 3} \le 0$, do not multiply across by $x + 3$, whose sign you do not know. Mark the zero of the numerator ($x = 1$) and of the denominator ($x = -3$), test the sign in each interval, and exclude the denominator's zero for good: the solution is $-3 < x \le 1$.

Absolute values

Form	Rule	Example
$\lvert A \rvert < k$ (with $k > 0$)	sandwich: $-k < A < k$	$\lvert 2x - 5 \rvert < 7 \Rightarrow -2 < 2x < 12 \Rightarrow -1 < x < 6$
$\lvert A \rvert > k$ (with $k > 0$)	split: $A > k$ or $A < -k$	$\lvert 3x + 1 \rvert > 8 \Rightarrow x > \frac{7}{3}$ or $x < -3$

If $k$ is negative, $\lvert A \rvert < k$ has no solution and $\lvert A \rvert > k$ holds for every $x$ where $A$ is defined.

Systems and remainders

A line and a parabola meet where their $y$-values are equal. For $y = 2x + 3$ and $y = x^2$: $x^2 - 2x - 3 = 0$, so $(x - 3)(x + 1) = 0$ and the points are $(3, 9)$ and $(-1, 1)$. The discriminant of that equation tells you in advance whether there are two, one or no meeting points.

Remainder conditions are lists in disguise. "Leaves 2 when divided by 10" means the last digit is 2; "leaves 1 when divided by 7" means the number is 1 more than a multiple of 7. List the candidates of the stricter condition, then test them against the other.

Domains

Structure	Condition	Example
Fraction $\frac{P(x)}{Q(x)}$	$Q(x) \neq 0$	$\frac{1}{x - 5}$: $x \neq 5$
Even root $\sqrt[2n]{g(x)}$	$g(x) \ge 0$	$\sqrt{x - 3}$: $x \ge 3$
Logarithm $\log_b h(x)$	$h(x) > 0$, with $b > 0$, $b \neq 1$	$\ln(4 - x)$: $x < 4$

When a function has several pieces, every condition must hold at once: $f(x) = \frac{\sqrt{x - 3}}{x - 5}$ needs $x \ge 3$ and $x \neq 5$. An even root in a denominator needs its radicand strictly positive.

Inverses and composition

To invert $y = f(x)$, swap $x$ and $y$ and solve for $y$. For $f(x) = \frac{2x - 3}{5}$: $x = \frac{2y - 3}{5}$, so $5x + 3 = 2y$ and $f^{-1}(x) = \frac{5x + 3}{2}$. Check by composing: $f(f^{-1}(x)) = \frac{(5x + 3) - 3}{5} = x$.

In $f(g(x))$ the inner function acts first. With $f(x) = x^2 + 1$ and $g(x) = 3x - 2$: $g(2) = 4$, so $f(g(2)) = f(4) = 17$. The order matters: $g(f(2)) = g(5) = 13$.

Exponentials

Write both sides as powers of the same base and compare exponents: $2^{x + 1} = 32 = 2^5$ gives $x = 4$. In an inequality the direction depends on the base. With a base greater than 1, $a^u > a^v$ means $u > v$; with a base between 0 and 1 the function is decreasing, so the inequality flips. Remember also that $a^x > 0$ for every $x$, so an equation like $2^x = -4$ has no solution.

Logarithms

Law	Statement
Product	$\log_b(xy) = \log_b x + \log_b y$
Quotient	$\log_b \frac{x}{y} = \log_b x - \log_b y$
Power	$\log_b(x^k) = k \log_b x$
Change of base	$\log_a b = \frac{\ln b}{\ln a}$

The laws hold for positive $x$ and $y$. To condense $2\log x + \log y - 3\log z$, push the coefficients into exponents and merge: $\log\frac{x^2 y}{z^3}$. Also keep $\log_b 1 = 0$, $\log_b b = 1$ and $b^{\log_b x} = x$ to hand. After solving a log equation, check every root against the domain of the original equation.