Algebraic expressions: powers, radicals, polynomials and identities

Why this matters without a calculator

Without a calculator, the way through heavy-looking numbers is to simplify the expression first. A question that looks like heavy arithmetic, such as $\frac{2^{10} \cdot 4^3}{8^4}$ or $101^2 - 99^2$, is often a question about a rule. Your job is to spot the rule before you multiply anything.

Powers

For any base $a \ne 0$ and integers $m, n$:

Rule	Example
$a^m \cdot a^n = a^{m+n}$	$x^3 \cdot x^5 = x^8$
$\frac{a^m}{a^n} = a^{m-n}$	$\frac{5^7}{5^4} = 5^3$
$(a^m)^n = a^{mn}$	$(2^3)^4 = 2^{12}$
$(ab)^n = a^n b^n$	$(3x)^2 = 9x^2$
$a^0 = 1$, $a^{-n} = \frac{1}{a^n}$	$2^{-3} = \frac{1}{8}$
$a^{\frac{m}{n}} = \sqrt[n]{a^m}$ (for $a > 0$)	$8^{\frac{2}{3}} = (\sqrt[3]{8})^2 = 4$

The rules only combine the same base. Before you use them, rewrite every number as a power of a common base: $4 = 2^2$, $8 = 2^3$, $27 = 3^3$, $\frac{1}{9} = 3^{-2}$. There is no rule for $a^m + a^n$; a sum of powers is simplified only by factoring, for example $2^{10} + 2^{10} = 2 \cdot 2^{10} = 2^{11}$.

Watch the minus sign: $-3^2 = -9$, because the power applies only to the 3, while $(-3)^2 = 9$.

Radicals

For $a, b \ge 0$, $\sqrt{ab} = \sqrt{a}\sqrt{b}$ and $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$ (with $b > 0$). To simplify a root, pull out the largest square factor: $\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}$. Roots with the same radicand combine like like terms: $3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}$.

Two facts cause most mistakes:

• $\sqrt{a + b} \ne \sqrt{a} + \sqrt{b}$. For instance $\sqrt{9 + 16} = 5$, but $\sqrt{9} + \sqrt{16} = 7$.
• $\sqrt{x^2} = |x|$, not $x$. A square root is never negative, so $\sqrt{(-4)^2} = 4$.

To rationalise a denominator, multiply top and bottom by what removes the root: by $\sqrt{a}$ for $\frac{1}{\sqrt{a}}$, and by the conjugate for a sum or difference, because $(\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b}) = a - b$.

Special products

These identities are true for every value of the letters, and they work in both directions: to expand and to factor.

$$(a + b)^2 = a^2 + 2ab + b^2 \qquad (a - b)^2 = a^2 - 2ab + b^2$$

$$(a + b)(a - b) = a^2 - b^2$$

$$a^3 + b^3 = (a + b)(a^2 - ab + b^2) \qquad a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$

They also turn arithmetic into algebra: $49 \times 51 = (50 - 1)(50 + 1) = 2500 - 1 = 2499$. And they let you find one symmetric expression from others without solving for the letters: since $(a + b)^2 = a^2 + b^2 + 2ab$, knowing $a + b$ and $ab$ gives $a^2 + b^2$.

Polynomials and the remainder theorem

When a polynomial $P(x)$ is divided by $x - c$, the remainder is $P(c)$. So $x - c$ is a factor of $P(x)$ exactly when $P(c) = 0$. This gives a fast test for factors and a way to find a missing coefficient: substitute the root and set the result to zero. Once one factor is found, divide (or match coefficients) to find the rest.

Simplifying fractions

You may cancel factors, never terms. In $\frac{x^2 - 9}{x - 3}$, factor first: $\frac{(x - 3)(x + 3)}{x - 3} = x + 3$, valid for $x \ne 3$, where the original is undefined. In $\frac{x + 6}{6}$ nothing cancels: the 6 on top is a term of a sum, not a factor.