Factor polynomials and simplify surds

Use the factor theorem and five identities

Test a proposed linear factor by substituting its root. For x+a, the test input is −a; the sign inside the factor changes.

A polynomial of degree n has form $a_0+a_1x+a_2x^2+a_3x^3+\cdots+a_nx^n$, with $a_n\ne0$. The factor theorem says

$$(x-a)\text{ is a factor of }f(x)\quad\Longleftrightarrow\quad f(a)=0.$$

A value a with $f(a)=0$ is called a root. Example 5 tests x+1 by substituting −1. The source lists all five useful factorisation identities:

$$\begin{aligned} a^2+2ab+b^2&=(a+b)^2,\\ a^2-2ab+b^2&=(a-b)^2,\\ a^2-b^2&=(a+b)(a-b),\\ a^3+b^3&=(a+b)(a^2-ab+b^2),\\ a^3-b^3&=(a-b)(a^2+ab+b^2). \end{aligned}$$

Continue with Tutorial Question 6.

Simplify surds while respecting the radical

Look for a square factor inside the radical before multiplying out. To rationalise a two-term denominator, multiply numerator and denominator by its conjugate.

The source revisits rational numbers as a/b with integer a,b and $b\ne0$, using $1$, $0.5$ and $1/3$. Irrational examples are $\sqrt2$, $-\sqrt5$ and $\pi$. A surd expresses an irrational number using a radical/root symbol. For positive a,b, the three printed properties are

$$\sqrt a\sqrt b=\sqrt{ab},\qquad \frac{\sqrt a}{\sqrt b}=\sqrt{\frac ab},\qquad \sqrt a\sqrt a=a.$$

The source warns against distributing a square root over addition or subtraction: $\sqrt{a+b}$ is not generally $\sqrt a+\sqrt b$, and $\sqrt{a-b}$ is not generally $\sqrt a-\sqrt b$. Where subtraction is used over the reals, its radicand must also be nonnegative. Example 6 retains the printed product and conjugate rationalisation. Continue with Tutorial Question 7.