# APEX Calculus: for University of Lethbridge

## Factors and Zeros of Polynomials.

Let $$p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$$ be a polynomial. If $$p(a)=0\text{,}$$ then $$a$$ is a $$zero$$ of the polynomial and a solution of the equation $$p(x)=0\text{.}$$ Furthermore, $$(x-a)$$ is a $$factor$$ of the polynomial.

## Fundamental Theorem of Algebra.

An $$n$$th degree polynomial has $$n$$ (not necessarily distinct) zeros. Although all of these zeros may be imaginary, a real polynomial of odd degree must have at least one real zero.

If $$p(x) = ax^2 + bx + c\text{,}$$ and $$0 \le b^2 - 4ac\text{,}$$ then the real zeros of $$p$$ are $$x=(-b\pm \sqrt{b^2-4ac})/2a$$

## Special Factors.

\begin{align*} x^2 - a^2 \amp = (x-a)(x+a)\\ x^3 - a^3 \amp= (x-a)(x^2+ax+a^2)\\ x^3 + a^3 \amp= (x+a)(x^2-ax+a^2)\\ x^4 - a^4 \amp= (x^2-a^2)(x^2+a^2)\\ (x+y)^n \amp=x^n + nx^{n-1}y+\frac{n(n-1)}{2!}x^{n-2}y^2+\cdots +nxy^{n-1}+y^n\\ (x-y)^n \amp=x^n - nx^{n-1}y+\frac{n(n-1)}{2!}x^{n-2}y^2-\cdots \pm nxy^{n-1}\mp y^n \end{align*}

## Binomial Theorem.

\begin{align*} (x+y)^2 \amp= x^2 + 2xy + y^2\\ (x-y)^2 \amp= x^2 -2xy +y^2\\ (x+y)^3 \amp= x^3 + 3x^2y + 3xy^2 + y^3\\ (x-y)^3 \amp= x^3 -3x^2y + 3xy^2 -y^3\\ (x+y)^4 \amp= x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4\\ (x-y)^4 \amp= x^4 - 4x^3y + 6x^2y^2 - 4xy^3 + y^4 \end{align*}

## Rational Zero Theorem.

If $$p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$$ has integer coefficients, then every $$rational$$ $$zero$$ of $$p$$ is of the form $$x=r/s\text{,}$$ where $$r$$ is a factor of $$a_0$$ and $$s$$ is a factor of $$a_n\text{.}$$

## Factoring by Grouping.

$$ac x^3 + adx^2 + bcx + bd = ax^2(cx+d)+b(cx+d)=(ax^2+b)(cx+d)$$

## Arithmetic Operations.

\begin{align*} ab+ac\amp=a(b+c) \amp \frac{a}{b}+\frac{c}{d} \amp= \frac{ad+bc}{bd} \amp \frac{a+b}{c} \amp = \frac{a}{c} + \frac{b}{c}\\ \frac{\left(\displaystyle\frac{a}{b}\right)}{\left(\displaystyle\frac{c}{d}\right)}\amp=\left(\frac{a}{b}\right)\left(\frac{d}{c}\right)=\frac{ad}{bc} \amp \frac{\left(\displaystyle\frac{a}{b}\right)}{c} \amp = \frac{a}{bc} \amp \frac{a}{\left(\displaystyle\frac{b}{c}\right)} \amp= \frac{ac}{b}\\ a\left(\frac{b}{c}\right)\amp= \frac{ab}{c}\amp \frac{a-b}{c-d}\amp=\frac{b-a}{d-c}\amp \frac{ab+ac}{a}\amp=b+c \end{align*}