\usepackage{multirow}
\usepackage{newclude}
\usepackage{pgfplots}
+\usepackage{polynom}
\usepackage{pst-plot}
\usepackage{standalone}
\usepackage{subfiles}
\newtcolorbox{cas}{colframe=cas!75!black, title=On CAS, left*=3mm}
\newtcolorbox{warning}{colback=white!90!black, leftrule=3mm, colframe=important, coltext=important, fontupper=\sffamily\bfseries}
+\newtcolorbox{theorembox}[1]{colback=green!10!white, colframe=blue!20!white, coltitle=black, fontupper=\sffamily, fonttitle=\sffamily, #1}
\begin{document}
\begin{enumerate} \tightlist
\item Write as matrices: \(\begin{bmatrix}p & q \\ r & s \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} a \\ b \end{bmatrix}\)
- \item Find determinant of first matrix: \(\Delta = ps-qr\)
- \item Let \(\Delta = 0\) for number of solutions \(\ne 1\)\\
- or let \(\Delta \ne 0\) for one unique solution.
- \item Solve determinant equation to find variable \\
+ \item Find \(\det(\text{first matrix}) = ps-qr\)
+ \item Let \(\det = 0\) for \(\{0,\infty\}\) solutions
+ or \(\det \ne 0\) for 1 solution
+ \item Solve to find variable \\ \\
\textbf{For infinite/no solutions:}
\item Substitute variable into both original equations
- \item Rearrange equations so that LHS of each is the same
- \item \(\text{RHS}(1) = \text{RHS}(2) \implies (1)=(2) \> \forall x\) (\(\infty\) solns)\\
- \(\text{RHS}(1) \ne \text{RHS}(2) \implies (1)\ne(2) \> \forall x\) (0 solns)
+ \item Rearrange so that LHS of each is the same
+ \item \(\begin{aligned}[t]
+ \infty \text{ solns: } & \text{RHS}(1) = \text{RHS}(2) \implies (1)=(2) \> \forall x \\
+ 0 \text{ solns: } & \text{RHS}(1) \ne \text{RHS}(2) \implies (1)\ne(2) \> \forall x
+ \end{aligned}\)
\end{enumerate}
- \colorbox{cas}{On CAS:} Matrix \(\rightarrow\) \texttt{det}
+ \begin{cas}
+ Action \(\rightarrow\) Matrix \(\rightarrow\) Calculation \(\rightarrow\) \texttt{det}
+ \end{cas}
\subsubsection*{Solving \(\protect\begin{cases}a_1 x + b_1 y + c_1 z = d_1 \\ a_2 x + b_2 y + c_2 z = d_2 \\ a_3 x + b_3 y + c_3 z = d_3\protect\end{cases}\)}
\section{Polynomials}
+ \subsection*{Factor theorem}
+
+ \begin{theorembox}{title=General form \(\beta x + \alpha\)}
+ If \(\beta x + \alpha\) is a factor of \(P(x)\), \\
+ \-\hspace{1em}then \(P(-\dfrac{\alpha}{\beta})=0\).
+ \end{theorembox}
+
+ \begin{theorembox}{title=Simple form \(x-a\)}
+ If \((x-a)\) is a factor of \(P(x)\), remainder \(R=0\). \\
+ \-\hspace{1em}\(\implies P(a)=0\)
+ \end{theorembox}
+
+ \subsection*{Remainder theorem}
+
+ \begin{theorembox}{}
+ When \(P(x)\) is divided by \(\beta x + \alpha\), the remainder is \(-\dfrac{\alpha}{\beta}\).
+ \end{theorembox}
+
+ \subsection*{Rational root theorem}
+ Let \(P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0\) be a polynomial of degree \(n\) with \(a_i \in \mathbb{Z} \forall a\). Let \(\alpha, \beta \in \mathbb{Z}\) such that their highest common factor is 1 (i.e. relatively prime).
+
+ If \(\beta x + \alpha\) is a factor of \(P(x)\), then \(\beta\) divides \(a_n\) and \(\alpha\) divides \(a_0\) .
+
+ \subsubsection*{Discriminant}
+ \[\begin{cases}
+ b^2-4ac > 0 & \text{two solutions} \\
+ b^2-4ac = 0 & \text{one solution} \\
+ b^2-4ac < 0 & \text{no solutions}
+ \end{cases}\]
+ \begin{warning}
+ Flip inequality sign when multiplying by -1
+ \end{warning}
+
+ \subsection*{Long division}
+
+ \[ \polylongdiv{x^2+2x+4}{x-1} \]
+
+ \begin{cas}
+ Action \(\rightarrow\) Transformation \(\rightarrow\) \texttt{propFrac}
+ \end{cas}
+
\subsection*{Linear equations}
\subsubsection*{Forms}
Distance: \(|\vec{AB}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
\subsection*{Quadratics}
+
\setlength{\abovedisplayskip}{1pt}
\setlength{\belowdisplayskip}{1pt}
+
+ \textbf{Linear factorisation}
\[ x^2 + bx + c = (x+m)(x+n) \]
\hfill where \(mn=c, \> m+n=b\)