2.3 Polynomials


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2.3 Contents


Learn: Paul's Online Notes

2.3.1 Cancellation

In a factored function, like terms in both a numerator and denominator as multipliers reduce to 1 $$\frac{a(x)·f(x)}{a(x)·g(x)}=\frac{f(x)}{g(x)}$$
Jump to Sum & Difference Identities
Examples
$$\frac{(x-4)·(x+2)}{(x-4)·(x-2)}=\frac{(x+2)}{(x-2)}$$ $$\frac{(x-4)·(x+2)}{(4-x)·(x-2)}=-\frac{(x+2)}{(x-2)}$$

2.3.2 Quadratic Equations Definition

Second degree trinomial equations in the standard form $$a·x^2+b·x+c=0$$ The following is a more useful general form $$y=a·x^2+b·x+y_∘$$
Jump to Quadratic Formula

2.3.3 Composite Quadratic Equations Factored

Quadratic equations can be thought of as the product of two linear equations and are always factorable into linear terms $$a·c·x^2 \pm (a·d+b·c)·x+b·d$$ $$=(a·x \pm b)(c·x \pm d)$$
Reference Plus or Minus Notation
Learn: Yoshiwara Books

2.3.4 Prime Quadratic Equations Factored (Completing the Square)

The leading coefficient must be 1, otherwise the equation must be multiplied for it to be 1. Using the general form $$x^2 \pm b·x+y_∘=y$$ Add ${(b/2)}^2$ and subtract $y_∘$ $$x^2 \pm b·x+ {\Big(\frac{b}{2} \Big)}^2=y-y_∘+{\Big(\frac{b}{2} \Big)}^2$$ Factor $$\frac{{(2·x \pm b)}^2}{4}=y-y_∘+\frac{b^2}{4}$$
Reference Plus or Minus Notation
Learn: Paul's Online Notes
Jump to Circles

2.3.5 Quadratic Formula

The general solutions of $y=a·x^2+b·x+y_∘$ are $$x=\frac{-b \pm \sqrt{b^2+4·a·(y-y_∘)}}{2·a},\medspace a≠0$$
Reference Plus or Minus Notation
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Proof
Given the general formula, subtract $y_∘$ from both sides $$y-y_∘=a·x^2+b·x$$ Multiply by $4·a$ $$4·a·(y-y_∘)=4·a^2·x^2+4·a·b·x$$ Add $b^2$, then factor the right, similar to completing the square $$b^2+4·a·(y-y_∘)={(2·a·x+b)}^2$$ Take the square root $$\pm \sqrt{b^2+4·a·(y-y_∘)}=2·a·x+b$$ Subtract $b$, and divide by $2·a$
Example
Solve for $x$ and $y$ $$x+y=1\medspace\land\medspace x·y=1$$ Isolate $y$ in the first equation $$y=1-x$$ Substitute for $y$ in the second equation $$x·(1-x)=1$$ Expand $$x-x^2=1$$ Rearrange into a standard quadratic equation $$-x^2+x-1=0$$ Negate $$x^2-x+1=0$$ Use the quadratic formula to find the zeros of $x$ $$x=\frac{-(-1) \pm \sqrt{(-1)^2-4·1·1}}{2·1}$$ Simplify in expanded form $$x=\frac{1}{2}\pm\frac{\sqrt{3}}{2}·i$$ Substitute for $x$ in the first equation with $y$ isolated $$y=1-\bigg(\frac{1}{2}\pm\frac{\sqrt{3}}{2}·i\bigg)$$ Simplify in expanded form $$y=\frac{1}{2}\mp\frac{\sqrt{3}}{2}·i$$ The solutions for $(x,y)$ are $$\bigg(\frac{1}{2}\pm\frac{\sqrt{3}}{2}·i,\frac{1}{2}\mp\frac{\sqrt{3}}{2}·i\bigg)$$

2.3.6 Discriminant of a Quadratic Formula

$$Δ=b^2-4·a·y_∘$$

2.3.7 Sum & Difference of Powers

Sum of Powers for Real Roots (Odd Exponents Only)
$$a^n+b^n=(a+b)(a^{n-1}-a^{n-2}·b+a^{n-3}·b^2-...+a^2·b^{n-3}-a·b^{n-2}+b^{n-1}), \forall 2·n-1 \isin \N$$
Difference of Powers for Real Roots
$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}·b+a^{n-3}·b^2+...+a^2·b^{n-3}+a·b^{n-2}+b^{n-1}), \forall 2·n \isin \N$$
Reference Plus or Minus Notation
Sum & Difference of Squares & Cubes
$$a^2+b^2=(a+i·b)(a-i·b)$$ $$a^2-b^2=(a+b)(a-b)$$ $$a^3 \pm b^3=(a \pm b)(a^2 \mp a·b+b^2)$$
Other Sums & Differences of Interest
$$a^4-b^4=(a+i·b)(a-i·b)(a+b)(a-b)$$ $$a^6+b^6=(a^2+b^2)(a^4-a^2·b^2+b^4)$$ $$a^{10}+b^{10}=(a^2+b^2)(a^8-a^6·b^2+a^4·b^4-a^2·b^6+b^8)$$

2.3.8 Pascal's Triangle

Each number in the given 'triangle' below the first is the sum of the numbers diagonally above it

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
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2.3.9 Binomials

Binomial Coefficient
$$ \binom{n}{k}=\frac{n!}{(n-k)!·k!}=\frac{n·(n-1)(n-2)...(n-k+1)}{k!}$$
Binomial Theorem
The coefficients in the expanded form match the line $n+1$ down from the top on Pascal’s triangle $${(x+y)}^n=\binom{n}{0}·x^n+\binom{n}{1}·x^{n-1}·y+\binom{n}{2}·x^{n-2}·y^2+...+\binom{n}{n}·y^n, \forall n \isin \Z$$
Reference Plus or Minus Notation
Jump to Heron's Formula
Binomials Squared & Cubed
$${(a \pm b)}^2=a^2 \pm 2·a·b+b^2$$
$${(a \pm b)}^3=a^3 \pm 3·a^2·b+3·a·b^2 \pm b^3$$

2.3.A Equating Coefficients (Coefficient Method of Solving)

In polynomial equations with matching factors, orders, and results, variable coefficients can be equated for solving $$a·x+b·y=13\medspace\land\medspace 4·x+3·y=13$$ $$a=4 \qquad b=3$$
$$a·x+b·y=m\medspace\land\medspace (n-1)·x+\frac{p}{2}·y=m$$ $$a=(n-1) \qquad b=p/2$$
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Jump to Ellipses
Jump to Parabolas

2.3.B Polynomial Long Division

Given the rational function $N(x)/D(x)$, the division results in the equality $N(x)=D(x)·Q(x)+R(x)$, or $$\begin{array}{r} Q(x)+R(x)\\ D(x){\overline{\smash{\big)}\,N(x)}}\phantom{+R(x)}\\ \end{array}$$
Jump to Synthetic Division
Solving by Example
Given $(x^3+2·x^2+12)/(x–2)$, express in the following form with the missing (zero coefficient) terms included $$\begin{array}{r} x-2{\overline{\smash{\big)}\,x^3+2·x^2+0·x+12}}\\ \end{array}$$ Multiply the divisor with the coefficient and variable order necessary to cancel the first term with subtraction, meaning the sign must match, in this case $x^2$ $$\begin{array}{r} x^2\phantom{+0·x+12}\\ x-2{\overline{\smash{\big)}\,x^3+2·x^2+0·x+12}}\\ \phantom{0}-x^3+2·x^2\phantom{+0·x+12}\\ {\overline{\phantom{-x^3+}4·x^2+0·x+12}}\\ \end{array}$$ Repeat the process $$\begin{array}{r} x^2+4·x\phantom{+12}\\ x-2{\overline{\smash{\big)}\,x^3+2·x^2+0·x+12}}\\ \phantom{0}-x^3+2·x^2\phantom{+0·x+12}\\ \overline{\phantom{-x^3+}4·x^2+0·x+12}\\ \phantom{0}-4·x^2+8·x\phantom{+12}\\ \overline{\phantom{-4·x^2+}8·x+12}\\ \end{array}$$ Keep repeating until the first term no longer applies $$\begin{array}{r} Q=x^2+4·x+\phantom{0}8\\ x-2{\overline{\smash{\big)}\,x^3+2·x^2+0·x+12}}\\ \phantom{0}-x^3+2·x^2\phantom{+0·x+12}\\ \overline{\phantom{-x^3+}4·x^2+0·x+12}\\ \phantom{0}-4·x^2+8·x\phantom{+12}\\ {\overline{\phantom{-4·x^2+}8·x+12}}\\ -8·x+16\\ R=\overline{\phantom{-8·x}+28}\\ \end{array}$$ Substitute the values for $N(x)=D(x)·Q(x)+R(x)$ $$x^3+2·x^2+12=(x-2)(x^2+4·x+8)+28$$

2.3.C Synthetic Division

Learn: Paul's Online Notes
The example in polynomial long division can be rewritten a different way, with the number to the left of the line as the negative of the constant in $D(x)$, and the numbers to the right being the coefficients and constant of $N(x)$. $$ \frac{(x^3+2·x^2+12)}{(x–2)}→ \begin{array}{c} 2\thickspace |\thickspace 1\thickspace 2\thickspace 0\thickspace 12 \\ \phantom{2}\thickspace |\underline{\thickspace \phantom{1\thickspace 2\thickspace 0\thickspace 12}} \end{array} $$ In this format, the first number is always copied to the last line $$ \begin{array}{c} 2\thickspace |\thickspace 1\thickspace 2\thickspace 0\thickspace 12 \\ \phantom{2}\thickspace |\underline{\thickspace \phantom{1\thickspace 2\thickspace 8\thickspace 16}} \\ \phantom{2\thickspace |}\thickspace 1\thickspace \phantom{4\thickspace 8\thickspace 28} \\ \end{array} $$ The number on the bottom is then multiplied by the divisor, and the result is added to the next column $$ \begin{array}{c} 2\thickspace |\thickspace 1\thickspace 2\thickspace 0\thickspace 12 \\ \phantom{2}\thickspace |\underline{\thickspace \phantom{1}\thickspace 2\thickspace \phantom{8\thickspace 16}} \\ \phantom{2\thickspace |}\thickspace 1\thickspace 4\thickspace \phantom{8\thickspace 28} \\ \end{array} $$ The process is repeated until every column is filled $$ \begin{array}{c} 2\thickspace |\thickspace 1\thickspace 2\thickspace 0\thickspace 12 \\ \phantom{2}\thickspace |\underline{\thickspace \phantom{1}\thickspace 2\thickspace 8\thickspace 16} \\ \phantom{2\thickspace |}\thickspace 1\thickspace 4\thickspace 8\thickspace 28 \\ \end{array} $$ The result is the coefficients of $Q(x)$ with the last number being $R(x)$

2.3.D Partial Fractions Solves

Partial Fraction Decomposition
Given the rational function $N(x)/D(x)$, expand by the smallest possible roots of the denominator

1) Linear roots are treated as follows $$\frac{N(x)}{(x+a)(x+b)}=\frac{A}{x+a}+\frac{B}{x+b}$$ 2) Quadratic roots are treated as follows $$\frac{N(x)}{a·x^2+b·x+c}=\frac{A·x+B}{a·x^2+b·x+c}$$ 3) Repeated roots are treated as follows $$\frac{N(x)}{{D(x)}^P}=\frac{A}{D(x)}+\frac{B}{{D(x)}^2}+...+\frac{C}{{D(x)}^P}$$ Example $$\frac{5·x}{(x-1)(x^2+2){(x+7)}^2}=\frac{A}{x-1}+\frac{B·x+C}{x^2+2}+\frac{D}{x+7}+\frac{E}{{(x+7)}^2}$$
Solving by Example
$$\frac{12·x}{x^2-10·x+16}$$ Factor $$\frac{12·x}{(x-8)(x-2)}$$ Decompose $$\frac{12·x}{(x-8)(x-2)}=\frac{A}{x-8}+\frac{B}{x-2}$$ Multiply by the factored denominator $$12·x=(x-2)·A+(x-8)·B$$ Enter values for $x$ to cancel terms $$x=2 \medspace\therefore\medspace 12·2=0·A-6·B \medspace\therefore\medspace B=-4$$ $$x=8 \medspace\therefore\medspace 12·8=6·A+0·B \medspace\therefore\medspace A=16$$ Substitute $A$ and $B$ to solve $$\frac{12·x}{(x-8)(x-2)}=\frac{16}{x-8}-\frac{4}{x-2}$$

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