2.3 Contents

1. Cancellation

2. Quadratic Equations Definition

3. Composite Quadratic Equations Factored

4. Prime Quadratic Equations Factored (Completing the Square)

5. Quadratic Formula

6. Discriminant of a Quadratic Formula

7. Cubic Formula 🔧

8. Sum & Difference of Powers

9. Pascal's Triangle

1. Binomials

2. Equating Coefficients (Coefficient Method of Solving)

3. Polynomial Long Division

4. Synthetic Division

5. Partial Fractions Solves

2.3.1 Cancellation

In a factored function, like terms in both a numerator and denominator as multipliers reduce to 1 $$\frac{a(x)\cdot f(x)}{a(x)\cdot g(x)}=\frac{f(x)}{g(x)}$$

Examples

$$\frac{(x-4)\cdot (x+2)}{(x-4)\cdot (x-2)}=\frac{(x+2)}{(x-2)}$$ $$\frac{(x-4)\cdot (x+2)}{(4-x)\cdot (x-2)}=-\frac{(x+2)}{(x-2)}$$

2.3.2 Quadratic Equations Definition

Second degree trinomial equations in the standard form $$a\cdot x^2+b\cdot x+c=0$$ The following is a more useful general form $$y=a\cdot x^2+b\cdot x+y_∘$$

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 $$a\cdot c\cdot x^2 \pm (a\cdot d+b\cdot c)\cdot x+b\cdot d$$ $$=(a\cdot x \pm b)(c\cdot x \pm d)$$

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 $$x^2 \pm b\cdot x+y_∘=y$$ Add ${(b/2)}^2$ and subtract $y_∘$ $$x^2 \pm b\cdot x+ {\Big(\frac{b}{2} \Big)}^2=y-y_∘+{\Big(\frac{b}{2} \Big)}^2$$ Factor $$\frac{{(2\cdot x \pm b)}^2}{4}=y-y_∘+\frac{b^2}{4}$$

2.3.5 Quadratic Formula

The general solutions of $y=a\cdot x^2+b\cdot x+y_∘$ are $$x=\frac{-b \pm \sqrt{b^2+4\cdot a\cdot (y-y_∘)}}{2\cdot a},\medspace a≠0$$

Proof

Given the general formula, subtract $y_∘$ from both sides $$y-y_∘=a\cdot x^2+b\cdot x$$ Multiply by $4\cdot a$ $$4\cdot a\cdot (y-y_∘)=4\cdot a^2\cdot x^2+4\cdot a\cdot b\cdot x$$ Add $b^2$, then factor the right, similar to completing the square $$b^2+4\cdot a\cdot (y-y_∘)={(2\cdot a\cdot x+b)}^2$$ Take the square root $$\pm \sqrt{b^2+4\cdot a\cdot (y-y_∘)}=2\cdot a\cdot x+b$$ Subtract $b$, and divide by $2\cdot a$

Example

Solve for $x$ and $y$ $$x+y=1\medspace\land\medspace x\cdot y=1$$ Isolate $y$ in the first equation $$y=1-x$$ Substitute for $y$ in the second equation $$x\cdot (1-x)=1$$ 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 $x$ $$x=\frac{-(-1) \pm \sqrt{(-1)^2-4\cdot 1\cdot 1}}{2\cdot 1}$$ Simplify in expanded form $$x=\frac{1}{2}\pm\frac{\sqrt{3}}{2}\cdot i$$ Substitute for $x$ in the first equation with $y$ isolated $$y=1-\bigg(\frac{1}{2}\pm\frac{\sqrt{3}}{2}\cdot i\bigg)$$ Simplify in expanded form $$y=\frac{1}{2}\mp\frac{\sqrt{3}}{2}\cdot i$$ The solutions for $(x,y)$ are $$\bigg(\frac{1}{2}\pm\frac{\sqrt{3}}{2}\cdot i,\frac{1}{2}\mp\frac{\sqrt{3}}{2}\cdot i\bigg)$$

2.3.6 Discriminant of a Quadratic Formula

$$Δ=b^2-4\cdot a\cdot y_∘$$

• If $Δ>0$, the equation has two real factorable roots
• If $Δ=0$, the equation has one factorable root squared
• If $Δ \lt 0$, the equation has two complex factorable roots

2.3.7 Cubic Formula 🔧

Discriminant

$$\Delta=b^2\cdotp c^2-4\cdotp a\cdotp c^3-4\cdotp b^3\cdotp d-27\cdotp a^2\cdotp d^2+18\cdotp a\cdotp b\cdotp c\cdotp d$$

• If $\Delta>0$, the equation has three real factorable roots
• If $\Delta=0$, the equation has one factorable root either squared or cubed
• If $\Delta\lt 0$, the equation has one real and two complex factorable roots

Proof

2.3.8 Sum & Difference of Powers

Sum of Powers for Real Roots (Odd Exponents Only)

$$a^n+b^n=(a+b)(a^{n-1}-a^{n-2}\cdotp b+a^{n-3}\cdotp b^2-...+a^2\cdotp b^{n-3}-a\cdotp b^{n-2}+b^{n-1}), \forall 2\cdotp n-1 \isin \N$$

Difference of Powers for Real Roots

$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}\cdotp b+a^{n-3}\cdotp b^2+...+a^2\cdotp b^{n-3}+a\cdotp b^{n-2}+b^{n-1}), \forall 2\cdotp n \isin \N$$

Sum & Difference of Squares & Cubes

$$a^2+b^2=(a+i\cdotp b)(a-i\cdotp b)$$ $$a^2-b^2=(a+b)(a-b)$$ $$a^3 \pm b^3=(a \pm b)(a^2\mp a\cdotp b+b^2)$$

Other Sums & Differences of Interest

$$a^4-b^4=(a+i\cdotp b)(a-i\cdotp b)(a+b)(a-b)$$ $$a^6+b^6=(a^2+b^2)(a^4-a^2\cdotp b^2+b^4)$$ $$a^{10}+b^{10}=(a^2+b^2)(a^8-a^6\cdotp b^2+a^4\cdotp b^4-a^2\cdotp b^6+b^8)$$

Example

$$x^3=8$$ Factor 8 to find the real solution $$x^3=2^3$$ Given the equality, subtract $2^3$ $$x^3-2^3=0$$ Substitute the difference of cubes $$(x-2)(x^2+2\cdot x+2^2)=0$$ Find the zeros of $x$ for the second term with the quadratic formula $$x=\frac{-2\pm\sqrt{4-4\cdot 1\cdot 4}}{2}$$ Simplify $$x=-1\pm\sqrt{-3}$$ The solutions for $x^3=8$ are $$\{2,-1+i\cdot\sqrt{3},-1-i\cdot\sqrt{3}\}$$

2.3.9 Pascal's Triangle

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

2.3.A Binomials

Binomial Coefficient

$$ \binom{n}{k}=\frac{n!}{(n-k)!\cdot k!}=\frac{n\cdot (n-1)(n-2)...(n-k+1)}{k!}$$

Binomial Theorem

The coefficients in the expanded form match the line $n+1$ down from the top on Pascal’s triangle $${(x+y)}^n=\binom{n}{0}\cdot x^n+\binom{n}{1}\cdot x^{n-1}\cdot y+\binom{n}{2}\cdot x^{n-2}\cdot y^2+...+\binom{n}{n}\cdot y^n, \forall n \isin \Z$$

Binomials Squared & Cubed

$${(a \pm b)}^2=a^2 \pm 2\cdot a\cdot b+b^2$$
$${(a \pm b)}^3=a^3 \pm 3\cdot a^2\cdot b+3\cdot a\cdot b^2 \pm b^3$$

2.3.B Equating Coefficients (Coefficient Method of Solving)

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

2.3.C Polynomial Long Division

Given the rational function $N(x)/D(x)$, the division results in the equality $N(x)=D(x)\cdot Q(x)+R(x)$, or $$\begin{array}{r} Q(x)+R(x)\\ D(x){\overline{\smash{\big)}\,N(x)}}\phantom{+R(x)}\\ \end{array}$$

Solving by Example

Given $(x^3+2\cdot x^2+12)/(x–2)$, express in the following form with the missing (zero coefficient) terms included $$\begin{array}{r} x-2{\overline{\smash{\big)}\,x^3+2\cdot x^2+0\cdot x+12}}\\ \end{array}$$ 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 $x^2$ $$\begin{array}{r} x^2\phantom{+0\cdot x+12}\\ x-2{\overline{\smash{\big)}\,x^3+2\cdot x^2+0\cdot x+12}}\\ \phantom{0}-x^3+2\cdot x^2\phantom{+0\cdot x+12}\\ {\overline{\phantom{-x^3+}4\cdot x^2+0\cdot x+12}}\\ \end{array}$$ Repeat the process $$\begin{array}{r} x^2+4\cdot x\phantom{+12}\\ x-2{\overline{\smash{\big)}\,x^3+2\cdot x^2+0\cdot x+12}}\\ \phantom{0}-x^3+2\cdot x^2\phantom{+0\cdot x+12}\\ \overline{\phantom{-x^3+}4\cdot x^2+0\cdot x+12}\\ \phantom{0}-4\cdot x^2+8\cdot x\phantom{+12}\\ \overline{\phantom{-4\cdot x^2+}8\cdot x+12}\\ \end{array}$$ Keep repeating until the first term no longer applies $$\begin{array}{r} Q=x^2+4\cdot x+\phantom{0}8\\ x-2{\overline{\smash{\big)}\,x^3+2\cdot x^2+0\cdot x+12}}\\ \phantom{0}-x^3+2\cdot x^2\phantom{+0\cdot x+12}\\ \overline{\phantom{-x^3+}4\cdot x^2+0\cdot x+12}\\ \phantom{0}-4\cdot x^2+8\cdot x\phantom{+12}\\ {\overline{\phantom{-4\cdot x^2+}8\cdot x+12}}\\ -8\cdot x+16\\ R=\overline{\phantom{-8\cdot x}+28}\\ \end{array}$$ Substitute the values for $N(x)=D(x)\cdot Q(x)+R(x)$ $$x^3+2\cdot x^2+12=(x-2)(x^2+4\cdot x+8)+28$$

2.3.D Synthetic Division

The example in polynomial long division can be formatted as follows, with the number to the left of the line as the negative of the constant in $D(x)$, and the numbers to the right $(1\cdot x-n)$ being the coefficients and constant of $N(x)$. $$ \frac{(x^3+2\cdot x^2+12)}{(x–2)}→ \begin{array}{c} 2\thickspace |\thickspace 1\thickspace 2\thickspace 0\thickspace 12 \\ \phantom{2}\thickspace |\underline{\thickspace \phantom{1\thickspace 2\thickspace 0\thickspace 12}} \end{array} $$ In this format, the first number is always copied to the last row $$ \begin{array}{c} 2\thickspace |\thickspace 1\thickspace 2\thickspace 0\thickspace 12 \\ \phantom{2}\thickspace |\underline{\thickspace \phantom{1\thickspace 2\thickspace 8\thickspace 16}} \\ \phantom{2\thickspace |}\thickspace 1\thickspace \phantom{4\thickspace 8\thickspace 28} \\ \end{array} $$ 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 $R(x)$

Example

Factor $x-1$ from $x^5-1$ $$ \begin{array}{c} 1\thickspace |\thickspace 1\enspace 0\enspace 0\enspace 0\enspace 0\thickspace -1 \\ \phantom{1}\thickspace |\underline{\thickspace \phantom{1}\enspace 1\enspace 1\enspace 1\enspace 1\thickspace \phantom{-}1} \\ \phantom{1\thickspace |} \thickspace 1\enspace 1\enspace 1\enspace 1\enspace 1\thickspace \phantom{-}0 \\ \end{array} $$ Starting with the coefficient for $x^4$, and ending with a remainder of $0$, $$x^5-1=(x-1)(x^4+x^3+x^2+x+1)$$

2.3.E 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 $$\frac{N(x)}{a\cdot x^2+b\cdot x+c}=\frac{A\cdot x+B}{a\cdot x^2+b\cdot x+c}$$ 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 $$\frac{5\cdot x}{(x-1)(x^2+2){(x+7)}^2}=\frac{A}{x-1}+\frac{B\cdot x+C}{x^2+2}+\frac{D}{x+7}+\frac{E}{{(x+7)}^2}$$

Solving by Example

$$\frac{12\cdot x}{x^2-10\cdot x+16}$$ Factor $$\frac{12\cdot x}{(x-8)(x-2)}$$ Decompose $$\frac{12\cdot x}{(x-8)(x-2)}=\frac{A}{x-8}+\frac{B}{x-2}$$ Multiply by the factored denominator $$12\cdot x=(x-2)\cdot A+(x-8)\cdot B$$ Enter values for $x$ to cancel terms $$x=2 \medspace\therefore\medspace 12\cdot 2=0\cdot A-6\cdot B \medspace\therefore\medspace B=-4$$ $$x=8 \medspace\therefore\medspace 12\cdot 8=6\cdot A+0\cdot B \medspace\therefore\medspace A=16$$ Substitute $A$ and $B$ to solve $$\frac{12\cdot x}{(x-8)(x-2)}=\frac{16}{x-8}-\frac{4}{x-2}$$