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Polynomials
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Sum of multiples of powers of a variable
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Degree/order refer to the highest power of the variable
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Polynomial can have more variables, (x,y,z) but we are limited to one (x)
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Solutions correspond to linear factors of the polynomial (converse is true)
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Now we are going to talk about higher degrees?!!!??!!
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Polynomial algebra
- Addition
- Subtraction
- Multiplication
- Division
- This is the hard one
- Long division
- Dividing polynomial: $\frac{q\text{(quotient)}}{d\text{(divisor)}}$
- It follows the remainder is always strictly smaller than the divisor for integers
- For: $\frac{x^{3}-x^{2}-3x-5}{x-3\text{(divisor)}}=x^{2}+2x+3\text{(quotient)} \text{ R }4$
- We supposed that the remainder is smaller than the divisor for integers.
- But for polynomials, it is similar but not the case
- Hence, "When a polynomial $P(x)$ is divided by a polynomial $D(x)$, the degree of the remainder $R(x)$ will be strictly smaller than the degree of $D(x)$."
- Using this we can write original polynomial in terms of the quotient and remainder.
- $P(x) = Q(x)D(x)+R(x)$
- Compared to numerical division: $100=4\times23+8$
- It follows if $\frac{100}{23}=4+\frac{8}{23}$ $\implies$ $\frac{P(x)}{D(x)} = Q(x)+\frac{R(x)}{D(x)}$
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Proper algebraic fraction: When the degree of numerator is strictly smaller than degree of denominator
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Improper algebraic fraction: When the degree of numerator is strictly larger than degree of denominator
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Important for partial fractions! We gotta convert improper fractions to proper ones. This is a skill for manipulating fractions that is important later on!
Two methods to convert improper to proper
- Using polynomial division
$$\begin{align}
\text{\{Insert polynomialing\}} \\
\frac{x^{2}+3x+4}{x+4}&=\frac{(x-1)(x+4)+8}{x+4} \text{ (still improper)} \\
&= x-1 + \frac{8}{x+4}\text{, now it is proper!}
\end{align}
$$
- The "algebraic juggling" method (eww)
Dividing by linear(specifically) polynomials
- If $P(x)$ and $(x-a)$ are polynomials then $P(x)=Q(x)(x-a)+r$
- This means for any polynomial and any linear factor then if the polynomial itself isn't divisible by that linear factor then there is a polynomial multiple of the linear factor that is only different by a constant r!
The Remainder Theorem and The Factor Theorem
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Most likely to be tested on content!?!?!?!?
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Straight forward, easy to prove theorems
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Remainder: For a polynomial $P(x)$ and a number $a$, the remainder when $P(x)$ is divided by $(x-a)$ is $P(a)$
$$\begin{align}
\text{Proving The Remainder Theorem.} \\
\text{Let } r \text{ be the remainder when }P(x) \text{ is divided by }(x-a).\\
\text{Then for some polynomial }Q(x), \\
P(x)=Q(x)(x-a)+r. \\
\text{Hence, } P(x)=Q(a)(a-a)+r=r\text{ as required.}
\end{align}
$$
- Factor: For a polynomial $P(x)$, $(x-a)$ is a factor of $P(x)$ if and only if $P(a)=0$.
$$\begin{align}
\text{Proving The Factor Theorem (if and only if)} \\
\text{Suppose } (x-a) \text{ is a factor of }P(x). \\
\text{ Then the remainder when }P(x)\text{ is divided by }(x-a)\text{ is 0.} \\
\text{ Hence, by the Remainder Theorem, }P(a)=0. \\
\text{Conversely, suppose that }P(a)=0. \\
\text{ Then the remainder when }P(x)\text{ is divided by }(x-a)\text{ is 0. } \\
\text{Hence}(x-a)\text{ is a factor of }P(x).
\end{align}
$$
e.g. Find $k$ given that $(x-2)$ is a factor of $P(x)=x^3+kx^2-3x+6$. Hence, fully factorise $x^3+kx^2-3x+6$.
$$\begin{align}
\text{Let P(x)}&=x^3+kx^2-3x+6 \\
\text{Since }(x-2)&\text{ is a factor, by The Factor}\text{ theorem} \\
P(2) &= 0 \\
8 +4k -6+6&=0 \\
\text{So }k&=-2 \\
\therefore P(x)&= x^3+-2x^2-3x+6 \\
\therefore P(x) &= Q(x)(x-2) \text{ where Q has degree 2.} \\
\therefore x^3+-2x^2-3x+6&=(ax^{2}+bx+c)(x-2) \\
&\implies a=1, c=-3 \\
\therefore -2x^{2}&=-2x^{2}+bx^{2} \\
&\implies b=0 \\
\therefore P(x)&=(x^2-3)(x-2) \\
&=(x+\sqrt{ 3 })(x-\sqrt{ 3 })(x-2)
\end{align}
$$
