Univariate Polynomials

Before we move on to the relevant theorems we need to know what is a polynomial. Polynomial is an expression with several terms which might have one or more indeterminate. If a polynomial has only one variable, then we call it a Univariate Polynomial. In this lesson we are going to consider mainly about Univariate Polynomials, but there are Multivariate Polynomials as well. The general form of a Univariate Polynomial is as follows,

$f(x)=a_0x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+...+a_{n-1}x+a_{n}$

Here $x$ is the variable with $a_0,a_1,a_2,....,a_n \in \mathbb{R}$ coefficients.
We call that this is a $n^{th} degree$ polynomial if $a_0 \ne 0$. 

For example,

• $f(x)=x+2$ is a polynomial of $1^{st}$ degree
• $f(x)=-5x^6+3x^4-2x+8$ is a polynomial of $6^{th}$ degree

If two polynomials are equal, then their degree and all the corresponding coefficients must be equal. 

Note: Following algebraic expansions might help you in further studies.

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• $(a+b)^2=a^2+2ab+b^2$
• $(a-b)^2=a^2-2ab+b^2$
• $(a+b)^3=a^3+3a^2b+3ab^2+b^3$
• $(a-b)^3=a^3-3a^2b+3ab^2-b^3$
• $a^2-b^2=(a-b)(a+b)$
• $a^3+b^3=(a+b)(a^2-ab+b^2)$
• $a^3-b^3=(a+b)(a^2+ab+b^2)$

It is possible to use Pascal Triangle for further expansions. 
$\begin{matrix} & & & & & &1& & & & & &\\\mathit{1^{st}}& & & & &1& &1& & & & &\\\mathit{2^{nd}}& & & &1& &2& &1& & & &\\\mathit{3^{rd}}& & &1& &3& &3& &1& & &\\\mathit{4^{th}}& &1& &4& &6& &4& &1& &\\\mathit{5^{th}}&1& &5& &10& &10& &5& &1&\\ & &\vdots& &\vdots& &\vdots& &\vdots& &\vdots& &\end{matrix}$

As we can see each raw of the pascal triangle is related to the corresponding coefficients of the terms of expansion.

Ex.-
If we need to expand $(a-b)^4$, then we need to use $4^{th}$ raw of the pascal triangle.

$(a-b)^4={\color{red}1}a^4-{\color{red}4}a^3b+{\color{red}6}a^2b^2-{\color{red}4}ab^3+{\color{red}1}b^4$

Here the power of $a$ has reduced from $4$ to $0$ and the power of $b$ has increased from $0$ to $4$. As here we have a minus$(-)$ sign we have to use it alternatively.
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Divide a polynomial by another

Here are some examples how to divide polynomials using long division.

1) $\frac{2x^3+3x+5}{x-3}$

$\begin{equation}\cfrac{}{x-3 |}\cfrac{2x^2+6x+21}{\begin{array}[b]{r}2x^3+3x+5\\ \cfrac{{2x}^3-6x^2}{\begin{array}.6x^2+3x+5\\\cfrac{6x^2-18x}{\begin{array}.21x+5\\\cfrac{21x-63}{ 68}\end{array}}\end{array}}\end{array}}\end{equation}$

So the quotient is $ {2x}^2+6x+21 $ and remainder is $ 68 $.

2)$\frac{5x^4+2x-4}{x^2-3}$

$\begin{equation}\cfrac{}{x^2-3 |}\cfrac{5x^2+15}{\begin{array}[b]{r}5x^4+2x-4\\ \cfrac{{5x}^4-15x^2}{\begin{array}.15x^2+2x-4\\\cfrac{15x^2-45}{2x+41}\end{array}}\end{array}}\end{equation}$

In this example quotient is $5x^2+15$ and remainder is $2x+41$.

There are certain conclusions that we can take from these examples and we need to complete some other example to ascertain it.

• If the divisor is a linear factor (that is a first degree polynomial) then the remainder is a constant. 
• If the divisor is a higher degree polynomial, then the remainder can be a polynomial with a degree less than the divisor or can be a constant.

Exercise 01

Find the quotient and the remainder of the following fractions using long division.
1) $\cfrac{5x^3-3x+2}{x+1}$     2) $\cfrac{-3x^2+3x+1}{2x-1}$     3) $\cfrac{2x^5-3}{x^2+2}$     4) $\cfrac{x^4-3x^3-2}{x-4}$     5) $\cfrac{4x^6-3x-7}{x^3-2}$

Remainder Theorem

Let $f(x)$ be a univariate polynomial. The remainder, after dividing $f(x)$ by the linear factor $x-a$ is $f(a)$. Here $a \in \mathbb{R}$

Proof

Now we know that the remainder must be a constant, as the divisor is a linear factor. Let $\Phi(x)$ be the quotient and $R$ be the remainder. Then we can write $f(x)$ as follows,

$f(x)=\Phi(x)(x-a)+R$

Let's substitute $x=a$ in the above equation.

$f(a)=\Phi(a)(a-a)+R$

As $a-a=0$,

$R=f(a)$ 


Examples

1) Find the remainder of following fractions.

i) $\cfrac{3x^2+2x-3}{x-5}$     ii) $\cfrac{2x^3-4x-2}{x+2}$     iii) $\cfrac{-2x^3+5x+4}{2x-1}$

i) $3\times 5^2+2\times5-3=82$
ii) $2\times (-2)^3-4\times(-2)-2=-10$
iii) $-2\times{\left(\frac{1}{2}\right)}^3+5\times\frac{1}{2}+4=\frac{25}{4}$


2) If the remainder is $2$ after dividing $2x^2-ax+3$ by the linear factor $x-2$, then find the value of $a$.

$f(x)=2x^2-ax+3$
As the remainder is $2$ after dividing $f(x)$ by $x-2$,
$f(2)=2$
$2\times2^2-a\times2+3=2$
$\therefore a=\cfrac{9}{2}$

Factor Theorem 

If there is a real number $a$, which is as $f(a)=0$, then $(x-a)$ is a factor of $f(x)$.

We can see that this theorem has also derived from the Remainder Theorem. This theorem says what happen when the remainder is going to be zero, as $f(a)$ is the remainder of polynomial which has divided by $(x-a)$. 

Proof

Now we know that remainder of the polynomial $f(x)$ which has divided by the linear factor $(x-a)$, is $f(a)$. Then,

$f(x)=\Phi(x)(x-a)+f(a)$ where $\Phi(x)$ is the quotient.

If the corresponding $a$ satisfies the requirement $f(a)=0$, then,

$f(x)=\Phi(x)(x-a)$

So $(x-a)$ is a factor of $f(x)$.

To be continued...