Algebraic Basic Operations

Highest Common Factor

• The Highest Common Factor (H.C.F) is also known as Greatest Common Divisor (G.C.D).
• H.C.F or G.C.D of two or more two polynomials is the product of their common factors.
• The H.C.F exactly divides each of the two given polynomials.
  
  

The following two methods for finding the H.C.F of given polynomials.
1. Factor method
2. Divisor Method

Factor method

• For finding the H.C.F of the given polynomials, first, we find their factors.
• Then, take the product of their common factors.
• This product of common factor is known as H.C.F or G.C.D.

Example:
Find the H.C.F of $12x^2$ $-xy$ $-5y^2$, $30x^2$ $+11xy$ $-30y^2$, and $6x^2$ $+xy-$ $5y^2$ by a factor method.

Solution:
First, we find the factors of the given expressions

First expression:
$=6x^2$ $-16xy$ $+5y^2$
$=12x^2$ $-10xy$ $-6xy$ $+5y^2$
$=2x(6x-5y)$ $-y(6x-5y)$
$=(6x-5y)(2x-y)$
$12x^2$ $-xy$ $-5y^2$ $=(6x-5y)$ $(2x-y)$

Second expression:
$30x^2$ $+11xy$ $-30y^2$
$=30x^2$ $+36xy$ $-25xy$ $-30y^2$
$=6x(5x+6y)$ $-5y(5y+6y)$
$=(5x+6y)$ $(5x-6y)$
$30x^2$ $+11xy$ $-30y^2$ $=(5x+6y)$ $(5x-6y)$

Third Expression:
$6x^2$ $+xy-$ $5y^2$
$=6x^2$ $+6xy$ $-5xy$ $-5y^2$
$=6x(x+y)$ $-5y(x+y)$
$=(x+y)$ $(6x-5y)$
$6x^2$ $+xy$ $-5y^2$ $=(x+y)$ $(6x-5y)$

Here, $(6x-5y)$ is the only highest common factor in all

So,

H.C.F $=(6x-5y)$

Division Method

• In this method, we divide a polynomial with a greater degree by the other polynomial and proceed as in the case of H.C.F of numbers.

Example:
Find the H.C.F of the polynomial by division method $x^2$ $+xy$ $-2y^2$, $x^2$ $+3xy$ $+2y^2$ and $x^3$ $+2x^2y$ $+xy^2$ $+2y^3$

Solution:
The degree of the polynomial $x^2$ $+xy$ $-2y^2$ is two and the degree of $x^2$ $+3xy$ $+2y^2$ is also two

So,
$\left( x^2+3xy+2y^2 \right)$ $\left(x^2+xy-2y^2\right)$

image

Here, Remainder $=2xy$ $+4y^2$ $=2y$ $\left(x+2y\right)$

Note that: The $2y$ is a common factor so, we ignore it and after ignoring the degree of the remainder will be less than the divisor so the new remainder will be the new divisor and the old divisor will be a dividend.

image

Here,

$\left( x+2y \right)$ is H.F.C. of $\left( x^2+3xy+2y^2 \right)$ and $\left(x^2+xy-2y^2\right)$

Now divide the left expression
$\left(x^3+2x^2y+xy^2+2y^3 \right)$ by $\left(x+2y \right)$

image

Hence, The H.C.F. of $\left(x^2+xy-2y^2 \right)$, $\left( x^2+3xy+2y^2 \right)$ and $\left(x^3+2x^2y+xy^2+2y^3 \right)$ is $(x+2y)$

H.C.F $=(x+2y)$

Least Common Multiple (L.C.M) of polynomials

• If a polynomial is exactly divisible by another polynomial then the first polynomial is called a multiple of the second polynomial.

Example:
$(x+2)$ exactly divides $x^2$ $+4x$ $+4$
so $x^2$ $+4x$ $+4$ is a multiple of $x$ $+2$

Common Multiple

• If a polynomial is exactly divisible by two or more polynomials, then it is called the common multiple of these polynomials.

Example:
$x^2$ $+3x$ $+2$ is exactly divisible by the polynomials $(x+1)$ and $(x+2)$.
Therefore, $x^2$ $+3x$ $+2$ is the common multiple of $(x+1)$ and $(x+2)$.

Similarly,

• $\left( x^3+2x^2-x-2 \right)$ is a common multiple of $\left( x^2+3x+2 \right)$ and $\left(x^2-1 \right)$
• It is also a common multiple of $\left(x^2+3x+2 \right)$ and $\left(x^2+x-2 \right)$
• There exist many common multiples of given polynomials.
• L.C.M is the polynomial of the least degree among all common multiples.
  
  

There are two methods to find the least common multiple (L.C.M)
$1.$  By Factorization method
$2.$  By finding H.C.F

1st. Method by Factorization 

• First of all, we find the factors of the given polynomials. Then, the product of all the common and non-common factors of polynomials are known as the L.C.M of the given polynomials.
  
  

Example:
Find the L.C.M of the polynomials by factorization method $6x^2$ $+11x$ $+3$, $2x^2$ $-5x-11$ and $3x^2$ $-11x$ $-4$
Solution:
$6x^2$ $+11x$ $+3$ $=6x^2$ $+9x$ $+2x$ $+3$
 $=3x(2x+3)$ $+1(2x+3)$
 $=(2x+3)$ $(3x+1)$
$6x^2$ $+11x$ $+3$ $=(2x+3)$ $(3x+1)$

$2x^2$ $-5x$ $-11$ $=2x^2$ $-8x$ $+3x$ $-12$
 $=2x(x-4)$ $+3(x-4)$
 $=(x-4)$ $(2x+3)$
$2x^2$ $-5x$ $-11$ $=(x-4)$ $(2x+3)$

$3x^2$ $-11x$ $-4$ $=3x^2$ $-12x$ $+x$ $-4$
 $=3x(x-4)$ $+1(x-4)$
 $=(x-4)$ $(3x+1)$
$3x^2$ $-11x$ $-4$ $=(x-4)$ $(3x+1)$

$6x^2$ $+11x$ $+3$ $=(2x+3)$ $(3x+1)$
$2x^2$ $-5x$ $-11$ $=(x-4)$ $(2x+3)$
$3x^2$ $-11x$ $-4$ $=(x-4)$ $(3x+1)$

There is no common factor in all
The common factors in two polynomials $=(2x+3)$, $(x-4)$, $(3x+1)$

The common factors product is
$(2x+3)$ $(x-4)$ $(3x+1)$

There is no non-common so the L.C.M of the given polynomial is the product of common factors.

L.C.M $=(2x+3)$ $(x-4)$ $(3x+1)$

2nd Method By H.C.F
As we know that
H.C.F $\times$ L.C.M $=$ (First polynomial) $\times$ (second polynomial)
Therefore,
L.C.M $=\frac {(First polynomial)\times (second polynomial)}{H.C.F}$

• It may be noted that this result is true only for two polynomials.
• It may also be noted that

L.C.M $=$ (product of common factors) $\times$ (product of non-common factors)

Simplification of algebraic fractions

• An expression of the form $\frac {P}{Q};$ where $P$ and $Q$ are polynomials and $Q\neq 0$ is known as an algebraic fraction.
• Although every rational expression is a polynomial, not all rational expressions are polynomials.
  
  

Equivalent fractions
We know that
$\frac{1}{2}$, $\frac{2}{4}$, $\frac{3}{6}$, $\frac{4}{8}$, $\frac{5}{10}$,... are all equivalent fractions

Similarly,
$\frac{p(x)}{q(x)}$, $\frac{p(x)\times r(x)}{q(x)\times r(x)}$, $\frac{p(x)\times s(x)}{q(x)\times s(x)}$,.... are equivalent rational polynomials.

$\frac{1}{x-1}$, $\frac{x+1}{x^2-}$, $\frac{x^2+x+1}{x^3-1}$ are equivalent rational polynomials.

Note: Simplifications of algebraic fractions mean to find an equivalent fraction such that the degree of the denominator is as least as possible.

Simplify $1.$ $\frac{a^3-8a^2+11a+20}{a^3-6a^2-7a+60}$, $\forall$ $a:a^3$ $-6a^2$ $-7a$ $+60$ $\neq 0$

Solution:
First of all, we find the factors of the numerator and denominator
Let $P(a)$ $=a^3$ $-8a^2$ $+11a$ $+20$   $(i)$
Possible factors of $20$ are: $\pm 1$, $\pm 2$, $\pm 4$, $\pm 5$, $\pm 10$, $\pm 20$

Try for $a+1$ $=0$ then $a=-1$ put in $(i)$

$P(-1)$ $=(-1)^3$ $-8(-1)$ $+11(-1)$ $+20$
 $=-1$ $-8$ $-11$ $+20$
 $=20$ $-20$
$P(-1)=0$

Hence $(a+1)$ is one of the factors of the $P(a)$

$P(a)$ $=a^3$ $-8a^2$ $+11a$ $+20$
As $(a+1)$ is a factor of $P(a)$ so we take the common $(a+1)$ in such a way

$=a^3$ $+a^2$ $-9a^2$ $-9a$ $+20a$ $+20$
$=a^2(a+1)$ $-9a(a+1)$ $+20(a+1)$
$=(a+1)$ $\left(a^2-9a+20 \right)$
$=(a+1)$ $\left (a^2-4a-5a+20 \right)$
$=(a+1)$ $\left (a(a-4)-5(a-4) \right)$
$P(a)$ $=(a+1)$ $(a-4)$ $(a-5)$
$a^3$ $-8a^2$ $+11a$ $+20$ $=(a+1)$ $(a-4)$ $(a-5)$

Now find the factors of $a^3$ $-6a^2$ $-7a$ $+60$
let
$P(a)$ $=a^3$ $-6a^2$ $-7a$ $+60$    $(ii)$

All possible factors of $60$ are: $\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, $\pm 5$, $\pm 6$, $\pm 10$, $\pm 12$, $\pm 15$, $\pm 20$, $\pm 30$, $\pm 60$
Try for $a+3=0$ then $a=-3$ put in $(ii)$

$P(-3)$ $=\left ( -3 \right)^3$ $-6(-3)^2$ $-7(-3)$ $+60$
$P(-3)$ $=-27$ $-6(9)$ $+21$ $+60$
$P(-3)$ $=-27$ $-54$ $+21$ $+60$
$P(-3)$ $=-81$ $+81$
$P(-3)$ $=0$

Hence, $P(-3)= 0$ so $(a+3)$ is a factor of expression $(2)$

$P(a)$ $=a^3$ $-6a^2$ $-7a$ $+60$
We set the expression in such a way that after taking the $(a+3)$ common in all

$a^3$ $+3a^2$ $-9a^2$ $-27a$ $+20a$ $+60$
$a^2(a+3)$ $-9a(a+3)$ $+20(a+3)$
$(a+3)$ $(a^2-9a+20)$
$(a+3)$ $(a^2-4a-5a+20)$
$(a+3)$ $\left( a(a-4)-5(a-4) \right)$
$(a+3)$ $(a-4)(a-5)$

$\left (a^3-6a^2-7a+60 \right)$ $=(a+3)$ $(a-4)$ $(a-5)$

$\frac {a^3-8a^2+11a+20}{a^3-6a^2-7a+60}$ $=\frac {(a+1)(a-4)(a-5)}{(a+3)(a-4)(a-5)}$

$=\frac {(a+1)}{(a+3)}$ this is the simplest form

Addition and Subtraction of algebraic Fractions

In these types of problems, an easier method is to take the L.C.M of their denominators.
First, we make common factors of denominators.

Example:
$\frac {x^2-\left ( 2y-3z \right)^2}{\left( x+3z \right)^2-4y^2}$ $-\frac {4y^2-\left ( x-3z^2 \right)^2}{\left ( x+2y \right)^2-9z^2}$ $+\frac {\left ( (x-2y)^2-9z^2 \right)}{\left (  \right)}$