Multiplying Rational Expressions

Multiplying rational expressions I (math)

A rational expression  is an algebraic expression  which is in the  form P/Q, where P and Q are simpler expressions P and Q are usually polynomials.  The denominator Q is not zero. It is the quotient of two polynomials. It is important to remember that the divisor cannot be zero. A quotient is the answer to a division problem. Polynomials are expressions which has the sum of powers with at least one variable.  This variable is multiplied by coefficients.

Examples of rational expressions are :

2/5 ,    3x + 9    ,        x^2 + 6x + 3
X + 3               x^2 +x +2

Multiplication Rational Expressions

When  we multiply the rational-expressions the numerators  and the denominators  are to be multiplied.   If P,Q,R  and S polynomials   then

P / Q   .  R / S  =   PR / QS  When Q is not equal to zero and S is not equal to 0.

Here we multiply the  numerators and the denominators.   Before multiplying it has to be seen if they can be reduced.   If reduction is possible then  it has to be done before multiplication.

Multiplying Rational Expressions Examples

Multiply the rational expression given below

18a3       x     5b4

20b2                   6a

Before multiplying the above expression it can be simplified.  Here we can cancel the numerator with the denominator or denominator with the numerator


18 a3       x     5 b4

20 2                   6a

18 can be divided by 6 and we get 3 and   a3 divided by a  = a2   and  b4 /b2 =b2 .  Thus we have 3a2 b2 in the numerator and 4 in the denominator.



3a2        x   b2                                           3a2  b2
=                 4
4               1

Multiply 3a2 with b2 and 4 with 1.  We get 3a2b2  in the  numerator and 4 in the denominator.  In the above problem first the expression was simplified then it was multiplied.  It can also be done vice versa.  First the rational-expressions can be  multiplied  that is numerator with numerator 4xy^2  with 2x and denominator with denominator 3y with 4y.
Then it can  be simplified.

4xy^2         2x
x
3y           4y

4x y^2. 2 x      =                4 . 2x^2 y^2
3y. .4y                             4. 3 y^2

Understanding formula for quadratic equation is always challenging for me but thanks to all math help websites to help me out.

When multiplying the numerator we get 8x^2 y^2   and when multiplying the denominator we get 12 y^2


4.2 x^2 y^2

4.3y^2

The 4 in the numerator and denominator gets cancelled. The y^2 in the numerator and denominator also gets cancelled. And we are left out with

2 x^2
3

Which gives the final answer.

Practice Abstract Algebra Problems

Introduction to practice abstract algebra problems:

In algebra, we use a, b,c.... x, y and z to denote numbers. Performing addition, subtraction, multiplication, and division of roots on this symbols and real numbers, we obtain what are called algebraic expressions

The following are some examples for algebraic equations:

2x + 3 = x + 6,

Symbols in an algebraic expression are called  variables in Algebra

In Algebra  the word can be understood by a simple example. In the equation x + 5 = 9, the left hand side is the addition (sum) of two parts x and 5. If we add  (–5) to each side of the equation, we get

(x + 5) + (–5) = 9 + (–5) or x + [5 + (–5)] = 9 – 5 or x + 0 = 4 or x = 4.

Here 9 and −5 are reunited to get 4. It is said to be  algebra.  

Practice Problems in Algebra

Abstract algebra problems using Subtract 2x^3 – 3x^2 + 1 from x^3 + 5x^2 – 4x – 6 for practice.

Solution:  Associative and Distributive properties, we have

(x^3 + 5x^2 – 4x – 6) – (2x^3 – 3x^2 + 1)

= x^3 + 5x^2 – 4x – 6 – 2x^3 + 3x^2 - 1

= x^3 – 2x^3 + 5x^2 + 3x^2 – 4x – 6 - 1

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

= –x^3 + 8x^2 – 4x – 7.

Abstract algebra problems using find the product of x^3 – 2x^2 + 4 and 2x^2 + 3x – 1 .

Solution: (x^3 – 2x^2 +4) (2x^2 + 3x – 1)

= x^3 (2x^2 + 3x – 1) + (– 2x^2) (2x^2 + 3x – 1) + (+ 4) (2x^2 + 3x – 1)

= (2x5 + 3x4 – x^3) + (– 4x4 – 6x^3 + 2x^2) + (+ 8x^2 + 12x - 4)

= 2x5 + 3x4 – x^3 – 4x4 – 6x^3 + 2x^2 + 8x^2 + 12x - 4

= 2x5 + (3x4 – 4x4) + (–x^3 – 6x^3) + (2x^2 + 8x^2) + (+ 12x) -4

= 2x5 – x4 – 7x^3 + 10x^2 + 12x - 4.

Algebra is widely used in day to day activities watch out for my forthcoming posts on algebra 2 online help and answers to algebra 2 problems. I am sure they will be helpful.

Problem Using the Identities for Practice:

Abstract algebra problems using Identities (a + b + c)2 – 2(ab + bc + ca) = a2 + b2 + c2 for practice.

Solution:

Abstract algebra problems (i) (2x + y + 2z)2

(2x + y + 2z)2 = [(2x) + y + (2z)]2 = (2x)2 + y2 + (2z)2 + 2(2x)y + 2y(2z) + 2(2z)(2x)

= 4x^2 + y2 + 4z2 + 4xy + 4yz + 8zx.

Abstract algebra problems (ii) (x – 2y + z)2

(x – 2y + z)2 = [x + (–2y) + z]2= x^2 + (–2y)2 + z2 + 2x(–2y) +2 (–2y)z + 2zx

= x^2 + 4y2 + z2 – 4xy – 4yz + 2zx.

Abstract algebra problems (iii) (2p – 3q – r)2

(2p – 3q – r)2 = [(2p) + (–3q) + (–r)]2

= (2p)2 + (–3q)2 + (–r)2 + 2(2p) (–3q) + 2(–3q) (–r) + 2(–r)(2p).

= 4p2 + 9q2 + r2 – 12pq + 6qr – 4rp.

Abstract algebra problems (iv) (2a + 3b − 2c)2

= (2a + 3b – 2c)2 = [(2a) + (3b) + (–2c)]2

= (2a)2 + (3b)2 + (–2c)2 + 2(2a)(3b) + 2(3b)(–2c) + 2(–2c)(2a)

= 4a2 + 9b2 + 4c2 + 12ab – 12bc – 8ca.