Free Algebra Tutorials!

Try the Free Math Solver or Scroll down to Tutorials!

 Depdendent Variable

 Number of equations to solve: 23456789
 Equ. #1:
 Equ. #2:

 Equ. #3:

 Equ. #4:

 Equ. #5:

 Equ. #6:

 Equ. #7:

 Equ. #8:

 Equ. #9:

 Solve for:

 Dependent Variable

 Number of inequalities to solve: 23456789
 Ineq. #1:
 Ineq. #2:

 Ineq. #3:

 Ineq. #4:

 Ineq. #5:

 Ineq. #6:

 Ineq. #7:

 Ineq. #8:

 Ineq. #9:

 Solve for:

 Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:

# Simplifying Expressions Containing only Monomials

Monomials are algebraic expressions consisting of a single term (though this term may contain subexpressions which do contain more than one term). So here, we look at the simplification of simple products, quotients, and powers – in effect, we are reviewing and illustrating the laws of exponents, but using algebraic expressions rather than simple numerical expressions.

Symbolically, we have five distinct rules for combining powers:

Usually simplification involves combining rules (iv) and (v) with one or more of rules (i), (ii), or (iii). When division is involved, rule (ii) brings in something like cancelling common factors between the numerator and the denominator of a fraction. We’ll show a few examples here, but our detailed descriptions of methods for simplifying fractions must wait until a later document in this series.

Remember that multiplication with simple numbers or symbols representing simple numbers is commutative – the order of the factors doesn’t matter:

a Â· b = b Â· a

Example 1:

Simplify (3x 2 )(5x 3 )

solution:

We can rewrite this expression in detail as

(3x 2 )(5x 3 ) = (3)(x 2 )(5)(x 3 )

the product of four distinct factors. Now

(3)(x 2 )(5)(x 3 ) = (3)(5)(x 2 )(x 3 )

by rearranging the order of the factors, which leaves the result unchanged since the multiplication here is commutative. Now

(3)(x 2 )(5)(x 3 ) = (3 Â· 5)(x 2 Â· x 3 ) = 15x 2 + 3 = 15x 5

So, we conclude that

(3x 2 )(5x 3 ) = 15x 5

Example 2:

Simplify (2x 3 )2 (3x 3 )4

solution:

Using property (iv)

(2x 3 )2 = (2)2 (x 3 )2 = 4x 3 Â· 2 = 4x 6

and

(3x 3 )4 = (3)4 (x 3 )4 = 81x 3 Â· 4 = 81x 12

So,

(2x 3 )2 (3x 3 )4 = (4x 6 )(81x 12) = (4 Â· 81)(x 6 Â· x 12 ) = 324x 6 + 12 = 324x 18

Example 3:

Simplify (-3x 2 )5 (2x 4 )2

solution:

Care must be taken with the minus sign here. Again, applying property (iv), we get

(-3x 2 )5 = (-3)5 (x 2 )5 = -243x 10

and

(2x 4 )2 = (2)2 (x 4 )2 = 4x 8

So

(-3x 2 )5 (2x 4 )2 = (-243x 10 )(4x 8 ) = -972x 10+8 = -972x 18