Multiplying by a Monomial
The main aim of this document is to revisit the meaning of
brackets (or parentheses) when used in algebraic expressions.
No matter how difficult an aplication of brackets shall be,
what stands behind it is the socalled distributive law
a(b + c) = ab + ac
What we mean by placing ‘b + c’ inside the brackets
and multiplying the bracketed expression by ‘a’, is
that every term inside the brackets is to be multiplied by
‘a’. That is, we are distributing the factor
‘a’ to all of the terms inside the brackets.
The process of eliminating a pair of brackets as shown in the
distributive law above is called expanding the
brackets..
Example 1:
Simplify: 3t + 7 – (2t – 5) .
solution:
3t + 7 – (2t – 5)
= 3t + 7 + (1)(2t – 5)
= 3t + 7 + (1)(2t) + (1)(5)
= 3t + 7 – 2t + 5
= (3 – 2)t + 7 + 5
= t + 12
Of course, if a bracketed expression is preceded only
by a minus sign, the brackets can simple be dropped as long as
the sign of every term originally inside those brackets is
reversed. For the present example, this gives
3t + 7 – (2t – 5) = 3t + 7 – 2t + 5
= (3 – 2)t + 7 + 5
= t + 12
When the brackets were dropped in going from the first to the
second line here, the +2t was reversed to a 25 and the 5 was
reversed to a +5 because the brackets were preceded by a minus
sign.
Example 2:
Simplify: 2 – {6 – [(2c + 5) – (3c – 4)] }
+ 7c – 6 .
solution:
This is a very complicated expressions, which we will be able
to simplify greatly, but the simplification must be done with
great care. Recall that the rule for nested pairs of brackets is
that we begin with the innermost pair and work our way outwards.
Here, the innermost pair of brackets is
[(2c + 5) – (3c – 4)] = (2c + 5) – (3c –
4)
= 2c + 5 + (1)(3c) + (1)(4)
= 2c + 5 – 3c + 4
= c + 9.
The brackets around ‘2c + 5’ had no effect, so they
could simply be deleted. However, to remove the brackets around
‘3c – 4,’ we had to distribute the minus sign in
front. Once both pairs of brackets were removed, the remaining
simplification by collection of like terms was easy.
So, now the original expression has been reduced to
2 – {6 – [(2c + 5) – (3c – 4)] } + 7c
– 6
= 2 – {6 – (c + 9) } + 7c – 6
= 2 – {6 + (1)(c) + (1)(9) } + 7c – 6
= 2 – {6 + c – 9} + 7c – 6
= 2 – {c – 3} + 7c – 6
= 2 + (1)(c) + (1)(3) + 7c – 6 }
= 2 – c + 3 + 7c – 6
= (7 – 1)c + 2 + 3 – 6
= 6c – 1
as our final result.
Here we just continued to carry out the process of removing
the innermost pair of brackets and simplifying the result. You
can see that the final result is very, very much simpler than the
original expression. Study this example carefully to make sure
you understand each step well so that you would be able to
explain how you can tell that the changes made from one step to
the next are correct.
Here’s one more example that you can use as a practice
problem if you like.
Example 3:
Simplify: 3t + {4 – 5 [2t – (4 – 3t) ] – 7
} + 6.
solution:
We’ll just show the steps here in detail. The strategy we
are using is to work from the innermost set of brackets outwards.
3t + {4 – 5 [2t – (4 – 3t) ] – 7 } + 6
= 3t + {4 – 5 [2t + (1)(4) + (1)(3t) ] 7 } + 6
= 3t + {4 – 5 [2t – 4 + 3t] – 7 } + 6
= 3t + {4 – 5 [5t – 4] – 7 } + 6
= 3t + {4 + (5)(5t) + (5)(4) – 7} + 6
= 3t + {4 – 25t + 20 – 7} + 6
= 3t + {25t + 17} + 6
= 3t – 25t + 17 + 6
= 22t + 23
as the final answer.
It is awfully tempting to start out here by doing ‘4
– 5 = 1’ just inside the curly brackets in the initial
expression. However, this would be doing a subtraction before
doing a multiplication involving one of the two numbers –
the 5 is multiplied onto something else. Doing subtraction before
multiplication violates the operation priority rules, and so
would be an error.
