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# 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 so-called 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