✖

Retrieved from

Before we get into the reproof, let's discuss multiplying parentheses, because that is very related to the reproof itself. You could even say that reproofing is a completely opposite function to bracket multiplication.

Bracket multiplication

Bracket multiplication is a simple operation, which we will demonstrate with an example: let's compute the expression 3 · (5 + 8). We have two ways to proceed. We can add the expression in the bracket and get 3 · 13, because 5 + 8 = 13.

The second way is that we multiply the parenthesis. That is, we take the number before the parenthesis, the three, and multiply the two numbers in the parenthesis by that number. The plus sign in the parenthesis will remain. So we get:

$$ 3 \cdot (5 + 8) = (3 \cdot 5 + 3 \cdot 8) $$

Both procedures lead to the same result: (3 · 5 + 3 · 8) = 15 + 24 = 39.

Why bracket multiplication works

Why can we afford this procedure, why does multiplication work? Imagine the product as an expanded sum. For example, if we write 3 · 7, we are saying that we want to add the number seven three times, i.e. 3 · 7 = 7 + 7 + 7.

But we can break down the bracketed example in the same way: 3 · (5 + 8) is the same as (5 + 8) + (5 + 8) + (5 + 8). Since order doesn't matter in addition, we can remove the brackets and move the fives and eights next to each other like this: 5 + 5 + 5 + 8 + 8 + 8. And here we can see that we can rewrite this expression using multiplication like this: 3 · 5 + 3 · 8, which is what we did for multiplication.

Examples for multiplication

  1. Multiply 5 · (4 + 7). Again, just multiply the numbers in parentheses by five to get 5 · (4 + 7) = 5 · 4 + 5 · 7.
  • Multiply 6 · (x + 4). Let's not be scared and just multiply it by six, too. We get: 6 · (x + 4) = 6x + 6 · 4.
  • Multiply x · (4 + 173). Now we have the unknown in front of the parenthesis, so we multiply both numbers inside the parenthesis by x. We get x · (4 + 173) = 4x + 173x.
  • Multiply by: 4x · (2x + 7). Now we have an unknown before and inside the parenthesis, but we also have a more complex expression before the parenthesis. That's okay again, we multiply the area of the parenthesis by 2x like this: 4x · (2x + 7) = 4x · 2x + 4x · 7 = 8x2 + 28x.

Simple enumeration

The exclamation is the reverse of multiplying a parenthesis. For example, at the beginning we have the expression 10x + 5. Now, if we find a number that we can somehow reasonably divide the two addends by, we can dial that number, so to speak. In our expression, this will be the number 5, because 10x / 5 = 2x and 5 / 5 = 1. We can thus write that

$$ 10x + 5 = 5 \cdot (2x + 1). $$

When we backward multiply the parenthesis 5 · (2x + 1), we get 10x + 5 again. The goal of the exclusion is usually to simplify the expression or to get it into product form so that we can truncate one addend in a fraction.

Deduction of the unknown

Often the unknown itself is reproached, we don't have to reproach just the number. So if we have an expression 3x2 + 7x, we can extract x, i.e. divide both expressions by x, add parentheses and multiply the parentheses by x. We get 3x2 + 7x = x · (3x + 7).

Usually we try to output as many "things" as possible, so if we had the expression 8x2 + 12x, we could output only x and get 8x2 + 12x = x · (8x + 12), but we can see that we could also output the number 4 from the parentheses. We can do so now, additionally, and get: x · (4 · (2x + 3)). We can remove the outer brackets and get: 4x · (2x + 3).

However, we would get the same result if we were to extract the expression 4x directly from the original expression 8x2 + 12x. If we were to break this down, we would get

$$ 8x^2 + 12x = 4x \cdot (8x^2 / 4x + 12x / 4x) = 4x \cdot (2x + 3). $$

Spelling out more complex expressions

So far, we have only output from expressions that contain two addends. But we can also reproach from longer expressions. For example, from 7x3 + 5x2 + 2x we can weed out x and get x · (7x2 + 5x + 2).

Also, we don't always have to reproach only some nice number, we can reproach colons. For example, from the expression 6x + 7 we can corollary 2, but we get an ugly result because we have to divide both expressions by two. For 6x, we get a nice result because 6x / 2 = 3x, but for the number 7, we don't get a nice result: 7 / 2 = 3,5. We get the resulting expression: 6x + 7 = 2 · (3x + 3,5).

Solved examples

  • Somehow print 25x + 45. At first glance, we can see that we can print the number 5, so we get 5 · (5x + 9).

  • Simplify the fraction

    $$ \frac{3x^2+7x}{x}. $$

    The first thing we output in the numerator of the fraction is x. So we get x · (3x + 7). Now we can truncate x throughout the fraction .

    $$ \frac{3x^2+7x}{x} = \frac{x \cdot (3x + 7)}{x} = 3x + 7 $$

  • Somehow print out −2x2 − 8x. Here we have negative numbers for the first time. That's okay, just like we reproached positive numbers, we can reproach negative numbers. So we can just go to −2x. We'll get

    $$ -2x^2-8x = -2x (x + 4). $$