Let's explore how to rewrite expressions by grouping terms. We'll use this balance scale to see what's happening.
On the left side, we have two squares labeled S and six triangles labeled T.
This represents the expression 2 S + 6 T. The right side shows two identical boxes, which tells us that the items on the left can be split into two identical groups. To figure out what's in each group, let's look for a common factor in 2S and 6T. Both 2 and 6 are divisible by two. This means we can pull a two outside the parenthesis.
If we divide 2s by two, we're left with 1 s. If we divide 6 t by 2, we get 3t.
So, we have two groups and each group contains s + 3 t.
Let's look at a different problem. On the left, we have a total weight of 8 made from four weights of 2 plus four circles labeled c. The expression is 8 + 4 C. The equation shows that this balances with four identical groups.
This tells us we can divide each term by four. Let's divide each part of the expression by four. 8 / 4 is 2 and 4 C / 4 is just C. This means each of the four groups contains 2 + C. So 8 + 4 C is the same as four groups of 2 + C.
We can represent this visually. 8 + 4 C has been packed into four identical boxes, each containing 2 + C.
In this example, the left side of the scale has six triangles T and three circles C, representing the expression 6 T + 3 C. We need to figure out how many identical groups we can make. This is the same as finding the greatest common factor of the coefficients 6 and 3. The largest number that divides both 6 and three is three. So we can make three identical groups. Now let's find out what's in each group. We divide 6 T by 3 to get 2 T. Then we divide 3 C by 3 to get C. So we have three groups and each group contains 2 T + C.
For our last problem, the left side has six squares labeled S and a single weight with a weight of 15. The expression is 6 S + 15. The equation tells us we're making three identical groups, so we need to pull out a three.
Let's check if this is possible. 6 is divisible by 3 and 15 is also divisible by three. So, this works. To find what's inside each group, we divide each term by three. 6 / three groups gives us 2 S per group. 15 divided into three groups gives five per group. This means each of the three identical groups contains 2 S + 5.
Arranging terms into identical groups is a way to visualize factoring. This skill of repackaging an expression is very useful when simplifying and solving equations.
