Q1.
Suppose 7x is split as 2x + 5x; can a similar rectangular arrangement be formed? Consider other possibilities and check.
Answer
No — the split 2x + 5x does not work. If two of the x-tiles go to the right of the x²-tile and five go below it, then the unit tiles must fill a 2 by 5 corner, which holds only 10 units. We have 12, so two tiles are left over and the shape is not a rectangle.
There are only three ways to split 7x into two whole-number pieces. Check each:
| Split of 7x | Corner the units must fill | Units needed | Units available | Rectangle? |
|---|---|---|---|---|
| 1x + 6x | 1 by 6 | 6 | 12 | No |
| 2x + 5x | 2 by 5 | 10 | 12 | No |
| 3x + 4x | 3 by 4 | 12 | 12 | Yes |
Why only one split can work: putting a x-tiles on one side and b on the other builds a rectangle of sides (x + a) and (x + b). Its area is (x + a)(x + b) = x² + (a + b)x + ab. Matching this with x² + 7x + 12 forces two conditions at once: a + b = 7 and ab = 12. Splitting 7x is easy — keeping the product at 12 at the same time is the real constraint, and only 3 and 4 satisfy both.
Tip: This is exactly why the ‘splitting the middle term’ method of Section 4.6 lists the factor pairs of the constant term first, and only then checks which pair adds to the middle coefficient. The tiles make the two conditions visible: one is a length, the other is an area.