Suppose 7x is split as 2x + 5x; can a similar rectangular arrangement be formed? Consider other possibilities and check.

Step-by-step solution

Idea: The arrangement being copied is: the x²-tile in the corner, some x-tiles along its right side, the rest along its bottom, and the unit tiles filling the remaining corner. If p tiles go on the right and q below, the corner is a p × q block, so it needs exactly p × q unit tiles. We need p + q = 7 and p × q = 12.

1. Put the x²-tile down. Place 2 x-tiles along its right side and 5 x-tiles along its bottom.½ mark
2. The empty corner is 2 units wide and 5 units tall, so it holds exactly 2 × 5 = 10 unit tiles.½ mark
3. But we have 12 unit tiles. After filling the corner, 2 unit tiles are left over and cannot be placed without breaking the rectangle. So 2x + 5x does not work.½ mark
4. Other possibilities (pairs of whole numbers adding to 7):
   1 + 6: corner = 1 × 6 = 6 unit tiles ✗
   2 + 5: corner = 2 × 5 = 10 ✗
   3 + 4: corner = 3 × 4 = 12 ✓ (7 + 0 would need no units ✗)1 mark
5. Only 3x + 4x uses all 12 unit tiles, giving a rectangle with sides x + 3 and x + 4. So x² + 7x + 12 = (x + 3)(x + 4).½ mark

Check: (x + 2)(x + 5) = x² + 7x + 10, not + 12, confirming 2 + 5 fails. (x + 3)(x + 4) = x² + 7x + 12 ✓.

Answer to write in the exam

Common mistakes that cost marks

• Thinking any split of 7x works because the x-terms always add to 7x. The constant term must also match: the two numbers must multiply to 12.
• Squeezing the extra unit tiles in somewhere. Then the shape is no longer a rectangle, so its sides are not factors.
• Forgetting that the order does not matter: 3 + 4 and 4 + 3 give the same rectangle turned round.

How this can come in the exam

Try one yourself

Using the same reasoning, how should 8x be split to arrange tiles for x² + 8x + 15 into a rectangle? Write its sides.

More questions like this

• Algebra tiles can be used to represent products and find factors.
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  2. Lay out algebra tiles for x² + 11x + 30 in such a way that you will see its factors.
• We have seen that (x + 3)(x + 4) = x² + 7x + 12.
  Also (x + 6)(x + 7) = x² + 13x + 42.
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• Now consider the case where we have a rectangle of sidelengths 2x + 3 and 3x + 1, as shown in the figure. What can you say about its area (2x + 3) (3x + 1)?
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  (px + a) (qx + b) = (_____)x² + (_____)x + _____ .
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• Let us begin with x² + 7x + 12 = x² + (a + b)x + ab.

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