Skip to content
Exit presentation

Algebra Play

Every term of one meets every term of the other

Two brackets sit side by side: (x + 3)(x + 5). Neither bracket has a value until x is known.

Grid: rows x, 3; columns x, 5.
Step 1 of 4
  1. Grid: rows x, 3; columns x, 5.
  2. x by x → x².
  3. x by 5 and 3 by x → 5x and 3x.
  4. 3 by 5 → 15. Total: x² + 8x + 15.

Rule: every term of one bracket meets every term of the other. Then join the like terms.

This works because (x + 3)(x + 5) is x(x + 5) + 3(x + 5).

At x = 2 the brackets give 5 × 7 = 35. And x² + 8x + 15 is 4 + 16 + 15 = 35.

A minus inside works the same way. The four pairings are x², −2x, 3x and −6. Then −2x + 3x = x, so (x + 3)(x − 2) = x² + x − 6.

Watch out!

The mistake pairs first with first: x by x. It pairs last with last: 3 by 5. That gives x² + 15. This answer matches at x = 0, by chance. At x = 5 it gives 25 + 15 = 40. The brackets give 8 × 10 = 80.

Slide x. Find where the short answer breaks.

Loading lab…

Jump to lesson

This track

  1. 0 Start here
  2. 1 What a term is built from
  3. 2 Joining terms of one kind
  4. 3 One term times another
  5. 4 Multiplying a bracket by one term
  6. 5 Every term of one meets every term of the other
  7. 6 Checkpoint: parts, joining and multiplying
  8. 7 A square cut into four pieces
  9. 8 The easy way round an awkward product
  10. 9 Taking out what every term shares
  11. 10 Putting a number in to find out
  12. 11 From a situation to a line
  13. 12 Checkpoint: patterns, factors and checking
  14. 13 The big picture
  15. 14 Track quiz

Other tracks