J23

Expanding algebraic expressions
Consider the product of (a + b) and (c + d). This product can be written as
(a + b)(c + d).
Suppose we let (c + d) = u
Then, (a + b)(c + d) = (a + b) u
= au + bu
= a(c + d) + b(c + d)
= ac + ad + bc + bd
We see that each term in the first brackets is multiplied by each term in the
second brackets.
Example
Expand (3x + 2)(5 + 4y).
Solution
Let (5 + 4y) = u
(3x + 2)(5 + 4y) = (3x + 2) u
= 3xu + 2u
= 3x(5 + 4y) + 2(5 + 4y)
= 15x + 12xy + 10 + 8y
(3x + 2)(5 + 4y) = 15x + 12xy + 10 + 8y.
Example
Expand (2a – 5)(b + 9).
Solution
(2a - 5)(b + 9) = 2a(b + 9) – 5(b + 9)
= 2ab + 18a – 5b – 45
We have multiplied each term in the second brackets by -5 and not just 5.
Terms in an expression are identified with the signs before them.
Example
Expand (x + 5 3 )(y - 4 1 )
Solution
(x + 3 5
)(y - 1 4
) = xy - 1 4
x + 3 5
y -
3
20
Exercise
Expand.
1. (x + 1)(4 + y) 2. (6a + 1)(c + 3)
3. (2a + b)(c + 3d) 4. (x - y)(4 - z)

5. (a – 3d)(4c - b) 6. (8x + y)(a + 1)
7. (2u - v)(3x - y) 8. (p + q)(s - t)
9. (x + y)(c - 5) 10. (x + 1)(y - 1)
11. (x + 2 1 )(y + 4
1
) 12. (u + w)(a + 2)
13. (5a + 4)(b - 3) 14. (x + y)(
1 a - c)
2