Back to Course

JSS3: MATHEMATICS - 1ST TERM

0% Complete
0/0 Steps
  1. Binary Number System I | Week 1
    5Topics
    |
    1 Quiz
  2. Binary Number System II | Week 2
    6Topics
    |
    1 Quiz
  3. Word Problems I | Week 3
    4Topics
    |
    1 Quiz
  4. Word Problems with Fractions II | Week 4
    1Topic
    |
    1 Quiz
  5. Factorisation I | Week 5
    4Topics
    |
    1 Quiz
  6. Factorisation II | Week 6
    3Topics
    |
    1 Quiz
  7. Factorisation III | Week 7
    3Topics
    |
    1 Quiz
  8. Substitution & Change of Subject of Formulae | Week 8
    2Topics
    |
    1 Quiz
  9. Simple Equations Involving Fractions | Week 9
    3Topics
    |
    1 Quiz
  10. Word Problems | Week 10
    1Topic
    |
    1 Quiz
Lesson Progress
0% Complete

Multiplying out brackets

To multiply out two binomial expressions such as (x + 3)  (x + 3), make sure that each term in the second bracket is multiplied by each term in the  first bracket.

(x + 3)  (x + 3) = x(x + 3) + 3(x + 3)

= x2 + 3x + 3x + 9

= x2 + 6x + 9

Expand the brackets then simplify your expressions where possible.

This expansion can easily remembered by the word FOIL.

F = First two terms

O = Outer two terms

I  =  Inner two terms

L = Last two terms

Using Foil Method

(x + 4)(x + 2)

Step 1: According to the FOIL Rule, the first step is to Multiply the First two terms, x, and x.

\( \scriptsize x \; \times \; x = x^2 \)

foil2 e1606237953898

Step 2: The second step is to Multiply the Outer terms, that is, x, and 2

\( \scriptsize x \; \times \; 2 = 2x \)

outer terms e1606238090127

Step 3: The third step is to Multiply the Inner terms, that is, 4, and x

\( \scriptsize 4 \: \times \: x = 4x \)

inner terms e1606238567280

Step 4: The fourth step is to Multiply the Last terms, that is, 4, and 2

\( \scriptsize 4 \: \times \: 2 = 8\)
last terms e1606238657901
FOIL
x22x4x8

= x2 + 2x + 4x + 8

= x2 + 6x + 8

Example 1

Expand the brackets then simplify your expression where possible.

a. (x + 5)(x + 2)

x(x + 2) + 5 (x + 2)

x2 + 2x + 5x  + 10

x2 + 7x  + 10

b. (x + 3)(x + 4)

= x(x + 4) + 3 (x + 4)

= x2 + 4x + 3x  + 12

= x2 + 7x  + 12

c. (a + 3) (a +5)

= a(a + 5 )  + 3( a + 5) 

a2 + 5a + 3a + 15

(Add Like Terms)

a2 +8a + 15

d. (x – 2)(x + 3)

= x(x + 3) – 2 (x + 3)

x2 + 3x – 2x  – 6

x2 + x  – 6

e. (2x – 5)(x + 3)

= 2x(x + 3) – 5(x + 3)

2x2 + 6x – 5x  – 15

2x2 + x  – 15

f. (3x – 2)(3x + 3)

= 3x(3x + 3) – 2(3x + 3)

9x2 + 9x – 6x  – 6

9x2 + 3x  – 6

back-to-top
error: