Expanding Two Brackets

Four products, one from each pair of terms.

Two brackets make a rectangle

To expand (x + 2)(x + 3), draw a rectangle x + 3 wide and x + 2 tall. Its area is the width times the height, which is exactly (x + 2)(x + 3).

x3x2(x + 2)(x + 3)

A rectangle x + 3 wide and x + 2 tall has area (x + 2)(x + 3).

Cut it both ways: four pieces

Cut the width where x ends and 3 begins, and cut the height where x ends and 2 begins. The two cuts split the rectangle into four pieces: x × x = x², x × 3 = 3x, 2 × x = 2x, and 2 × 3 = 6.

Every term in the first bracket has multiplied every term in the second. The first bracket has two terms and the second has two, so there are 2 × 2 = 4 products.

x²3x2x6x3x2(x + 2)(x + 3)

The four pieces are x², 3x, 2x and 6.

Collect the middle

The area is the four pieces added: x² + 3x + 2x + 6. The two middle pieces, 3x and 2x, are both lots of x, so they are like terms and join into 5x. So (x + 2)(x + 3) = x² + 5x + 6.

Look at where the numbers went. The x term, 5x, comes from adding the two numbers in the brackets, 2 + 3 = 5. The number at the end, 6, comes from multiplying them, 2 × 3 = 6.

Check with a number. When x = 10, (x + 2)(x + 3) is 12 × 13 = 156, and x² + 5x + 6 is 100 + 50 + 6 = 156.

x²3x2x6x3x2x² + 5x + 6

The two colored pieces are both lots of x, so 3x and 2x join into 5x.

x²1x2x2x1

the middle term is the sum of the two strips: 1x + 2x = 3x

Make the middle term 7x

Drag the corner to change the numbers in the two brackets. The two strips are the x terms, and the middle term is their sum. Make the middle term 7x.

Do not forget the middle

The usual mistake is to multiply only the first terms and the last terms, x × x and 2 × 3, and write (x + 2)(x + 3) = x² + 6. That takes only the two corner pieces of the rectangle and leaves out the two strips, 3x and 2x. Every term must multiply every term, which always gives four products.

With x = 10 the difference shows: x² + 6 is 106, but the rectangle is 12 × 13 = 156.

x²3x2x6x3x2x² + 6 misses two pieces

x² + 6 counts only the two colored corner pieces. The strips 3x and 2x are left out.

A minus sign in a bracket

Expand (x + 3)(x − 1). The second bracket's terms are x and −1, and each keeps its sign when it multiplies. The four products are x × x = x², x × (−1) = −x, 3 × x = 3x, and 3 × (−1) = −3.

Collect the x terms: −x + 3x = 2x. So (x + 3)(x − 1) = x² + 2x − 3. Check with x = 10: 13 × 9 = 117, and 100 + 20 − 3 = 117.

Worked example: A Path of Constant Width Round a Rectangular Lawn

Question A rectangular lawn is x m wide and (x + 5) m long. A path 1 m wide runs all the way round it. (a) Find the area of the path as a simplified expression in x. (b) Find the area of the path when x = 8.

  1. 1.The path adds 1 m at both ends of each side, so the outer rectangle is (x + 2) m wide and (x + 7) m long.

    x + 5xlawnx + 7x + 2the path is 1 m wide
    x + 5xlawnx + 7x + 2the path is 1 m wide
    The path adds 1 m at both ends, so the outer rectangle is (x + 2) m by (x + 7) m.

  2. 2.The area of the path is the outer rectangle minus the lawn: (x + 2)(x + 7) − x(x + 5).

    x + 5xlawnx + 7x + 2the path is 1 m wide(x + 2)(x + 7) − x(x + 5)
    x + 5xlawnx + 7x + 2the path is 1 m wide(x + 2)(x + 7) − x(x + 5)
    The path is the outer rectangle minus the lawn: (x + 2)(x + 7) − x(x + 5).

  3. 3.Expand each product: (x + 2)(x + 7) = x2 + 7x + 2x + 14 = x2 + 9x + 14, and x(x + 5) = x2 + 5x.

    x + 5xlawnx + 7x + 2the path is 1 m wide(x + 2)(x + 7) − x(x + 5)(x2+ 9x + 14) − (x2+ 5x)expand each product
    x + 5xlawnx + 7x + 2the path is 1 m wide(x + 2)(x + 7) − x(x + 5)expand each product(x2+ 9x + 14) − (x2+ 5x)
    (x + 2)(x + 7) = x2 + 9x + 14 and x(x + 5) = x2 + 5x.

  4. 4.Subtract every term of the lawn, so both of its terms change sign: x2 + 9x + 14 − x2 − 5x.

    x + 5xlawnx + 7x + 2the path is 1 m wide(x + 2)(x + 7) − x(x + 5)(x2+ 9x + 14) − (x2+ 5x)expand each productx2+ 9x + 14 − x2− 5xsubtract every term
    x + 5xlawnx + 7x + 2the path is 1 m wide(x + 2)(x + 7) − x(x + 5)expand each product(x2+ 9x + 14) − (x2+ 5x)subtract every termx2+ 9x + 14 − x2− 5x
    Subtract every term of the lawn: x2 + 9x + 14 − x2 − 5x.

  5. 5.(a) Collect like terms: x2 − x2 = 0 and 9x − 5x = 4x. The area of the path is (4x + 14) m2.

    x + 5xlawnx + 7x + 2the path is 1 m wide(x + 2)(x + 7) − x(x + 5)(x2+ 9x + 14) − (x2+ 5x)expand each productx2+ 9x + 14 − x2− 5xsubtract every term4x + 14collect like terms
    x + 5xlawnx + 7x + 2the path is 1 m wide(x + 2)(x + 7) − x(x + 5)expand each product(x2+ 9x + 14) − (x2+ 5x)subtract every termx2+ 9x + 14 − x2− 5xcollect like terms4x + 14
    (a) x2 − x2 = 0 and 9x − 5x = 4x, so the area of the path is (4x + 14) m2.

  6. 6.(b) When x = 8 the area is 4 × 8 + 14 = 46 m2. Check: the outer rectangle is 10 × 15 = 150 m2 and the lawn is 8 × 13 = 104 m2, and 150 − 104 = 46.

    x + 5 = 13x = 8lawnx + 7 = 15x + 2 = 10the path is 1 m wide(x + 2)(x + 7) − x(x + 5)(x2+ 9x + 14) − (x2+ 5x)expand each productx2+ 9x + 14 − x2− 5xsubtract every term4x + 14collect like terms4 × 8 + 14 = 46 m2
    x + 5 = 13x = 8lawnx + 7 = 15x + 2 = 10the path is 1 m wide(x + 2)(x + 7) − x(x + 5)expand each product(x2+ 9x + 14) − (x2+ 5x)subtract every termx2+ 9x + 14 − x2− 5xcollect like terms4x + 144 × 8 + 14 = 46 m2
    (b) When x = 8 the area of the path is 4 × 8 + 14 = 46 m2.

Answer: (a) (4x + 14) m2; (b) 46 m2

Common mistakes

  • Making the outer rectangle (x + 1) m by (x + 6) m. The path lies on both sides of the lawn, so each side of the outer rectangle is 2 m longer than the lawn, not 1 m.
  • Subtracting only the first term of the lawn, as in x2 + 9x + 14 − x2 + 5x. The whole area x2 + 5x is taken away, so both of its terms change sign.

More algebraic expressions problems, worked step by step →

Practice Expanding Two Brackets in the app