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Exploring Algebraic Identities - Factorisation Using Algebra Tiles

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Algebra tiles are geometric models used to represent polynomials. A large square represents x2x^2, a rectangle represents xx, and a small square represents a unit 11.

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Factorisation using tiles involves arranging a given set of tiles representing an expression into a perfect rectangle.

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The area of the rectangle formed represents the algebraic expression, while the length and width of the rectangle represent its factors.

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For trinomials of the form x2+bx+cx^2 + bx + c, we use one x2x^2 tile, bb number of xx tiles, and cc number of unit tiles.

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To factorise expressions like a2−b2a^2 - b^2, we use the concept of adding 'zero pairs' (adding and subtracting the same term) to complete a rectangular shape.

📐Formulae

(x+a)(x+b)=x2+(a+b)x+ab(x + a)(x + b) = x^2 + (a + b)x + ab

(a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2

(a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2

a2−b2=(a+b)(a−b)a^2 - b^2 = (a + b)(a - b)

💡Examples

Problem 1:

Factorise the quadratic trinomial x2+5x+6x^2 + 5x + 6 using algebra tiles.

Solution:

  1. Take one x2x^2 tile, five xx tiles, and six unit tiles.
  2. Arrange the x2x^2 tile at the top-left.
  3. Arrange the xx tiles and unit tiles to form a rectangle. This is achieved by placing 3x3x tiles horizontally and 2x2x tiles vertically (or vice versa) to accommodate the 66 unit tiles in a 2×32 \times 3 grid.
  4. The resulting rectangle has a length of (x+3)(x + 3) and a width of (x+2)(x + 2).
  5. Therefore, x2+5x+6=(x+3)(x+2)x^2 + 5x + 6 = (x + 3)(x + 2).

Explanation:

We split the middle term 5x5x into 3x3x and 2x2x because 3+2=53 + 2 = 5 and 3×2=63 \times 2 = 6 (the constant term). The tiles form a rectangle with dimensions corresponding to the factors.

Problem 2:

Factorise x2−4x^2 - 4 using the identity for the difference of two squares.

Solution:

  1. We have one x2x^2 tile and four negative unit tiles −4-4.
  2. To form a rectangle, we need to add 'zero pairs' of xx tiles. We add +2x+2x and −2x-2x tiles.
  3. Arrange the x2x^2 tile, the +2x+2x tiles, the −2x-2x tiles, and the −4-4 unit tiles into a square/rectangle.
  4. The horizontal dimension becomes (x+2)(x + 2) and the vertical dimension becomes (x−2)(x - 2).
  5. Thus, x2−4=(x+2)(x−2)x^2 - 4 = (x + 2)(x - 2).

Explanation:

By adding 2x−2x=02x - 2x = 0, we do not change the value of the expression, but we provide the necessary tiles to complete the rectangular dimensions required for factorisation.

Problem 3:

Use the identity (a+b)2(a+b)^2 to find the area of a square with side (x+4)(x + 4).

Solution:

The area is given by: Area=(x+4)2Area = (x + 4)^2 Using the identity (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2, where a=xa = x and b=4b = 4: Area=x2+2(x)(4)+42Area = x^2 + 2(x)(4) + 4^2 Area=x2+8x+16Area = x^2 + 8x + 16 Using tiles, this would be one x2x^2 tile, eight xx tiles, and sixteen unit tiles arranged in a square of side (x+4)(x+4).

Explanation:

The identity (a+b)2(a+b)^2 represents the area of a square composed of one a×aa \times a square, two a×ba \times b rectangles, and one b×bb \times b square.