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Algebra - Expansions

Grade 9ICSE

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

🔑Concepts

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Expansion is the algebraic process of removing brackets by multiplying terms. Visually, the expansion (a+b)(c+d)(a+b)(c+d) can be represented as the area of a large rectangle with sides (a+b)(a+b) and (c+d)(c+d), which is equal to the sum of the areas of four smaller internal rectangles: ac,ad,bc,ac, ad, bc, and bdbd.

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Squaring a binomial (a+b)2(a+b)^2 results in a trinomial. Geometrically, this represents a square of side a+ba+b being divided into four sections: a square of area a2a^2, a square of area b2b^2, and two identical rectangles each of area abab. The result is a2+2ab+b2a^2 + 2ab + b^2.

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The product of a sum and a difference, (a+b)(a−b)(a+b)(a-b), leads to the 'Difference of Two Squares' identity. Visually, if you take a square with side aa (area a2a^2) and remove a smaller square of side bb (area b2b^2) from the corner, the remaining shape can be sliced and rearranged into a rectangle with dimensions (a+b)(a+b) and (a−b)(a-b).

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Expansion of a trinomial (a+b+c)2(a+b+c)^2 involves the sum of the squares of each term plus twice the product of every possible pair of terms. This expands into 6 terms: a2+b2+c2+2ab+2bc+2caa^2 + b^2 + c^2 + 2ab + 2bc + 2ca.

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Cubing a binomial (a+b)3(a+b)^3 represents the volume of a cube with side length a+ba+b. This volume is composed of a smaller cube a3a^3, another cube b3b^3, three rectangular prisms with dimensions a×a×ba \times a \times b, and three rectangular prisms with dimensions a×b×ba \times b \times b.

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Expansion of reciprocals involves expressions like (x+1x)2(x + \frac{1}{x})^2. Because the product of a number and its reciprocal is 1 (x⋅1x=1x \cdot \frac{1}{x} = 1), the middle term of the expansion becomes a constant integer, which is a unique property used to solve higher-order power problems.

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Conditional identities are specific results that hold true under given constraints. A common one is: if a+b+c=0a + b + c = 0, then a3+b3+c3=3abca^3 + b^3 + c^3 = 3abc. This simplifies calculations for sum of cubes significantly when the sum of the bases is zero.

📐Formulae

(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

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

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

(a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

(a+b)3=a3+b3+3ab(a+b)(a+b)^3 = a^3 + b^3 + 3ab(a+b) or a3+3a2b+3ab2+b3a^3 + 3a^2b + 3ab^2 + b^3

(a−b)3=a3−b3−3ab(a−b)(a-b)^3 = a^3 - b^3 - 3ab(a-b) or a3−3a2b+3ab2−b3a^3 - 3a^2b + 3ab^2 - b^3

a3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a+b)(a^2 - ab + b^2)

a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a-b)(a^2 + ab + b^2)

a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)

(x+1x)2=x2+1x2+2(x + \frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2

(x−1x)2=x2+1x2−2(x - \frac{1}{x})^2 = x^2 + \frac{1}{x^2} - 2

💡Examples

Problem 1:

Expand (2a−5b)2(2a - 5b)^2 and find the value if a=2,b=1a=2, b=1.

Solution:

  1. Use the identity (x−y)2=x2−2xy+y2(x-y)^2 = x^2 - 2xy + y^2.
  2. Let x=2ax = 2a and y=5by = 5b.
  3. (2a−5b)2=(2a)2−2(2a)(5b)+(5b)2(2a - 5b)^2 = (2a)^2 - 2(2a)(5b) + (5b)^2
  4. =4a2−20ab+25b2= 4a^2 - 20ab + 25b^2
  5. Substitute a=2,b=1a=2, b=1: 4(2)2−20(2)(1)+25(1)24(2)^2 - 20(2)(1) + 25(1)^2
  6. =4(4)−40+25=16−40+25=1= 4(4) - 40 + 25 = 16 - 40 + 25 = 1.

Explanation:

We apply the square of a binomial formula. Note that both the coefficient and the variable must be squared inside the parentheses.

Problem 2:

If x+1x=6x + \frac{1}{x} = 6, find the value of x2+1x2x^2 + \frac{1}{x^2} and x4+1x4x^4 + \frac{1}{x^4}.

Solution:

  1. Square the given equation: (x+1x)2=62(x + \frac{1}{x})^2 = 6^2
  2. x2+2(x)(1x)+1x2=36x^2 + 2(x)(\frac{1}{x}) + \frac{1}{x^2} = 36
  3. x2+2+1x2=36  ⟹  x2+1x2=34x^2 + 2 + \frac{1}{x^2} = 36 \implies x^2 + \frac{1}{x^2} = 34
  4. Square the result: (x2+1x2)2=342(x^2 + \frac{1}{x^2})^2 = 34^2
  5. x4+2(x2)(1x2)+1x4=1156x^4 + 2(x^2)(\frac{1}{x^2}) + \frac{1}{x^4} = 1156
  6. x4+2+1x4=1156  ⟹  x4+1x4=1154x^4 + 2 + \frac{1}{x^4} = 1156 \implies x^4 + \frac{1}{x^4} = 1154.

Explanation:

By squaring the sum of a term and its reciprocal, the middle term simplifies to the constant 2, allowing us to find the sum of squares by simple subtraction.