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We Distribute, Yet Things Multiply - Special Cases of the Distributive Property

Grade 8CBSE

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

🔑Concepts

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Multiplying a binomial by another binomial involves applying the distributive property twice: (a+b)(c+d)=a(c+d)+b(c+d)=ac+ad+bc+bd(a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd.

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The identity for the square of a sum is derived from distribution: (a+b)2=(a+b)(a+b)=a2+2ab+b2(a + b)^2 = (a + b)(a + b) = a^2 + 2ab + b^2.

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The identity for the square of a difference is: (a−b)2=(a−b)(a−b)=a2−2ab+b2(a - b)^2 = (a - b)(a - b) = a^2 - 2ab + b^2.

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The product of the sum and difference of two terms results in the difference of their squares: (a+b)(a−b)=a2−b2(a + b)(a - b) = a^2 - b^2.

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When binomials have a common first term, we use the identity: (x+a)(x+b)=x2+(a+b)x+ab(x + a)(x + b) = x^2 + (a + b)x + ab.

📐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

💡Examples

Problem 1:

Evaluate (103)2(103)^2 using algebraic identities.

Solution:

We can write 103103 as (100+3)(100 + 3). Using the identity (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2, where a=100a = 100 and b=3b = 3: (100+3)2=1002+2(100)(3)+32(100 + 3)^2 = 100^2 + 2(100)(3) + 3^2 =10000+600+9=10609= 10000 + 600 + 9 = 10609

Explanation:

By breaking the number into a sum of a round number and a small integer, we simplify the calculation using the square of a sum identity.

Problem 2:

Simplify (2x+5y)(2x−5y)(2x + 5y)(2x - 5y).

Solution:

Using the identity (a+b)(a−b)=a2−b2(a + b)(a - b) = a^2 - b^2, where a=2xa = 2x and b=5yb = 5y: (2x+5y)(2x−5y)=(2x)2−(5y)2(2x + 5y)(2x - 5y) = (2x)^2 - (5y)^2 =4x2−25y2= 4x^2 - 25y^2

Explanation:

This is a special case of the distributive property where the middle terms +10xy+10xy and −10xy-10xy cancel each other out.

Problem 3:

Find the product of 98×10398 \times 103 using the identity (x+a)(x+b)(x + a)(x + b).

Solution:

We can write 9898 as (100−2)(100 - 2) and 103103 as (100+3)(100 + 3). Here, x=100x = 100, a=−2a = -2, and b=3b = 3. Using (x+a)(x+b)=x2+(a+b)x+ab(x + a)(x + b) = x^2 + (a + b)x + ab: (100−2)(100+3)=1002+(−2+3)100+(−2)(3)(100 - 2)(100 + 3) = 100^2 + (-2 + 3)100 + (-2)(3) =10000+(1)100−6= 10000 + (1)100 - 6 =10000+100−6= 10000 + 100 - 6 =10100−6=10094= 10100 - 6 = 10094 Vertical subtraction for the final step: 10100−610094\begin{array}{r} 10100 \\ - 6 \\ \hline 10094 \end{array}

Explanation:

The identity (x+a)(x+b)(x+a)(x+b) allows us to multiply numbers near a common base (like 100) efficiently.