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We Distribute, Yet Things Multiply - This Way or That Way, All Ways Lead to the Bay

Grade 8CBSE

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

🔑Concepts

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The Distributive Law: Multiplying a monomial by a polynomial involves multiplying the monomial with each term of the polynomial, such as a(b+c)=ab+aca(b + c) = ab + ac.

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Multiplying a Binomial by a Binomial: Each term in the first binomial must multiply every term in the second binomial. This can be done using the Horizontal Method or the Vertical (Column) Method.

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Horizontal Method: To solve (a+b)(c+d)(a+b)(c+d), we expand it as a(c+d)+b(c+d)=ac+ad+bc+bda(c+d) + b(c+d) = ac + ad + bc + bd.

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Standard Identities: These are equalities that hold true for any value of the variables involved. They act as shortcuts for multiplication.

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Identity I: Square of a sum, (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2.

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Identity II: Square of a difference, (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2.

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Identity III: Product of sum and difference, (a+b)(a−b)=a2−b2(a + b)(a - b) = a^2 - b^2.

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Identity IV: Product of two binomials with a common first term, (x+a)(x+b)=x2+(a+b)x+ab(x + a)(x + b) = x^2 + (a + b)x + ab.

📐Formulae

a(b+c)=ab+aca(b + c) = ab + ac

(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:

Multiply (2x+3y)(2x + 3y) and (3x+4y)(3x + 4y) using the horizontal method.

Solution:

(2x+3y)(3x+4y)=2x(3x+4y)+3y(3x+4y)=6x2+8xy+9xy+12y2=6x2+17xy+12y2(2x + 3y)(3x + 4y) = 2x(3x + 4y) + 3y(3x + 4y) = 6x^2 + 8xy + 9xy + 12y^2 = 6x^2 + 17xy + 12y^2

Explanation:

We distribute the first term of the first binomial over the second binomial, then distribute the second term, and finally combine the like terms 8xy8xy and 9xy9xy.

Problem 2:

Evaluate 103×97103 \times 97 using a suitable identity.

Solution:

103×97=(100+3)(100−3)=1002−32=10000−9=9991103 \times 97 = (100 + 3)(100 - 3) = 100^2 - 3^2 = 10000 - 9 = 9991

Explanation:

We use the identity (a+b)(a−b)=a2−b2(a + b)(a - b) = a^2 - b^2 where a=100a = 100 and b=3b = 3.

Problem 3:

Multiply (x+5)(x + 5) and (x+2)(x + 2) using the vertical method.

Solution:

x+5×x+22x+10x2+5x+00x2+7x+10\begin{array}{r} x + 5 \\ \times x + 2 \\ \hline 2x + 10 \\ x^2 + 5x \phantom{+ 00} \\ \hline x^2 + 7x + 10 \end{array}

Explanation:

First, multiply (x+5)(x + 5) by 22 to get 2x+102x + 10. Then multiply (x+5)(x + 5) by xx to get x2+5xx^2 + 5x. Align the like terms vertically and add them.

Problem 4:

Expand (4p−3q)2(4p - 3q)^2 using the standard identity.

Solution:

(4p−3q)2=(4p)2−2(4p)(3q)+(3q)2=16p2−24pq+9q2(4p - 3q)^2 = (4p)^2 - 2(4p)(3q) + (3q)^2 = 16p^2 - 24pq + 9q^2

Explanation:

Using Identity II: (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2, where a=4pa = 4p and b=3qb = 3q.