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Practice the questions given in the Worksheet on Factorization. Factorization of expressions of the form a^3 ± b^3 1. Factorize: (i) 8x^3 + 27y^3 (ii) 216a^3 + 1 (iii) a^6 + 1 (iv) x^3 + \(\frac{1}{x^{3}}\) (v) a^3 + 8b^6

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Here we will learn the process of On Factorizations of expressions of the Form a^3 + b^3 + c^3 – 3abc. We have, a^3 + b^3 + c^3 – 3abc = (a + b + c)(a^2 + b^2 + c^2 – bc – ca – ab). [Verify by actual multiplication.] Example: Factorize: x^3 + y^3 – 3xy + 1. Solution: Here

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Here we will learn the process of On Factorization of expressions of the Form a^3 + b^3 + c^3 , where a + b + c = 0. We have, a^3 + b^3 + c^3 = a^3 + b^3 – (-c)^3 = a^3 + b^3 – (a + b)^3, [Since, a + b + c = 0] = a^3 + b^3 – {a^3 + b^3 + 3ab(a + b)} = -3ab(a + b) = -3ab(-c

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Here we will solve different types of Miscellaneous Problems on Factorization. 1. Factorize: x(2x + 5) – 3 Solution: Given expression = x(2x + 5) – 3 = 2x^2 + 5x – 3 = 2x^2 + 6x – x – 3, [Since, 2(-3) = - 6 = 6 × (-1), and 6 + (-1) = 5] = 2x(x + 3) – 1(x + 3)

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Here we will learn the process of Factorization of Expressions of the Form a^3 - b^3. We know that (a - b)^3 = a^3 - b^3 - 3ab(a - b), and so a^3 - b^3 = (a - b)^3 + 3ab(a - b) = (a - b){(a - b)^2 + 3ab} Therefore, a^3 - b^3 = (a - b)(a^2 + ab + b^2) Example: 1. Factorize:

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Here we will learn the process of Factorization of Expressions of the Form a^3 + b^3. We know that (a + b)^3 = a^3 + b^3 + 3ab(a + b), and so a^3 + b^3 = (a + b)^3 – 3ab(a + b) = (a + b){(a + b)^2 – 3ab} Therefore, a^3 + b^3 = (a + b)(a^2 – ab + b^2). Example: 1. Factorize:

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Practice the questions given in the worksheet on factorization of the trinomial ax^2 + bx + c. 1. Factorization of a perfect-square trinomial. (i) a^2 + 6a + 9 (ii) a^2 + a + \(\frac{1}{4}\) (iii) 25x^2 – 10x + 1 (iv) 4x^2 – 4xy + y^2 2. Factorization of expressions of the

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Here we will solve different types of Problems on Factorization of Expressions of the Form x^2 + (a + b)x + ab. 1. Factorize: a^2 + 25a - 54 Solution: Here, constant term = -54 = (27) × (-2), and 27 + (-2) = 25 (= coefficient of a). Therefore, a^2 + 25a – 54

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The below examples show that the method of factorizing ax^2 + bx + c by breaking the middle term involves the following steps. Steps: 1.Take the product of the constant term and the coefficient of x^2, i.e., ac. 2. Break ac into two factors p, q whose sum is b,

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Here we will learn the process of Factorization of Expressions of the Form x^2 + (a + b)x + ab. We know, (x + a)(x + b) = x^2 + (a + b)x + ab. Therefore, x^2 + (a + b)x + ab = (x + a)(x + b). 1. Factorize: a^2 + 7a + 12. Solution: Here, constant term = 12 = 3 × 4, and 3 + 4

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Here we will learn the process of Factorization of a Perfect-square Trinomial. A trinomial of the form a^2 ± 2ab + b^2 = (a ± b)^2 = (a ± b)(a ± b) Solved examples on Factorization of a Perfect-square Trinomial 1. Factorize: x^2 + 6x + 9 Solution: Here, given expression

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Problems on Factorization using a^2 – b^2 = (a + b)(a – b) Here we will solve different types of Problems on Factorization using a^2 – b^2 = (a + b)(a – b). 1. Factorize: 4a^2 – b^2 + 2a + b Solution: Given expression = 4a^2 – b^2 + 2a + b = (4a^2 – b^2) + 2a + b

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Here we will solve different types of Problems on Factorization of expressions of the form a^2 – b^2. 1. Resolve into factors: 49a^2 – 81b^2 Solution: Given expression = 49a^2 – 81b^2 = (7a)^2 – (9b)^2 = (7a + 9b)(7a – 9b). 2.Factorize: (x + y)^2 – 4(x - y)^2 Solution: Given

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Here we will solve different types of Problems on Factorization by grouping of terms. 1. Factorize: a^2 – (b – 5)a – 5b. Solution: Given expression = a^2 – (b – 5)a – 5b = a^2 – ba + 5a – 5b = a(a - b) + 5(a - b) = (a – b)(a + 5). 2. Factorize: a^2 + b^2 + a + b + 2ab

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We will discuss here about the introduction to factorization. The method of expressing a given polynomial as a product of two or more polynomials is called factorization. The polynomials whose product is the given polynomial are called its factors. You are already familiar

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Here we will solve different types of application problems on expanding of (a ± b)\(^{3}\) and its corollaries. 1. Expanding the following: (i) (1 + x)\(^{3}\) (ii) (2a – 3b)\(^{3}\) (iii) (x + \(\frac{1}{x}\))\(^{3}\) Solution: (i) (1 + x)\(^{3}\) = 1\(^{3}\) +

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Practice the questions given in the worksheet on application problems on expansion of powers of binomials and trinomials. 1. Use (a ± b)^2 = a^2 ± 2ab + b2 to evaluate the following: (i) (3.001)^2 (ii) (5.99)^2 (iii) 1001 × 999 (iv) 5.63 × 5.63 + 11.26 × 2.37 + 2.37 × 2.37

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Here we will solve different types of application problems on expansion of powers of binomials and trinomials. 1. Use (x ± y)^2 = x^2 ± 2xy + y^2 to evaluate (2.05)^2. Solution: (2.05)^2 = (2 + 0.05)^2 = 2^2 + 2 × 2 × 0.05 + (0.05)^2 = 4 + 0.20 + 0.0025 = 4.2025.

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We will discuss here about the expansion of (x + a)(x + b)(x + c). (x + a)(x + b)(x + c) = (x + a){(x + b)(x + c)} = (x + a){x\(^{2}\) + (b + c)x + bc} = x{x\(^{2}\) + (b + c)x + bc} + a{x\(^{2}\) + (b + c)x + bc} = x\(^{3}\) + (b + c)x\(^{2}\) + bcx + ax\(^{2}\) + a(b + c)x

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We will discuss here about the expansion of (a + b + c)(a\(^{2}\) + b\(^{2}\) + c\(^{2}\) – ab – bc – ca). (a + b + c)(a\(^{2}\) + b\(^{2}\) + c\(^{2}\) – ab – bc – ca) = a(a\(^{2}\) + b\(^{2}\) + c\(^{2}\) – ab – bc – ca) + b(a\(^{2}\) + b\(^{2}\) + c\(^{2}\) –ab – bc – ca)

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We will discuss here about the expansion of (a ± b)(a\(^{2}\) ∓ ab + b\(^{2}\)). (a + b)(a\(^{2}\) - ab + b\(^{2}\)) = a(a\(^{2}\) - ab + b\(^{2}\)) + b(a\(^{2}\) - ab + b\(^{2}\)) = a\(^{3}\) - a\(^{2}\)b + ab\(^{2}\) + ba\(^{2}\) - ab\(^{2}\) + b\(^{3}\) =

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We will discuss here about the expansion of (a ± b)\(^{3}\). (a + b)\(^{3}\) = (a + b) ∙ (a + b)\(^{2}\) = (a + b)(a\(^{2}\) + 2ab + b\(^{2}\)) = a(a\(^{2}\) + 2ab + b\(^{2}\)) + b(a\(^{2}\) + 2ab + b\(^{2}\))=a\(^{3}\)+2a\(^{2}\)b+ab\(^{2}\)+ba\(^{2}\)+2ab\(^{2}\)+b\(^{3}\)

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Here we will express a^2 + b^2 + c^2 – ab – bc – ca as sum of squares. If a, b, c are real numbers then (a – b)^2, (b – c)^2 and (c – a)^2 are positive as square of every real number is positive. So, a^2 + b^2 + c^2 – ab – bc – ca is always positive.

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Practice the questions given in the worksheet on simplification of (a + b)(a – b). 1. Simplify by applying standard formula. (i) (5x – 9)(5x + 9) (ii) (2x + 3y)(2x – 3y) (iii) (a + b – c)(a – b + c) (iv) (x + y – 3)(x + y + 3) (v) (1 + a)(1 – a)(1 + a^2)

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Practice the questions given in the worksheet on completing square. Write the following as a perfect square. (i) 4X^2 + 4X + 1 (ii) 9a^2 – 12ab + 4b^2 (iii) 1 + 6/a + 9/a^2 2. Indicate the perfect squares among the following. Express each of the perfect squares as the square

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Here we will learn how to completing a square.Problems on completing a square 1. What should be added to the polynomial 4m^2 + 8m so that it becomes perfect square? Solution: 4m^2 + 8m = (2m)^2 + 2 ∙ (2m) ∙ 2

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Practice the questions given in the worksheet on expansion of (x ± a)(x ± b). 1. (i) Find the product using standard formula. (i) (x + 2)(x + 5) (ii) (a – 4)(a – 7) (iii) (x + 1)(x – 8) (iv) (a – 3)(a + 2) (v) (3x + 1)(3x + 2) (vi) (4x – y)(4x + 2y) 2. Find the product.

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Practice the questions given in the worksheet on expanding of (a ± b ± c)^2 and its corollaries. 1. Expand the squares of the following trinomials. (i) a + 2b + 3c (ii) 2x + 3y + 4z (iii) x + 2y – 3z (iv) 3a – 4b – c (v) 1 – x - \(\frac{1}{x}\) (vi) 1 – a – a^2. 2. Simplify:

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Practice the questions given in the worksheet on expansion of (a ± b)^2 and its corollaries. 1. Expand the squares of the following: (i) 4x + y (ii) 5a + 3b (iii) 2x + \(\frac{1}{x}\) 2. Expand the following: (i) (x – 2y)^2 (ii) (3y – 2z)^2 (iii) (3x - \(\frac{1}{3x}\))^2

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We will discuss here about the Simplification of (a + b)(a – b). (a + b)(a – b) = a(a – b) + b(a – b) = a\(^{2}\) - ab + ba - b\(^{2}\) = a\(^{2}\) - b\(^{2}\) Thus, we have (a + b)(a - b) = a\(^{2}\) - b\(^{2}\) Solved Examples on Simplification of (a + b)(a – b) 1.

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We will discuss here about the expansion of (x ± a)(x ± b) (x + a)(x + b) = x(x + b) + a (x + b) = x^2 + xb + ax + ab = x^2 + (b + a)x + ab (x - a)(x - b) = x(x - b) - a (x - b) = x^2 - xb - ax + ab = x^2 - (b + a)x + ab (x + a)(x - b) = x(x - b) + a (x - b) = x^2 - xb

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We will discuss here about the expansion of (a ± b ± c)^2. (a + b + c)^2 = {a + (b + c)}^2 = a^2 + 2a(b + c) + (b + c)^2 = a^2 + 2ab + 2ac + b^2 + 2bc + c^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) = sum of squares of a, b, c + 2(sum of the products of a, b, c taking two at a ti

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Here we will prove that a rhombus is a parallelogram whose diagonals meet at right angles. Given: PQRS is a rhombus. So, by definition, PQ = QR = RD = SP. Its diagonals PR and QS intersect at O. To prove: (i) PQRS is a parallelogram. (ii) ∠POQ = ∠QOR = ∠ROS = ∠SOP = 90°.

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Here we will discuss about one of the important geometrical property of parallelogram. A quadrilateral is a parallelogram if one pair of opposite sides are equal and parallel Given: PQRS is a quadrilateral in which PQ = SR and PQ ∥ SR. To prove: PQRS is a parallelogram.

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Here we will discuss about a quadrilateral is a parallelogram if its diagonals bisect each other. Given: PQRS is a quadrilateral whose diagonals PR and QS bisect each other at O, i.e., OP = OR and OQ = OS. To prove: PQRS is a parallelogram. Proof: In ∆OPQ and ∆ORS, OP = OR

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Here we will discuss about the diagonals of a parallelogram bisect each other. In a parallelogram, diagonals bisect each other and each diagonal bisects the parallelogram into two congruent triangles. Given: PQRS is a parallelogram in which PQ ∥ SR and PS ∥ QR. Its diagonals

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A binomial is an algebraic expression which has exactly two terms, for example, a ± b. Its power is indicated by a superscript. For example, (a ± b)2 is a power of the binomial a ± b, the index being 2. A trinomial is an algebraic expression which has exactly three terms

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Here we will discuss about the opposite angles of a parallelogram are equal. In a parallelogram, each pair of opposite angles are equal. Given: PQRS is a parallelogram in which PQ ∥ SR and QR ∥ PS To prove: ∠P = ∠R and ∠Q = ∠S Construction: Join PR and QS. Proof: Statement:

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Here we will discuss about the opposite sides of a parallelogram are equal in length. In a parallelogram, each pair of opposite sides are of equal length. Given: PQRS is a parallelogram in which PQ ∥ SR and QR ∥ PS. To prove: PQ = SR and PS = QR. Construction: Join PR

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Here we will discuss about the concept of parallelogram. Quadrilateral: A rectilinear figure enclosed by four line segments is called a quadrilateral. In the adjoining figures, we have two quadrilaterals PQRS, each enclosed by four line segments PQ, QR, RS and SP which

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Here we will prove that if each diagonal of a quadrilateral divides it in two triangles of equal area then prove that the quadrilateral is a parallelogram. Solution: Given: PQRS is a quadrilateral whose diagonals PR and QS cut at O such that ar(∆PQR) = ar(∆PSR), and

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Here we will prove that the area of a rhombus is equal to half the product of its diagonals. Solution: Given: PQRS is a rhombus whose diagonals are PR and QS. The diagonals intersect at O. To prove: ar(rhombus PQRS) = 1/2 ×PR × QS. Statement ar(∆RSQ) = 1/2 ×Base × Altitude

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Here we will prove that the area of the triangle formed by joining the middle points of the sides of a triangle is equal to one-fourth area of the given triangle. Solution: Given: X, Y and Z are the middle points of sides QR, RP and PQ respectively of the triangle PQR.

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Here we will learn how to solve different types of problems on finding area of triangle and parallelogram. 1. In the figure, XQ ∥ SY, PS ∥ QR, XS ⊥ SY, QY ⊥ SY and QY = 3 cm. Find the areas of ∆MSR and parallelogram PQRS. Solution: ar(∆MSR) = 1/2 × ar(rectangle of SR

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Here we will prove that triangles with equal areas on the same base have equal corresponding altitudes (or are between the same parallels). Given: PQR and SQR are two triangles on the same base QR, and ar(∆PQR) = ar(∆SQC). Also, PN and SM are their corresponding altitudes.

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Here we will prove that triangles on the same base and between the same parallels are equal in area. Given: PQR and SQR are two triangles on the same base QR and are between the same parallel lines QR and MN, i.e., P and S are on MN. To prove: ar(∆PQR) = ar(∆SQR)

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Here we will prove that the area of a triangle is half that of a parallelogram on the same base and between the same parallels. Given: PQRS is a parallelogram and PQM is a triangle with the same base PQ, and are between the same parallel lines PQ and SR.

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Here we will prove that the area of a parallelogram is equal to that of a rectangle on the same base and of the same altitude, that is between the same parallel lines. Given: PQRS is a parallelogram and PQ MN is a rectangle on the same base PQ and between the same parallel

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Here we will prove that parallelogram on the same base and between the same parallel lines are equal in area. Given: PQRS and PQMN are two parallelograms on the same base PQ and between same parallel lines PQ and SM. To prove: ar(parallelogram PQRS) = ar(parallelogram PQMN).

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