NCERT Class 9 Solutions: Polynomials (Chapter 2) Exercise 2.5 – Part 1

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Q-1 Use suitable identities to find the following products:

  1. (x+4)(x+10)

  2. (x+8)(x10)

  3. (3x+4)(3x5)

  4. (y2+32)(y232)

  5. (32x)(3+2x)

Solution:

Most common algebraic identities

Common Algebraic Identities

Most common algebraic identities

  1. Using identify (x+a)(x+b)=x2+(a+b)x+ab

    (x+4)(x+10)=x2+(4+10)x+(4×10)

    (x+4)(x+10)=x2+14x+40

  2. Using identify (x+a)(x+b)=x2+(a+b)x+ab

    (x+8)(x10)=x2+(810)x+(8×(10))

    (x+8)(x10)=x22x80

  3. Using identify (x+a)(x+b)=x2+(a+b)x+ab

    (3x+4)(3x5)=9x2+(15+12)x+(4×(5))

    (3x+4)(3x5)=9x23x20

  4. Using identify (x+y)(xy)=x2y2

    (y2+32)(y232)=(y2)2+(32)2

    (y2+32)(y232)=y494

  5. Using identify (x+y)(xy)=x2y2

    (32x)(3+2x)=32(2x)2

    (32x)(3+2x)=94x2

Q-2 Evaluate the following products without multiplying directly:

  1. 103×107

  2. 95×96

  3. 104×96

Solution:

  1. 103×107

    =(100+3)×(100+7)

    =(100)2×(3+7)×100+(3×7)

    =10000×1000+21

    =11021

  2. 95×96

    =(1005)×(1004)

    =(100)2(5+4)×100+(5×4)

    =10000900+20

    =9120

  3. 104×96

    =(100+4)×(1004)

    =(100)2(4)2

    =1000016

    =9984

Q-3 Factories the following using appropriate identities:

  1. 9x2+6xy+y2

  2. 4y24y+1

  3. x2y2100

Understanding the geometric proof of (a + b) squared

Proof of (A + B) Squared

Understanding the geometric proof of (a + b) squared

(a+b)²=a²+ab+ba+b²

Understanding the geometric proof of (a - b) squared

Proof of (A Minus B) Squared

Understanding the geometric proof of (a - b) squared

(ab)²=(a²+b²)2ab

Understanding the geometric proof of a squared - b squared

Geometric Proof of a Squared - B Squared

Understanding the geometric proof of a squared - b squared

a²b²=(a+b)(ab)

Solution:

  1. 9x2+6xy+y2

    =(3x)2+2(3x)(y)+(y)2

    =(3x+y)2 (Applying a2+2ab+b2=(a+b)2 )

    =(3x+y)(3x+y)

  2. 4y24y+1

    =(2y)22(2y)(1)+(1)2

    =(2y1)2 (Applying a22ab+b2=(ab)2 )

    =(2y1)(2y1)

  3. x2y2100

    =x2(y10)2

    =(x+y10)(xy10) (Applying a2b2=(ab)(a+b) )

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