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

Q-4 Expand each of the following using suitable identities:

  1. (x+2y+4z)2

  2. (2xy+z)2

  3. (2x+3y+2z)2

  4. (3a7bc)2

  5. (2x+5y3z)2

  6. [14a12b+1]2

Solution:

Mathmatical identities for squares of trinomials

Trinomial Square Formulas

Mathmatical identities for squares of trinomials

  1. (x+2y+4z)2

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

    Here a=x,b=2yandc=4z

    =x2+(2y)2+(4z)2+2(x)(2y)+2(2y)(4z)+2(4z)(x)

    =x2+4y2+16z2+4xy+16yz+8zx

  2. (2xy+z)2

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

    Here a=2x,b=yandc=z

    =(2x)2+(y)2+(z)2+2(2x)(y)+2(y)(z)+2(z)(2x)

    =4x2+y2+z24xy2yz+4zx

  3. (2x+3y+2z)2

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

    Here a=2x,b=3yandc=2z

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

    =4x2+9y2+4z212xy+12yz8zx

  4. (3a7bc)2

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

    Here a=3a,b=7bandc=c

    =(3a)2+(7b)2+(c)2+2(3a)(7b)+2(7b)(c)+2(c)(3a)

    =9a2+49b2+c242ab+14bc6ac

  5. (2x+5y3z)2

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

    Here a=2x,b=5yandc=3z

    =(2x)2+(5y)2+(3z)2+2(2x)(5y)+2(5y)(3z)+2(3z)(2x)

    =4x2+25y2+9z220xy30yz+12zx

  6. [14a12b+1]2

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

    Here a=14a,b=12bandc=1

    =(14a)2+(12b)2+(1)2+2(14a)(12b)+2(12b)(1)+2(1)(14a)

    =116a2+14b2+114abb+12a

    =a216+b24+1ab4b+a2

Q-5 Factorize:

  1. 4x2+9y2+16z2+12xy24yz16xz

  2. 2x2+y2+8z2+22xy+42yz8xy

Solution:

  1. 4x2+9y2+16z2+12xy24yz16xz

    =(2x)2+(3y)2+(4z)2+2(2x)(3y)2(3y)(4z)2(2x)(4z)

    =(2x)2+(3y)2+(4z)2+2(2x)(3y)2(3y)(4z)2(2x)(4z)

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

    =(2x+3y4z)2

    =(2x+3y4z)(2x+3y4z)

  2. 2x2+y2+8z2+22xy+42yz8xy

    =(2x)2+(y)2+(22z)2+2(2x)(y)2(y)(22z)2(2x)(22z)

    =(2x)2+(y)2+(22z)2+2(2x)(y)2(y)(22z)2(2x)(22z)

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

    =(2x+y+22z)2

    =(2x+y+22z)(2x+y+22z)

Q-6 Write the following cubes in expanded form:

  1. (2x+1)3

  2. (2a3b)3

  3. [32x+1]3

  4. [x23y]3

Solution:

Important identities with cubes

Most Important Identities Involving Cubes

Important identities with cubes

  1. (2x+1)3

    Using identify (x+y)3=x3+y33xy(x+y)

    =(2x)3+(1)3+3(2x)(1)(2x+1)

    =8x3+1+6x(2x+1)

    =8x3+1+12x2+6x

  2. (2a3b)3

    Using identify (xy)3=x3y33xy(xy)

    =(2a)3+(3b)3+3(2a)(3b)(2a3b)

    =8a327b318ab(2a3b)

    =8a327b336a2b+54ab2

  3. [32x+1]3

    Using identify (x+y)3=x3+y33xy(x+y)

    =(3x2)3+(1)3+3(3x2)(1)(3x2+1)

    =27x38+1+9x2(3x2+1)

    =27x38+27x249x2+1

  4. [x23y]3

    Using identify (xy)3=x3y33xy(xy)

    =x3+(2y3)3+3(x)(2y3)(x2y3)

    =x38y3272xy(x2y3)

    =x38y3272x2y+4xy23

Q-7 Evaluate the following using suitable identities:

  1. (99)3

  2. (102)3

  3. (998)3

Geometric interpretation of a + b cubed

Illustrating a + B Cubed

Geometric interpretation of a + b cubed

Solution:

  1. (99)3

    =(1001)3

    Using identity (xy)3=x3y33xy(xy)

    =(100)3+(1)33(100)(1)(1001)

    =10000001300(99)

    =1000000129700

    =970299

  2. (102)3

    =(100+2)3

    Using identity (x+y)3=x3+y3+3xy(x+y)

    =(100)3+(2)33(100)(2)(100+2)

    =1000000+8+600(102)

    =1000000+8+61200

    =1061208

  3. (998)3

    =(10002)3

    Using identity (xy)3=x3y33xy(xy)

    =(1000)3+(2)33(1000)(2)(10002)

    =100000000086000(998)

    =100000000085988000

    =10000000005988008

    =994011992

Q-8 Factorize each of the following:

  1. 8a3+b3+12a2b+6ab2

  2. 8a3b312a2b+6ab2

  3. 27125a3135a+225a2

  4. 64a327b3144a2b+108ab2

  5. 27p3121692p2+14p

Solution:

  1. 8a3+b3+12a2b+6ab2

    =(2a)3+b3+3(2a)2b+3(2a)b2

    (x+y)3=x3+3x2y+3xy2+y3

    =(2a+b)3

  2. 8a3b312a2b+6ab2

    =(2a)3b33(2a)2b+3(2a)b2

    (xy)3=x33x2y+3xy2y3

    =(2ab)3

  3. 27125a3135a+225a2

    =33(5a)33(3)3(5a)+3(3)(5a)2

    (xy)3=x33x2y+3xy2y3

    =(35a)3

  4. 64a327b3144a2b+108ab2

    =(4a)3(3b)33(4a)2(3b)+3(4a)(3b)2

    (xy)3=x33x2y+3xy2y3

    =(4a3b)3

  5. 27p3121692p2+14p

    =(3p)3(16)33(3p)2(16)+3(3p)(16)2

    (xy)3=x33x2y+3xy2y3

    =(3p16)3

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