Tag: Exercise 2.4 Chapter 2 Polynomials NCERT Maths Solutions

  • Exercise-2.4, Class 9th, Maths, Chapter 2, NCERT

    1. Use suitable identities to find the following products:

    (i) (x+4)(x+10)
    Answer:   x2+14x+40

    (ii) (x+8)(x10)
    Answer:   x22x80

    (iii) (3x+4)(3x5)
    Answer:   9x23x20

    (iv) (y2+32)(y232)
    Answer:   y494

    (v) (32x)(3+2x)
    Answer:   94x2


    2. Evaluate the following products without multiplying directly:

    (i) 103×107
    Write as (100+3)(100+7)=1002+(3+7)100+37
    Answer: 11021

    (ii) 95×96
    Write as (1005)(1004)=1002(5+4)100+54
    Answer: 9120

    (iii) 104×96
    Write as (100+4)(1004)=100242
    Answer: 9984


    3. Factorize using appropriate identities:

    (i) 9x2+6xy+y2
    This is (3x+y)2

    (ii) 4y24y+1
    This is (2y1)2

    (iii) x2y2100
    Write as x2(y10)2=(xy10)(x+y10)


    4. Expand each using suitable identities:

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

    (i) (x+2y+4z)2=x2+4y2+16z2+4xy+16yz+8xz

    (ii) (2xy+z)2=4x2+y2+z24xy2yz+4xz

    (iii) (2x+3y+2z)2=4x2+9y2+4z212xy+12yz8xz

    (iv) (3a7bc)2=9a2+49b2+c242ab+14bc6ca

    (v) (2x+5y3z)2=4x2+25y2+9z220xy30yz+12xz

    (vi) (14a12b+1)2=116a2+14b2+114abb+12a


    5. Factorise:

    (i) 4x2+9y2+16z2+12xy24yz16xz
    Recognize as (2x+3y4z)2

    (ii) 2x2+y2+8z222xy+42yz8xz
    Write as (2x+y22z)2


    6. Write the following cubes in expanded form:

    Use (x+y)3=x3+y3+3xy(x+y) and (xy)3=x3y33xy(xy)

    (i) (2x+1)3=8x3+12x2+6x+1

    (ii) (2a3b)3=8a327b336a2b+54ab2

    (iii) (32x+1)3=278x3+274x2+92x+1

    (iv) (x23y)3=x3827y32x2y+43xy2


    7. Evaluate the following using identities:

    (i) 993=(1001)3=10000001300(99)=970299

    (ii) 1023=(100+2)3=1000000+8+600(102)=1061208

    (iii) 9983=(10002)3=100000000086000(998)=994011992


    8. Factorise each of the following (cubic-type factorizations):

    (i) 8a3+b3+12a2b+6ab2=(2a+b)3

    (ii) 8a3b312a2b+6ab2=(2ab)3

    (iii) 27125a3135a+225a2=(35a)3

    (iv) 64a327b3144a2b+108ab2=(4a3b)3

    (v) 27p3121692p2+14p=(3p16)3


    9. Verify identities

    (i) x3+y3=(x+y)(x2xy+y2). Verified by expanding RHS.

    (ii) x3y3=(xy)(x2+xy+y2)

    Verified by expanding RHS.

    (standard verifications using cube identities.)


    10. Factorise sums/differences of cubes:

    (i) 27y3+125z3=(3y+5z)(9y215yz+25z2)

    (ii) 64m3343n3=(4m7n)(16m2+28mn+49n2)


    11. Factorise

    27x3+y3+z39xyz

    Write as (3x)3+y3+z33(3x)yz. Using the identity x3+y3+z33xyz=(x+y+z)(x2+y2+z2xyyzzx) we get

    (3x+y+z)(9x2+y2+z23xyyz3xz)


    12. Verify

    x3+y3+z33xyz=12(x+y+z)[(xy)2+(yz)2+(zx)2]

    Verified by expanding RHS (standard identity; equivalently derived from (x+y+z)(x2+y2+z2xyyzzx)


    13. If x+y+z=0, show x3+y3+z3=3xyz

    From identity x3+y3+z33xyz=(x+y+z)(x2+y2+z2xyyzzx)

    If x+y+z=0 RHS is 0, hence x3+y3+z3=3xyz


    14. Without actually calculating the cubes, find:

    (i) (12)3+73+53. Since 12+7+5=0, use result of Q13: value =3(12)(7)(5)=1260

    (ii) 283+(15)3+(13)3. Since 281513=0, value =328(15)(13)=16380


    15. Give possible expressions for length and breadth of rectangles whose areas are:

    (i) Area =25a235a+12. Factorize:

    25a235a+12=(5a3)(5a4)

    Possible dimensions: 5a4 and 5a3.

    (ii) Area =35y2+13y12. Factorize:

    35y2+13y12=(7y3)(5y+4)

    Possible dimensions: 7y3 and 5y+4.


    16. Possible expressions for dimensions of cuboids with given volumes:

    (i) Volume =3x212x=3x(x4). Possible dimensions: 3,  x,  x4

    (ii) Volume =12ky2+8ky20k=4k(3y2+2y5)=4k(3y+5)(y1). Possible dimensions: 4k,  3y+5,  y1