NCERT Class 9 Solutions: Polynomials (Chapter 2) Exercise 2.4 – Part 2

Q-3 Find the value of k, if x1 is a factor of p(x) in each of the following cases:

  1. p(x)=x2+x+k

  2. p(x)=2x2+kx+2

  3. p(x)=kx22x+1

  4. p(x)=kx23x+k

Solution:

  • According to factor theorem, if x1 is a factor of p(x) , then p(1)=0

  1. p(x)=x2+x+k

    p(1)=12+1+k

    p(1)=0 Then,

    1+1+k=0

    2+k=0

    k=2

  2. p(x)=2x2+kx+2

    p(1)=2(1)2+k(1)+2

    p(1)=0 Then,

    2+k+2=0

    k=22

    k=(2+2)

  3. p(x)=kx22x+1

    p(1)=k(1)22(1)+1

    p(1)=0 Then,

    k2+1=0

    k=21

  4. p(x)=kx23x+k

    p(1)=k(1)23(1)+k

    p(1)=0 Then,

    k3+k=0

    2k=3

    k=32

Q-4 Factorize:

  1. 12x27x+1

  2. 2x2+7x+3

  3. 6x2+5x6

  4. 3x2x4

FInding the monomial factors of a polynomial

Facttoring a Polynomial Into Monomials

FInding the monomial factors of a polynomial

Solution:

  1. 12x27x+1

    =12x24x3x+1

    =4x(3x1)1(3x1)

    =(4x1)(3x1)

  2. 2x2+7x+3

    =2x2+6x+x+3

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

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

  3. 6x2+5x6

    =6x2+9x4x6

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

    =(2x+3)(3x2)

  4. 3x2x4

    =3x2+3x4x4

    =3x(x+1)4(x+1)

    =(x+1)(3x4)

Q-5 Factorize:

  1. x32x2x+2

  2. x33x29x5

  3. x3+13x2+32x+20

  4. 2y3+y22y1

Solution:

Image describing the factorization of a polynomial.

Finding Factors of a Polynomial

Image describing the factorization of a polynomial.

  1. p(x)=x32x2x+2

    Let us guess a factor (xa) and choose value of a arbitrarily as 1.

    Now, putting this value in p(x)

    1+21+2=0

    So, (x1) is a factor of p(x)

    Now, x32x2x+2=x3x2+x2x+2

    =x2(x1)x(x1)2(x1)

    =(x1)(x2x2)

    =(x1)(x22x+x2)

    =(x1)(x(x2)+1(x2))

    =(x1)(x2)(x+1)

    So, x32x2x+2=(x1)(x2)(x+1)

  2. p(x)=x33x29x5

    Let us guess a factor (xa) . A should be a factor of 5

    Therefore, ±1or±5

    For (x1),x=1

    p(1)=(1)33(1)29(1)5

    p(1)=1395

    p(1)=16

    So, (x1) is not a factor of p(x)

    For (x5),x=5

    p(5)=(5)33(5)29(5)5

    p(5)=12575455

    p(5)=0

    So, (x5) is a factor of p(x)

    Now, x33x29x5=x35x2+2x210x+x5

    =x2(x5)+2x(x5)+1(x5)

    =(x5)(x2+2x+1)

    =(x5)(x+1)2

    =(x5)(x+1)(x+1)

    So, x33x29x5=(x5)(x+1)(x+1)

  3. p(x)=x3+13x2+32x+20

    Let us guess a factor (xa) . A should be a factor of 20

    Therefore, ±1,±2,±4,±5,±10

    For (x1),x=1

    p(1)=(1)3+13(1)2+32(1)+20

    p(1)=1+13+32+20

    p(1)=660

    So, (x1) is not a factor of p(x)

    For (x+1),x=1

    p(1)=(1)3+13(1)2+32(1)+20

    p(1)=1+1332+20

    p(1)=33+33=0

    So, (x+1) is a factor of p(x)

    Now, x3+13x2+32x+20=x3+x2+12x2+20x+20

    =x2(x+1)+12x(x+1)+20(x+1)

    =(x+1)(x2+12x+20)

    =(x+1)(x2+10x+2x+20)

    =(x+1)(x(x+10)+2(x+10))

    So, x3+13x2+32x+20=(x+2)(x+1)(x+10)

  4. p(x)=2y3+y22y1

    Factors of 2 are ±1,±2

    For (y1),x=1

    p(1)=2(1)3+(1)22(1)1

    p(1)=2+121

    p(1)=0

    So, (y1) is not a factor of p(y)

    Now, 2y3+y22y1=2y32y2+3y23y+y1

    =2y2(y1)+3y(y1)+1(y1)

    =(y1)(2y2+3y+1)

    =(y1)(2y2+2y+y+1)

    =(y1)(2y(y+1)+1(y+1))

    So, 2y3+y22y1=(y1)(2y+1)(y+1)

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