NCERT Class 9 Solutions: Polynomials (Chapter 2) Exercise 2.4 – Part 1

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Q-1 Determine which of the following polynomials has (x+1) a factor:

  1. x3+x2+x+1

  2. x4+x3+x2+x+1

  3. x4+3x3+3x2+x+1

  4. x3x2(2+2)x+2

Solution:

Understanding relation between polynomial and its roots

Polynomials and Roots

Understanding relation between polynomial and its roots

If (x+1) is a factor of a polynomial p(x) , then we can write, p(x)=(x+1)g(x). Since zero multiplied by any number is zero. Therefore, this polynomial (x+1)g(x) would become zero for all the values of x where either (x+1) becomes zero or g(x) becomes zero. We don’t know anything about g(x) but we know that (x+1) becomes zero for x=1 . Therefore, p(1)=(1+1)g(1)=0. Therefore, if x + 1 is a factor of p(x), p(x) must become zero at x = -1. For each polynomial above we would test its value at x = -1. If it comes out to be zero (x+1) is a factor of that polynomial.

  1. Let p(x)=x3+x2+x+1

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

    p(1)=1+11+1

    p(1)=0

    (x+1) Is a factor of polynomial x3+x2+x+1

  2. Let p(x)=x4+x3+x2+x+1

    p(1)=(1)4+(1)3+(1)2+(1)+1

    p(1)=11+11+1

    p(1)=10

    (x+1) Is not a factor of polynomial x4+x3+x2+x+1

  3. Let p(x)=x4+3x3+3x2+x+1

    p(1)=(1)4+3(1)3+3(1)2+(1)+1

    p(1)=1+3(1)+3(1)+(1)+1

    p(1)=13+31+1

    p(1)=10

    (x+1) Is not a factor of polynomial x4+3x3+3x2+x+1

  4. Let p(x)=x3x2(2+2)x+2

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

    p(1)=11+2+2+2

    p(1)=220

    (x+1) Is not a factor of polynomial x3x2(2+2)x+2

Q-2 Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:

  1. p(x)=2x3+x22x1,g(x)=x+1

  2. p(x)=x3+3x2+3x+1,g(x)=x+2

  3. p(x)=x34x2+x+6,g(x)=x3

Solution:

  1. p(x)=2x3+x22x1,g(x)=x+1

    Zero of x+1 is 1 .

    If g(x) is a factor of p(x) then p must also have a zero at -1, i.e. p(1)=0

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

    p(1)=2+1+21

    p(1)=0

    g(x) Is a factor of polynomial p(x) .

  2. p(x)=x3+3x2+3x+1,g(x)=x+2

    Zero of x+2 is 2 .

    If g(x) is a factor of p(x) then p must also have a zero at -2, i.e. p(2)=0

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

    p(2)=8+126+1

    p(2)=10

    g(x) Is not a factor of polynomial p(x) .

  3. p(x)=x34x2+x+6,g(x)=x3

    Zero of x3 is 3 .

    If g(x) is a factor of p(x) then p must also have a zero at 3, i.e. p(3)=0

    p(3)=(3)34(3)2+3+6

    p(3)2736+3+6

    p(3)=0

    g(x) Is a factor of polynomial p(x) .

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