CBSE Class 9 (Maths) - Polynomials - Exercise 2.4

Factor Theorem

1. Let p(x) be a polynomial and a be any real number. If p(a) = 0, then (x–a) is a factor of p(x).

2. If (x–a) is a factor of p(x), then p(a) = 0

EXERCISE 2.4 (NCERT Solution)

Q1: Determine which of the following polynomials has (x + 1) a factor :

(i) x³ + x² + x + 1

(ii) x⁴ + x³ + x² + x + 1

(iii) x⁴ + 3x³ + 3x² + x + 1

(iv) x³ – x² – (2 + √2)x + √2

Answer: If (x+1) is a factor of a polynomial p(x), then p(-1) must be equal to zero.

(i) Let p(x) = x³ + x² + x + 1

    p(-1) = (-1)³ + (-1)² + (-1) + 1  = -1 + 1 -1 + 1 = 0

∴ (x-1) is a factor of polynomial x³ + x² + x + 1

(ii)  Let p(x) = x⁴ + x³ + x² + x + 1

       p(-1) =  (-1)⁴ + (-1)³ + (-1)² + (-1) + 1 = 1 - 1 + 1 -1 + 1 = 1 ≠ 0

∴ (x-1) is not a factor of this polynomial.

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

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

                = 1 + 3(-1) + 3(1) -1 + 1 = 1 - 3 + 3 - 1 = 1 ≠ 0

∴ (x-1) is not a factor of this polynomial.

(iv) Let p(x) = x³ – x² – (2 + √2)x + √2

       p(-1) = (-1)³ – (-1)² – (2 + √2)(-1) + √2

                = -1  - 1 + 2 +  √2 + √2 =  2√2  ≠ 0

∴ (x-1) is not a factor of this polynomial.

Q2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:

(i)  p(x) = 2x³ + x² – 2x – 1, g(x) = x + 1

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

(iii) p(x) = x³ – 4x² + x + 6, g(x) = x – 3

Answer:

(i)  p(x) = 2x³ + x² – 2x – 1, g(x) = x + 1

Zero of x+1 is -1. If g(x) is a factor of p(x) then p(-1) = 0

⇒ p(-1) =  2(-1)³ + (-1)² – 2(-1) – 1  = -2 + 1 + 2 - 1 = 0

∴ g(x) is a factor of  polynomial p(x).

(ii) 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(-2) = 0

⇒ p(-2)  = (-2)³ + 3(-2)² + 3(-2) + 1  = -8 + 12 - 6 + 1 = -1 ≠ 0

∴ g(x) is not a factor of  polynomial p(x).

(iii)  p(x) = x³ – 4x² + x + 6, g(x) = x – 3

Root of x-2 is 3.  If g(x) is a factor of p(x) then p(3) = 0

⇒ p(3) = (3)³ – 4(3)² + 3 + 6  = 27 - 4(9) + 3 + 6 = 27 - 36  + 3 + 6 = 0

∴ g(x) is a factor of  polynomial p(x).

Q3: Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:

(i) p(x) = x² + x + k

(ii) p(x) = 2x² + kx + √2

(iii) p(x) = kx² – √2x + 1

(iv) p(x) = kx² – 3x + k

Answer:

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

(i) p(x) = x² + x + k

⇒  p(1) = (1)² + 1 + k = 0

⇒  1 + 1 + k = 0

⇒ k = -2

(ii) p(x) = 2x² + kx + √2

⇒  p(1) = 0

⇒  p(1) = 2(1)² + k(1) + √2 = 0

⇒  2 + k + √2 = 0

⇒ k = -2 - √2  = -(2 +√2)

(iii) p(x) = kx² – √2x + 1

⇒  p(1) = 0

⇒  p(1) = k(1)² – √2(1) + 1 = 0

⇒ k - √2 + 1

⇒  k = √2 - 1

(iv) p(x) = kx² – 3x + k

⇒  p(1) = 0

⇒  p(1) = k(1)² – 3(1) + k = 0

⇒   k - 3 + k = 0

⇒ 2k = 3

⇒ k = 3/2

Q4. Factorise:

(i) 12x² – 7x + 1

(ii) 2x² + 7x + 3

(iii) 6x² + 5x – 6

(iv) 3x² – x – 4

Answer:

(i) 12x² – 7x + 1 

Method I: By splitting method, let us find out two number p and q such that pq = 12 × 1

and p + q = -7

i.e. p = -4  and q = -3

⇒  = 12x² – 4x -3x + 1

     =  4x(3x - 1) - 1(3x -1)

     =  (4x -1)(3x -1)                              ...(answer)

Method II:  By factor theorem.

(ii) 2x² + 7x + 3

Let us find out two number p and q such that pq = 2 × 3 = 6 and p + q = 7

i.e. p = 6 and q = 1

⇒ = 2x² + 6x + x + 3 

    = 2x (x + 3) + 1(x + 3)

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

(iii) 6x² + 5x – 6

Let us find out two number p and q such that pq = 6 × -6 = -36 and p + q = 5

i.e. p = 9 and q = -4

∴  6x² + 5x – 6 = 6x² + 9x -4x – 6 

    = 3x(2x + 3) - 2(2x + 3)

     = (2x + 3)(3x - 2)                                          ... answer

(iv) 3x² – x – 4

Let us find out two number p and q such that pq = 3 × -4 = -12 and p + q = -1

i.e. p = 3 and q = -4

∴ 3x² – x – 4 = 3x² + 3x –4x – 4

   = 3x(x + 1) -4(x+1)

   = (x+1)(3x - 4)                                                  ... answer

Q5. Factorise:

(i) x³ – 2x² – x + 2

(ii) x³ – 3x²  – 9x – 5

(iii) x³ + 13x² + 32x + 20

(iv) 2y³ + y² – 2y – 1

Answer:

(i) Let p(x) = x³ – 2x² – x + 2

Here constant term is 2.  Possible factors of 2 are: ±1, ±2

By trial method, p(2) =  (2)³ – 2(2)² – 2 + 2 = 8 - 8 - 2 + 2 = 0

∴ (x -2 ) is factor of p(x).

P(x) ÷ (x -2) =

            x2 - 1 

           ▁▁▁▁▁▁▁▁▁▁▁▁▁
     x -2 ) x3 – 2x2 – x + 2
            x3 - 2x3 

             -   +
           ▔▔▔▔▔▔▔▔ 
                    -x + 2 
                    -x + 2
                     +  -
                   ▔▔▔▔▔
                      0

Since, Dividend = Divisor × Quotient + Remainder

∴ x³ – 2x² – x + 2 = (x-2)(x² – 1) + 0

  = (x - 2)( x² -x + x– 1)

  = (x -2) [x(x-1) + 1 (x-1)]

  = (x - 2) (x-1)(x+1)

(ii) Let f(x) = x³ – 3x²  – 9x – 5

Here constant is 5, Possible factors of 5 are ±1 and ±5

By trial method, p(5) =  (5)³ – 3(5)²  – 9(5) – 5 = 125 - 75 - 45 - 5 = 0

∴ (x - 5) is factor of polynomial f(x).

Let us find out quotient = f(x) ÷ (x -5)

             x2 + 2x + 1
           ▁▁▁▁▁▁▁▁▁▁
     x - 5) x3 – 3x2  – 9x – 5
             x3 - 5x2 

            -   +

            ▔▔▔▔▔▔

             0 + 2x2 - 9x - 5

                 2x2 - 10x 

                 -   +

                 ▔▔▔▔▔▔▔▔

                        x - 5

                        x - 5

                       ▔▔▔▔▔

                          0

Since, Dividend = Divisor × Quotient + Remainder

∴ x³ – 3x²  – 9x – 5 = (x - 5)(x² + 2x + 1)

Applying splitting method,

  = (x -5)( x² + x + x + 1)

  = (x -5)[x(x+1) +1(x+1)]

 = (x-5)(x+1)(x+1)

 = (x-5)(x+1)²

Method II:

x³ – 3x²  – 9x – 5 = x³ – 5x² + 2x²  – 9x – 5  

= x² (x – 5) + 2x²  – 10x + x – 5 

=  x² (x – 5) + 2x(x - 5) + 1 (x -5)

= (x - 5)( x² + 2x + 1)

= (x -5)[x(x+1) +1(x+1)]

 = (x-5)(x+1)(x+1)

 = (x-5)(x+1)²

(iii) Let p(x) = x³ + 13x² + 32x + 20

 Here constant term is 20. Its factors are: ±1, ±2, ±4, ±5 etc. 

 By trial method, p(-2) = (-2)³ + 13(-2)² + 32(-2) + 20

    = -8 + 52 -64 + 20 = 0

 ∴ x + 2 is a factor of polynomial p(x).

⇒ x³ + 13x² + 32x + 20 = x³ + 2x² + 11x²+ 22x + 10x + 20

   =   x²(x + 2) + 11x(x + 2) + 10(x+2)

 Taking (x + 2) common, we get

   = (x + 2)( x²+ 11x + 10)

(Note: you can use long division method also).

  = (x + 2)(x²+ x + 10x + 10)

  = (x+2)[x(x+1) + 10(x+1)]

  = (x+2)(x+10)(x+1)

(iv) Let p(y) = 2y³ + y² – 2y – 1

Here the constant term is 1. Possible factors are: ±1
By trial method, p(1) = 2(1)³ + (1)² – 2(1) – 1 = 2 + 1 - 2 - 1 = 0
So (y-1) is a factor of p(y)

Quotient = p(y) ÷ (y-1)

           2y2 + 3y + 1

         _____________________
  y -1  ) 2y3 + y2 – 2y – 1
           2y3 -2y2
          -    +
          ▔▔▔▔▔▔▔▔▔▔▔▔

               3y2 - 2y - 1

               3y2 - 3y 

                -    +

             ▔▔▔▔▔▔▔▔▔▔▔▔

                     y  - 1

                     y  - 1
                    -   +
             ▔▔▔▔▔▔▔▔▔▔▔▔
                           0

Since, Dividend = Divisor × Quotient + Remainder
2y³ + y² – 2y – 1 = (y -1)(2y² + 3y + 1)
= (y-1)(2y² + 2y + y + 1)
= (y-1)(2y(y+1) + 1(y + 1))
= (y-1)(y+1)(2y + 1)

Criterion to check if (x - 1) is a factor of polynomial p(x) = ax³ + bx² + cx + d, where a,b,c,d ∈ R, a≠0

Applying remainder theorem, if (x - 1) is factor of p(x) then p(1) = 0.
⇒ p(x) = ax³ + bx² + cx + d
⇒ p(1) = a(1)³ + b(1)² + c(1) + d = 0
⇒  a + b + c + d = 0
We can say that,

❝(x - 1) is a factor of polynomial p(x), if the sum of all the coefficients of polynomial p(x) is zero.❞

(✏ By by applying similar logic, can you find criterion to check if (x + 1) is a factor?)

Q6:Check if the polynomial 5x⁴ - 4x³ - 2x + 1 has (x - 1) its factor.

Answer: Adding the co-efficients i.e. 5 + (-4) + (-2) + 1 = 6 - 6 = 0
⇒ (x - 1) is a factor of the given polynomial.