NCERT Solutions Class 9 Mathematics

Exercise 2.4

Q.1. 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 + $\sqrt{2}$)x + $\sqrt{2}$

Q.2. 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 - 4 x2 + x + 6, g(x) = x - 3

Q.3. 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 +  $\sqrt{2}$ (iii) p(x) = kx2 - √2x + 1 (iv) p(x) = kx2 - 3x + k

Q.4. Factorise:

 (i) 12x2 + 7x + 1 (ii) 2x2 + 7x + 3 (iii) 6x2 + 5x - 6 (iv) 3x2 - x - 4

Q.5. Factorise:

 (i) x3 - 2x2 - x + 2 (ii) x3 - 3x2 - 9x - 5 (iii) x3 + 13x2 + 32x + 20 (iv) 2y3 + y2 - 2y - 1

NCERT Solutions Class 9 Mathematics

Exercise 2.4

Q.1. 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 + $\sqrt{2}$)x + $\sqrt{2}$

Solution:

If (x + 1) is a factor of p(x), p(-1) must be zero.

(i) p(x) = x3 + x2 + x + 1

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

= -1 + 1 - 1 + 1 = 0

Therefore, x + 1 is a factor of x3 + x2 + x + 1.

(ii) p(x) = x4 + x3 + x2 + x + 1

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

= 1 - 1 + 1 - 1 + 1 = 1≠ 0

Therefore, x + 1 is not a factor of x4 + x3 + x2 + x + 1.

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

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

= 1 - 3 + 3 - 1 + 1 = 1≠ 0

Therefore, + 1 is not a factor of x4 + 3x3 + 3x2 + x + 1.

(iv) p(x) = x3 - x2 - (2 + $\sqrt{2}$)x + $\sqrt{2}$

p(-1) =  (-1)3 -  (-1)2 -  (2 + $\sqrt{2}$) (-1) + $\sqrt{2}$

= -1 - 1 + 2 + $\sqrt{2}$ + $\sqrt{2}$

=2$\sqrt{2}$ ≠ 0

Therefore, x + 1 is not a factor of x3 - x2 - (2 + $\sqrt{2}$)x + $\sqrt{2}$.

Q.2. 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 - 4 x2 + x + 6, g(x) = x - 3

Solution:

If g(x) = (x - a) is a factor of p(x), p(a) must be zero.

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

Put g(x) = x + 1 = 0 ⇒ x = -1

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

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

Hence, g(x) = x + 1 is a factor of 2x3 + x2 - 2x - 1.

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

Put g(x) = x + 2= 0 ⇒ x = -2

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

= -8 + 12 - 6 + 1

= -1 ≠ 0

Hence g(x) = x + 2 is not a factor of x3 +3x2 + 3x + 1.

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

Put g(x) = x - 3 = 0 ⇒ x = 3

p(3) = (3)3 - 4(3)2 + 3 + 6

= 27 - 36 + 9 = 0

Therefore, g(x) = x - 3 is a factor of x3 - 4x2 + x + 6.

Q.3. 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 +  $\sqrt{2}$ (iii) p(x) = kx2 - √2x + 1 (iv) p(x) = kx2 - 3x + k

Solution:

If x - 1 is a factor of polynomial p(x), then p(1) should be 0.

(i) p(x) = x2 + x + k

⇒ (1)2 + 1 + k = 0

⇒ 2 + k = 0

⇒ k = - 2

Therefore, value of k is -2.

(ii) p(x) = 2x2 + kx +  $\sqrt{2}$

⇒ 2(1)2 + k(1) +  = 0

⇒ 2 + k + $\sqrt{2}$ = 0

⇒ k = -2 - $\sqrt{2}$= -(2 + $\sqrt{2}$)

Therefore, value of k is -(2 + $\sqrt{2}$).

(iii) p(x) = kx2 - $\sqrt{2}$x + 1

k(1)2 - $\sqrt{2}$(1) + 1 = 0

⇒ k - $\sqrt{2}$ + 1 = 0

⇒ k = $\sqrt{2}$ - 1

Therefore, value of k is $\sqrt{2}$ - 1.

(iv) p(x) = kx2 - 3x + k

k(1)2 + 3(1) + k = 0

⇒ k - 3 + k = 0

⇒ 2k - 3 = 0

⇒ k = $\frac{3}{2}$

Therefore, value of k is $\frac{3}{2}$.

Q.4. Factorise:

 (i) 12x2 + 7x + 1 (ii) 2x2 + 7x + 3 (iii) 6x2 + 5x - 6 (iv) 3x2 - x - 4

Solution:

Using midterm splitting for all the cases, we have,

(i) 12x2 + 7x + 1

=12x2 - 4x - 3x+ 1

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

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

(ii) 2x2 + 7x + 3

= 2x2 + 6x + x + 3

= 2x (x + 3) + 1 (x + 3)

= (x + 3) (2x + 1)

(iii) 6x2 + 5x - 6

= 6x2 + 9x - 4x - 6

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

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

(iv) 3x2 - x - 4

= 3x2 - 4x + 3x - 4

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

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

Q.5. Factorise:

 (i) x3 - 2x2 - x + 2 (ii) x3 - 3x2 - 9x - 5 (iii) x3 + 13x2 + 32x + 20 (iv) 2y3 + y2 - 2y - 1

Solution:

Using trial and error with factor theorem.

(i) Let p(x) = x3 - 2x2 - x + 2

Factors of 2 are ±1 and ± 2

By trial and error method, we find that,

p(-1) = (-1)3 - 2(-1)2 - (-1) + 2

= -1 -2 + 1 + 2 = 0

Therefore, (x+1) is a factor of p(x)

Now, we can write, using long division,

x3 - 2x2 - x + 2 =(x+1) (x2 - 3x + 2)

= (x+1) (x2 - x - 2x + 2) [by midterm splitting]

= (x+1) {x(x-1) -2(x-1)}

= (x+1) (x-1) (x+2)

(ii) Let p(x) = x3 - 3x2 - 9x - 5

Factors of 5 are ±1 and ±5

By trial and error method, we find that,

p(5) = (5)3 - 3(5)2 - 9(5) - 5

= 125 - 75 - 45 - 5 = 0

Therefore, (x-5) is the factor of p(x)

Now, we can write, using long division,

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

= (x-5) (x2 + x + x + 1) [by midterm splitting]

= (x-5) {x(x + 1) +1(x + 1)}

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

(iii) Let p(x) = x3 + 13x2 + 32x + 20

Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20

By trial and error method, we find that,

p(-1) = (-1)3 + 13(-1)2 + 32(-1) + 20

= -1 + 13 - 32 + 20 = 0

Therefore, (x+1) is a factor of p(x)

Now, we can write, using long division,

x3 + 13x2 + 32x + 20 = (x+1) (x2 + 12x + 20)

= (x+1) (x2 + 2x + 10x + 20) [by midterm splitting]

= (x-5) {x(x+2) +10(x+2)}

= (x-5) (x+2) (x+10)

(iv) Let p(y) = 2y3 + y2 - 2y - 1

Factors of ab = 2×(-1) = -2 are ±1 and ±2

By trial and error method, we find that,

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

= 2 +1 - 2 - 1 = 0

Therefore, (y-1) is a factor of p(y)

Now, we can write, using long division,

2y3 + y2 - 2y - 1 = (y-1) (2y2 + 3y + 1)

= (y-1) (2y2 + 2y + y + 1) [by midterm splitting]

= (y-1) {2y(y+1) +1(y+1)}

= (y-1) (2y+1) (y+1)