# Polynomials – Exercise 2.4 – Class IX

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 + √2) x + √2

Solution:

To get (x + 1) as the factor, we have to substitute x = -1 , it must give p(-1) = 0

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

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

= – 1 + 1 – 1 + 1

= 0

Thus, 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

Here remainder is not 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

Here remainder is not 0. Therefore, x + 1 is not a factor of x4+ 3x3+ 3x2+ x + 1

(iv) 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

Here remainder is not 0. Therefore, x + 1 is not a factor of x3 – x2 – (2 + √2) x + √2

1. 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

Solution:

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

g(x) = x + 1 , x = -1 to be substituted in p(x) = 2x3+ x2– 2x – 1

p (-1)= 2 (-1)3+ (-1)2– 2 (-1) – 1

= -2 + 1 + 2 – 1

= 0

So, g(x) is a factor of p(x).

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

g(x) = x + 2 , x = -2 to be substituted in p(x) = x3+ 3x2+ 3x + 1

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

= – 8 + 3*4 – 6 + 1

= -8 + 12 – 6 + 1

= -1

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

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

g(x) = x – 3, x = 3 to be substituted in p(x) = x3– 4x2+ x + 6

p(3) = 33– 4*32+ 3 + 6

= 27 – 4*9 + 3 + 6

= 27 – 36 + 3 + 6

= 0

Therefore, g(x) is a factor of  x3– 4x2+ x + 6

1. 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

Solution:

(x – 1) is a factor, so we substitute x = 1 in each case and solve for k by making p(1) equal to 0.

(i) p(x) = x2+ x + k

p(1) = 12 + 1 + k = 0

2 + k = 0

k = -2

(ii) p(x) = 2x2+ kx + √2

p(1) = 2*12 + k*1 + √2 = 0

2 + k + √2 = 0

k = -(2 + √2)

(iii) p(x) = kx2– √2x + 1

p(1) = k*12– √2*1 + 1 = 0

k – √2 + 1 = 0

k = √2 -1

(iv) p(x) = kx2– 3x + k

p(1) = k*12– 3*1 + k = 0

k – 3 + k = 0

2k – 3 = 0

k = 3/2

1. Factorise :

(i) 12x2– 7x + 1

(ii) 2x2+ 7x + 3

(iii) 6x2+ 5x – 6

(iv) 3x2 – x – 4

Solution:

(i) 12x2– 7x + 1

=12x2 – 4x – 3x + 1

= 4x(3x – 1) -1(3x – 1)

=(4x – 1)(3x – 1)

(ii) 2x2+ 7x + 3

= 2x2 + 6x + x + 3

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

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

(iii) 6x2+ 5x – 6

= 6x2 +9x – 4x  – 6

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

= (3x – 2)(2x + 3)

(iv) 3x2 – x – 4

=3x2 + 3x – 4x – 4

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

= (3x – 4)(x + 1)

1. Factorise :

(i) x3– 2x2– x + 2

(ii) x3– 3x2– 9x – 5

(iii) x3+ 13x2 + 32x + 20

(iv) 2y3+ y2– 2y – 1

Solution:

(i)p(x) =  x3– 2x2– x + 2

Let us guess a factor (x – a) and choose value of a and arbitrary as 1

Now, putting this value in p(x)

1 – 2 – 1 + 2 = 0

So, (x – 1) is a factor of p(x)

Now, x3– 2x2– x + 2 = x3 – x2 – x2 + x – 2x + 2

=x2 (x – 1) – x (x – 1) – 2(x – 1)

=(x – 1)(x2 – x – 2)

= (x – 1)(x2 – 2x + x – 2)

=(x – 1)(x(x – 2) +1(x – 2))

= (x – 1)(x – 2)(x + 1)

(ii) p(x) = x3– 3x2– 9x – 5

Let us guess a factor (x – a) and choose value of a and arbitrary as 1

Now, putting this value in p(x)

P(1) = 1 – 3 – 9 – 5 = -16

So, (x – 1), a = 1 is not a factor of p(x)

Let (x – a) and choose value of a as 5

Now, putting this value in p(x)

P(5) =53 – 3(52) – 9(5) – 5 = 125 – 3(25) – 45 – 5 = 0

So, (x – 5), a = 5 is a factor of p(x)

Now, x3– 3x2– 9x – 5 = x3 – 5x2 – 2x2 – 10x + x – 5

= x2(x – 5)-2x(x – 5) +1(x – 5)

= (x – 5)(x2 – 2x + 1)

=(x – 5)(x2 – x – x + 1)

=(x – 5)(x(x – 1)-1(x -1))

= (x – 5)(x – 1)(x – 1)

therefore, x3– 3x2– 9x – 5 = (x – 5)(x – 1)(x – 1)

(iii) x3+ 13x2 + 32x + 20

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

Let (x – a) and choose value of a as -1

Now, putting this value in p(x)

P(-1) = (-1)3 + 13(-1)2 + 32(-1) + 20 = -1 +13 – 32 + 20 = 0

So, (x – a), a = -1 is a factor of p(x)

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

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

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

(iv) 2y3+ y2– 2y – 1

p(y) = 2y3+ y2– 2y – 1

Let (y – a) and choose value of a as 1

Now, putting this value in p(1)

P(1) = 2(1)3 + (1)2 – 2(1) – 1 = 2 + 2 – 2 – 1 = 0

So, (y – a), a = 1 is a factor of p(x)

Now, 2y3+ y2– 2y – 1 = 2y3 – 2y2 + 3y2 – 3y + y – 1

= 2y2(y – 1) + 3y(y – 1) + 1(y – 1)

=(2y2 + 3y + 1)(y – 1)

=(2y2 + 2y + y + 1)(y – 1)

=[2y(y + 1)+1(y + 1)](y – 1))

=(2y + 1)(y + 1)( y – 1)

Therefore, 2y3+ y2– 2y – 1 = (2y + 1)(y + 1)( y – 1)