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

---

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– 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

---

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 + √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

---

4. 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)

---

5. 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)

---