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

**(i) x ^{3}+ x^{2}+ x + 1**

**(ii) x ^{4}+ x^{3}+ x^{2}+ x + 1**

**(iii) x ^{4}+ 3x^{3}+ 3x^{2}+ x + 1**

**(iv) x ^{3} – x^{2} – (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) = x^{3}+ x^{2}+ x + 1

p(-1) = (-1)^{3}+ (-1)^{2}+ (-1)+ 1

= – 1 + 1 – 1 + 1

= 0

Thus, x + 1 is a factor of x^{3}+ x^{2}+ x + 1

(ii) p(x) = x^{4}+ x^{3}+ x^{2}+ 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 x^{4}+ x^{3}+ x^{2}+ x + 1

(iii)p(x) = x^{4}+ 3x^{3}+ 3x^{2}+ 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 x^{4}+ 3x^{3}+ 3x^{2}+ x + 1

(iv) p(x) = x^{3} – x^{2} – (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 x^{3} – x^{2} – (2 + √2) x + √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) = 2x ^{3}+ x^{2}– 2x – 1, g(x) = x + 1**

**(ii) p(x) = x ^{3}+ 3x^{2}+ 3x + 1, g(x) = x + 2**

**(iii) p(x) = x ^{3}– 4x^{2}+ x + 6, g(x) = x – 3**

Solution:

(i) p(x) = 2x^{3}+ x^{2}– 2x – 1, g(x) = x + 1

g(x) = x + 1 , x = -1 to be substituted in p(x) = 2x^{3}+ x^{2}– 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) = x^{3}+ 3x^{2}+ 3x + 1, g(x) = x + 2

g(x) = x + 2 , x = -2 to be substituted in p(x) = x^{3}+ 3x^{2}+ 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) = x^{3}– 4x^{2}+ x + 6, g(x) = x – 3

g(x) = x – 3, x = 3 to be substituted in p(x) = x^{3}– 4x^{2}+ x + 6

p(3) = 3^{3}– 4*3^{2}+ 3 + 6

= 27 – 4*9 + 3 + 6

= 27 – 36 + 3 + 6

= 0

Therefore, g(x) is a factor of x^{3}– 4x^{2}+ x + 6

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

**(i) p(x) = x ^{2}+ x + k**

**(ii) p(x) = 2x ^{2}+ kx + √2**

**(iii) p(x) = kx ^{2}– √2x + 1**

**(iv) p(x) = kx ^{2}– 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) = x^{2}+ x + k

p(1) = 1^{2} + 1 + k = 0

2 + k = 0

k = -2

(ii) p(x) = 2x^{2}+ kx + √2

p(1) = 2*1^{2} + k*1 + √2 = 0

2 + k + √2 = 0

k = -(2 + √2)

(iii) p(x) = kx^{2}– √2x + 1

p(1) = k*1^{2}– √2*1 + 1 = 0

k – √2 + 1 = 0

k = √2 -1

(iv) p(x) = kx^{2}– 3x + k

p(1) = k*1^{2}– 3*1 + k = 0

k – 3 + k = 0

2k – 3 = 0

k = ^{3}/_{2}

**Factorise :**

**(i) 12x ^{2}– 7x + 1**

**(ii) 2x ^{2}+ 7x + 3**

**(iii) 6x ^{2}+ 5x – 6**

**(iv) 3x ^{2} – x – 4**

Solution:

(i) 12x^{2}– 7x + 1

=12x^{2} – 4x – 3x + 1

= 4x(3x – 1) -1(3x – 1)

=(4x – 1)(3x – 1)

(ii) 2x^{2}+ 7x + 3

= 2x^{2} + 6x + x + 3

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

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

^{ }

(iii) 6x^{2}+ 5x – 6

= 6x^{2} +9x – 4x – 6

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

= (3x – 2)(2x + 3)

^{ }

(iv) 3x^{2} – x – 4

=3x^{2} + 3x – 4x – 4

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

= (3x – 4)(x + 1)

**Factorise :**

**(i) x ^{3}– 2x^{2}– x + 2**

**(ii) x ^{3}– 3x^{2}– 9x – 5**

**(iii) x ^{3}+ 13x^{2} + 32x + 20**

**(iv) 2y ^{3}+ y^{2}– 2y – 1**

Solution:

(i)p(x) = x^{3}– 2x^{2}– 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, x^{3}– 2x^{2}– x + 2 = x^{3} – x^{2} – x^{2} + x – 2x + 2

=x^{2} (x – 1) – x (x – 1) – 2(x – 1)

=(x – 1)(x^{2} – x – 2)

= (x – 1)(x^{2} – 2x + x – 2)

=(x – 1)(x(x – 2) +1(x – 2))

= (x – 1)(x – 2)(x + 1)

(ii) p(x) = x^{3}– 3x^{2}– 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) =5^{3} – 3(5^{2}) – 9(5) – 5 = 125 – 3(25) – 45 – 5 = 0

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

Now, x^{3}– 3x^{2}– 9x – 5 = x^{3} – 5x^{2} – 2x^{2} – 10x + x – 5

= x^{2}(x – 5)-2x(x – 5) +1(x – 5)

= (x – 5)(x^{2} – 2x + 1)

=(x – 5)(x^{2} – x – x + 1)

=(x – 5)(x(x – 1)-1(x -1))

= (x – 5)(x – 1)(x – 1)

therefore, x^{3}– 3x^{2}– 9x – 5 = (x – 5)(x – 1)(x – 1)

(iii) x^{3}+ 13x^{2} + 32x + 20

p(x) = x^{3}+ 13x^{2} + 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, x^{3}+ 13x^{2} + 32x + 20 = x^{3} + x^{2} + 12x^{2} + 12x + 20x + 20

= x^{2}(x + 1) + 12x(x + 1) + 20(x + 1)

= (x + 1) (x^{2} + 12x + 20)

= (x + 1)(x^{2} + 10x + 2x + 20)

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

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

therefore, x^{3}+ 13x^{2} + 32x + 20 = (x + 1)(x+2)(x+10)

(iv) 2y^{3}+ y^{2}– 2y – 1

p(y) = 2y^{3}+ y^{2}– 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, 2y^{3}+ y^{2}– 2y – 1 = 2y^{3} – 2y^{2} + 3y^{2} – 3y + y – 1

= 2y^{2}(y – 1) + 3y(y – 1) + 1(y – 1)

=(2y^{2} + 3y + 1)(y – 1)

=(2y^{2} + 2y + y + 1)(y – 1)

=[2y(y + 1)+1(y + 1)](y – 1))

=(2y + 1)(y + 1)( y – 1)

Therefore, 2y^{3}+ y^{2}– 2y – 1 = (2y + 1)(y + 1)( y – 1)

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