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# TEXTBOOKS EXERCISE – 2.4

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

Sol. To have (x + 1) as a factor, substituting x = –1 must give p(–1) = 0.

(i) x3 + x2 + x + 1

= (–1)3 + (–1)2 + (–1) + 1

= – 1 + 1 – 1 + 1 = 0

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

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

= (–1)4 + (–1)3 + (–1)2 + (–1) + 1

= 1 – 1 + 1 – 1 + 1 = 1

Remainder is not 0, Therefore, (x + 1) is not its factor.

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

= (–1)4 + 3(–1)3 + 3(–1)2 + (–1) + 1

= 1 – 3 + 3 – 1 + 1 = 1. Remainder is not 0, therefore, (x + 1) is not its factor

(iv) x3 – x2 – (2 + √2 ) x + √2

= (–1)3 – (–1)2 – (2 + √2 ) (–1) + √2

= – 1 – 1 + 2 +√ 2 + √2 = 2 √2

Remainder is not 0, therefore (x + 1) is not a factor.

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

Sol. (i) 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) g(x) = x + 2, substitute x = – 2 in p(x)

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

p(–2) = (–2)3 + 3(–2)2 + 3(–2) + 1 = – 8 + 12 – 6 + 1 = –1.

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

(iii) g(x) = x – 3 substitute x = 3 in (x).

p(x) = x3 – 4x2 + x + 6

p(3) = (3)3 – 4(3)2 + 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 – √2 x + 1

(iv) p(x) = kx2 – 3x + k

Sol. (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) = 1 + 1 + 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 = – (2 + √2 )

(iii) p(x) = kx2 – √2 x + 1

p(1) = k – √2 + 1 = 0

⇒ k = √2 – 1

(iv) p(x) = kx2 – 3x + k

p(1) = 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) 3x2x – 4

Sol. (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) 3x2x – 4

= 3x2 – 4x + 3x – 4

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

5. Factorise :

(i) x3 – 2x2 – x + 2

(ii) x3 – 3x2 – 9x – 5

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

(iv) 2y3 + y2 – 2y – 1

Sol. (i) p(x) = x3 – 2x2 – x + 2

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

Now, putting this value in p(x).

p(1) = 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 + 1) (x – 2)

So, x3 – 2x2 – x + 2

(ii) p(x) = x3 – 3x2 – 9x – 5

Take a factor (x – a). a should be a factor of 5, i.e., ± 1 or ± 5.

For (x – 1), a = 1

p(1) = (1)3 – (–3) 12 – 9 × 1 – 5

= 1 – 3 – 9 – 5 = – 16.

So, (x – 1) is not a factor of p(x).

For a = 5

p(5) = (5)3 – 3(5)2 – 9 (5) – 5 = 125 – 75 – 45 – 5 = 0.

Therefore, (x – 5) is a factor of

x3 – 3x2 – 9x – 5.

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) (x + 1)2

= (x – 5) (x + 1) (x + 1)

So, x3 – 3x2 – 9x – 5 = (x – 5) (x + 1) (x + 1).

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

Let a factor be (x a). a should be a factor of 20 which are ± 1, ± 2, ± 4, ± 5, ± 10.

For x – 1 = 0 ⇒ x = 1

Now, p(1) = 1 + 13 + 32 + 20 = 66 ≠ 0

Hence, (x – 1) is not a factor of p(x).

Again, for x + 1 = 0 ⇒ x = – 1

Now, p (–1) = –1 + 13 – 32 + 20

= –33 + 33 = 0

Hence, (x + 1) is a factors of p(x).

Now, x3 + 13x2 + 32x + 20

= x3 + x2 + 12x2 + 12x + 20x + 20 = x2(x + 1) + 12 x (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 + 2) (x + 1) (x + 10)

(iv) p(y) = 2y3 + y2 – 2y – 1

Factors of –1 are ± 1, ± 2.

p(1) = 2 × 13 + 12 – 2 × 1 – 1 = 2 + 1 – 2 – 1 = 0.

Therefore, (y – 1) is a factor of p(y).

Now, 2y3 + y2 – 2y – 1

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

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

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