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Exercise 2.4

1.  Determine which of the following polynomials has (x + 1) a factor.
(i) x³ + x² + x + 1

Ans. –  If (x + 1 ) is a factor of p(x) = x³ + x² + x + 1 , then p(-1) = 0
Now, p(-1) = -1 + 1 -1 + 1 = 0.
Hence, (x+1) is a factor.

(ii) x4 + x³ + x² + x + 1

Ans. – If (x +1 )is a factor  of  p(x) = x+ x3 + xx + 1 , then p(-1) = 0.
Now, p(-1) = 1 -1 +1 – 1+ 1 = 1 ≠ 0.
Hnece, (x+1) is not factor.

(iii) x4 + 3x³ +3x² + x + 1

Ans. –  If (x + 1 ) is a factor of p(x) = x4 + 3x³ + x + 1 , then p(-1) = 0.
Now, p(-1) = 1 – 3 + 3 – 1 + 1 = 1 ≠ 0.
Hence, (x+1) is not a factor.

(iv) x³ – x² – (2 + √2)x + √2

Ans. – If (x + 1) is a factor of p(x) = x³ – x² – (2 + √2 )x + √2, then p(-1) = 0.
Now, p(-1) = (-1)³ – (-1)² – (2 + √2) (-1 ) + √2
= -1 -1 + 2 + √2  + √2
= 2√2 ≠ 0
Hence, (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) = 3x³ + x² – ex – 1 , g (x) = x + 1

Ans. – If g(x) is a factor of p(x) , then p (-1) = 0
Now, p(-1) = -2  + 1 + 2 – 1 = 0
Hence, g(x) is a factor of p(x).

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

Ans. – If g(x) is a factor  of p(x) , then p(-1) = 0 .
Now, p(-2)  = – 8 + 12 – 6 + 1 = 1 – 14 + 13 = -1 ≠ 0
Hence, g(x) is not a factor of p(x).

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

Ans. – If g(x)  is a factor of p(x) .
Now, p(3) = 27  – 36  + 3  + 6 = 36 – 36 = 0.
Hence, g(x) is a factor of p(x).

3. Find the value of ℜ, if x- 1 is a factor of p(x) in each of the following cases:
(i) p(x) = x² + x + ℜ

Ans.  – If (x-1) is a factor of p(x) , then p(1) = 0.
Now, p(1) ⇒  1 + 1 + ℜ = 0  ⇒ ℜ = – 2

(ii) p(x) = 2x² +  ℜx  +   √2

Ans.  If (x-1) is a factor of p(x) , then p(1) = 0
Now, p(1) = 0 ⇒ 2 + ℜ  + √2 = 0 ⇒ ℜ = – (2 + √2).

(iii) p(x)  = ℜx²  –   √2x  + 1

Ans. If (x-1) is a factor of p(x) , then p(1) = 0.
Now, p(1)  = 0  ⇒   ℜ  – $\sqrt 2$ + 1  = 0  ⇒ ℜ = √2 – 1 .

(iv) p(x) = ℜx²  – 3x   +    ℜ.

Ans.  If (x-1) is a factor of p(x) , then p(1)  = 0.
Now , p(1) = 0 ⇒ ℜ – 3 ℜ =  0   ⇒  ℜ =${3 \over 2}$.

4.  Factorise :
(i) 12x² – 7x + 1

Ans. – 12x² – 7x + 1 = 12x² = 4x – 3x + 1
.                                  = 4x(3x – 1) – 1(3x – 1) = (4x -1)(3x -1)

(ii) 2x² + 7x + 3

Ans. –  2x² + 7x + 3   = 2x²  + 6x   + x  +3
.                                     = 2x (x + 3) + 1 (x +3) = (2x + 1 ) (x + 3 )

(iii) 6x² + 5x – 6

Ans.  6x²   +  5x  – 6  =  6x²   + 9x   – 4x  – 6
.                                    = 3x (2x +3) – 2(2x + 3 ) = (3x -2 )(2x + 3).

(iv) 3x² – x – 4.

Ans.  3x² – x – 4   = 3x²  – 4x + 3x – 4
.                              = x(3x – 4 ) + 1 (3x – 4 ) = (x + 1 ) (3x – 4 )

5.  Factorise :
(i) x³ – 2x² – x + 2

Ans. –  Let p(x) = x³ – 2x² – x +2.
Possible factors of 2 are ± 1,  ± 2.
We notice p(1) = 1 – 2 – 1 + 2  = 0
⇒ x = 1 is a zero of polynomial p(x) or (x-1) is a factor of p(x).
Let us divide p(x)  by (x – 1) .

p(x) = x3 – 2x2 – x + 2
= (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).
Alternative Method:
x3 – 2x2 – x + 2 = x2(x – 2) – 1(x – 2)
= (x – 2)(x2 -1)
= (x – 2)(x2 – 12)
= (x – 2)(x + 1)(x – 1).

(ii) x³ – 3x² – 9x – 5

Ans. –
Let p(x) = x3 – 3x2 – 9x – 5.
Possible factors of 5 are ± 1, ± 5
We notice p(- 1) – 1 – 3 + 9 – 5 = 0
⇒ x  = – 1 is zero of polynomial p(x).
⇒ (x + 1) is a factor of p(x).
Let us divide p(x) by (x + 1 )

p(x) = x3 – 3x2 – 9x – 5
= (x + 1)(x2 – 4x – 5)
= (x + 1)(x2 – 5x + x – 5)
= (x + 1){x(x – 5) + 1(x – 5)}
= (x + 1)(x + 1)(x – 5).

(iii) x³ + 13x² + 32x +20

Ans. – Let p(x) = x3 + 13x2 + 32x + 20.
Possible factors of 20 are ± 1, ± 2, ± 4, ± 5, ± 10, ± 20.
We notice p(-1) =  – 1 +  13  – 32  + 20 = 0
⇒ x = – 1   is a zero of p(x)
⇒  (x + 1) is a factor p(x).
Let us divide p(x) by (x + 1 ).

∴ p(x) =  x³  + 13x2 + 32x + 20
= (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).

(iv) 2y³ + y² – 2y  – 1 .