NCERT Solutions - Polynomials (Exercise 2.3)

Q1. Determine which of the following polynomials has (x + 1) a factor:
(i) x³ + x² + x + 1
Ans: Let p(x) = x³ + x² + x + 1
The zero of x + 1 is -1. [x + 1 = 0 means x = -1]
p(-1) = (-1)³ + (-1)² + (-1) + 1
= -1 + 1 - 1 + 1
= 0
∴ By factor theorem, x + 1 is a factor of x³ + x² + x + 1

(ii) x⁴ + x³ + x² + x + 1
Ans: Let p(x) =  x⁴ + x³ + x² + x + 1
The zero of x + 1 is -1. . [x + 1= 0 means x = -1]
p(-1) = (-1)4 + (-1)³ + (-1)² + (-1) + 1
= 1 - 1 + 1 - 1 + 1
= 1 ≠ 0
∴ By factor theorem, x+1 is not a factor of x⁴ + x³ + x² + x + 1

(iii) x⁴ + 3x³ + 3x² + x + 1 
Ans: Let p(x)= x⁴ + x³ + x² + x + 1
The zero of x+1 is -1.
p(-1)=(-1)⁴+3(-1)³+3(-1)²+(-1)+1
=1-3+3-1+1
=1 ≠ 0
∴ By factor theorem, x + 1 is not a factor of x⁴ +3x³ + 3x² + x + 1

(iv) x³ - x²- (2+√2)x +√2
Ans: Let p(x) = x³-x²-(2+√2)x +√2
The zero of x+1 is -1.
p(-1) = (-1)³-(-1)²-(2+√2)(-1) + √2 = -1-1+2+√2+√2
= 2+ 2√2 ≠ 0
∴ By factor theorem, x+1 is not a factor of x³-x²-(2+√2)x +√2

Q2. 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
Ans: p(x) = 2x³+x²-2x-1, g(x) = x+1
g(x) = 0
⇒ x+1 = 0
⇒ x = -1
∴ Zero of g(x) is -1.
Now,
p(-1) = 2(-1)³+(-1)²-2(-1)-1
= -2 + 1 + 2 - 1
= 0
∴ By factor theorem, g(x) is a factor of p(x).

(ii) p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2
Ans: p(x) = x³+ 3x² 3x + 1, g(x) = x + 2
g(x) = 0
⇒ x + 2 = 0
⇒ x = -2
∴ Zero of g(x) is -2.
Now,
p(-2) = (-2)³+3(-2)²+3(-2)+1
= -8 + 12 - 6 + 1
= -1 ≠ 0
∴ By factor theorem, g(x) is not a factor of p(x).

(iii) p(x)= x³ - 4x² + x + 6, g(x) = x - 3
Ans: p(x) = x³- 4x² + x + 6, g(x) = x - 3
g(x) = 0
⇒ x-3 = 0
⇒ x = 3
∴ Zero of g(x) is 3.
Now,
p(3) = (3)³-4(3)² + (3) + 6
= 27 - 36 + 3 + 6
= 0
∴ By factor theorem, g(x) is a factor of p(x).

Q3. 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
Ans: If x - 1 is a factor of p(x), then p(1) = 0
By Factor Theorem
⇒ (1)²+(1)+k = 0
⇒ 1+1+k = 0
⇒ 2+k = 0
⇒ k = -2

(ii) p(x) = 2x² + kx + √2
Ans: If x-1 is a factor of p(x), then p(1)=0
⇒ 2(1)² + k(1) + √2 = 0
⇒ 2 + k + √2 = 0
⇒ k = -(2 + √2)

(iii) p(x) = kx²-√2x + 1
Ans: If x - 1 is a factor of p(x), then p(1)=0
By Factor Theorem
⇒ k(1)²-√2(1)+1=0
⇒ k = √2-1

(iv) p(x) = kx² - 3x + k
Ans: If x-1 is a factor of p(x), then p(1) = 0
By Factor Theorem
⇒ k(1)²-3(1)+k = 0
⇒ k-3+k = 0
⇒ 2k-3 = 0
⇒ k= 3/2

Q4. Factorize:
(i) 12x² - 7x + 1
Ans: Using the splitting the middle term method,
We have to find a number whose sum = -7 and product =1×12 = 12
We get -3 and -4 as the numbers [-3+-4=-7 and -3×-4 = 12]
12x²-7x+1= 12x²-4x-3x+1
= 4x(3x-1)-1(3x-1)
= (4x-1)(3x-1)

(ii) 2x² + 7x + 3
Ans: Using the splitting the middle term method,
We have to find a number whose sum = 7 and product = 2×3 = 6
We get 6 and 1 as the numbers [6+1 = 7 and 6×1 = 6]
2x²+7x+3 = 2x²+6x+1x+3
= 2x (x+3)+1(x+3)
= (2x+1)(x+3)

(iii) 6x² + 5x - 6 
Ans: Using the splitting the middle term method,
We have to find a number whose sum = 5 and product = 6×-6 = -36
We get -4 and 9 as the numbers [-4+9 = 5 and -4×9 = -36]
6x²+5x-6 = 6x²+9x-4x-6
= 3x(2x+3)-2(2x+3)
= (2x+3)(3x-2)

(iv) 3x²-x-4 
Ans: Using the splitting the middle term method,
We have to find a number whose sum = -1 and product = 3 × -4 = -12
We get -4 and 3 as the numbers [-4 + 3 = -1 and -4 × 3 = -12]
3x² - x - 4 = 3x² - x - 4
= 3x² - 4x + 3x - 4
= x(3x - 4) + 1(3x - 4)
= (3x - 4)(x + 1)

5. Factorize:
(i) x³- 2x² - x + 2
Ans: Let p(x) = x³-2x²-x+2
Factors of 2 are ±1 and ± 2
Now,
p(x) = x³-2x²-x+2
p(-1) = (-1)³-2(-1)²-(-1)+2
= -1-2+1+2
= 0
Therefore, (x+1) is the factor of p(x)

Now, Dividend = Divisor × Quotient + Remainder
Now by splitting the middle term method,
(x+1)(x²-3x+2) = (x+1)(x²-x-2x+2)
= (x+1)(x(x-1)-2(x-1))
= (x+1)(x-1)(x-2)

(ii) x³ - 3x² - 9x - 5
Ans: Let p(x) = x³-3x²-9x-5
Factors of 5 are ±1 and ±5
By trial method, we find that
p(x) = x³-3x²-9x-5
p(5) = (5)³-3(5)²-9(5)-5
= 125-75-45-5
= 0
Therefore, (x-5) is the factor of  p(x)

Now, Dividend = Divisor × Quotient + Remainder
(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)

(iii) x³ + 13x² + 32x + 20
Ans: Let p(x) = x³+13x²+32x+20
Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20
By trial method, we find that
p(x)= x³+13x²+32x+20
p(-1) = (-1)³+13(-1)²+32(-1)+20
= -1+13-32+20
= 0
Therefore, (x+1) is the factor of p(x)

Now, Dividend = Divisor × Quotient +Remainder
Now by splitting the middle term method,
(x+1)(x²+12x+20) = (x+1)(x²+2x+10x+20)
= (x+1)x(x+2)+10(x+2)
= (x+1)(x+2)(x+10)

(iv) 2y³ + y² - 2y - 1
Ans: Let p(y) = 2y³+y²-2y-1
Factors = 2×(-1)= -2 are ±1 and ±2
By trial method, we find that
p(y) = 2y³+y²-2y-1
p(1) = 2(1)³+(1)²-2(1)-1
= 2+1-2
= 0
Therefore, (y-1) is the factor of p(y)

Now, Dividend = Divisor × Quotient + Remainder
Now by splitting the middle term method,
(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)