NCERT Solutions for Class 10 Maths - 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 ≠ 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(5) = 0
So, (x-5) is factor of p(x)
Now,
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(-1) = 0
So, (x+1) is factor of p(x)
Now,
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−5)x(x+2)+10(x+2)
= (x−5)(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(1) = 0
So, (y-1) is factor of p(y)
Now,
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)