Q.1. Determine which of the following polynomials has (x + 1) a factor:
(i) x^{3}+x^{2}+x+1
Solution:
Let p(x) = x^{3}+x^{2}+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 x^{3}+x^{2}+x+1
(ii) x^{4}+x^{3}+x^{2}+x+1
Solution:
Let p(x)= x^{4}+x^{3}+x^{2}+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 x^{4}+ x^{3}+ x^{2}+x+1
(iii) x^{4}+3x^{3}+3x^{2}+x+1
Solution:
Let p(x)= x^{4}+ x^{3}+ x^{2}+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) x^{3} – x^{2}– (2+√2)x +√2
Solution:
Let p(x) = x^{3}–x^{2}–(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 x^{3}–x^{2}–(2+√2)x +√2
Q.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
Solution:
p(x) = 2x^{3}+x^{2}–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)=x^{3}+3x^{2}+3x+1, g(x) = x+2
Solution:
p(x) = x^{3}+3x^{2}+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)=x^{3}–4x^{2}+x+6, g(x) = x–3
Solution:
p(x) = x^{3}–4x^{2}+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).
Q.3. 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
Solution:
If x1 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) = 2x^{2}+kx+√2
Solution:
If x1 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) = kx^{2}–√2x+1
Solution:
If x1 is a factor of p(x), then p(1)=0
By Factor Theorem
⇒ k(1)^{2}√2(1)+1=0
⇒ k = √21
(iv) p(x)=kx^{2}–3x+k
Solution:
If x1 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
Q.4. Factorize:
(i) 12x^{2}–7x+1
Solution:
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]
12x^{2}–7x+1= 12x^{2}4x3x+1
= 4x(3x1)1(3x1)
= (4x1)(3x1)
(ii) 2x^{2}+7x+3
Solution:
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]
2x^{2}+7x+3 = 2x^{2}+6x+1x+3
= 2x (x+3)+1(x+3)
= (2x+1)(x+3)
(iii) 6x^{2}+5x6
Solution:
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]
6x^{2}+5x6 = 6x^{2}+9x–4x–6
= 3x(2x+3)–2(2x+3)
= (2x+3)(3x–2)
(iv) 3x^{2}–x–4
Solution:
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]
3x^{2}–x–4 = 3x^{2}–x–4
= 3x^{2}–4x+3x–4
= x(3x–4)+1(3x–4)
= (3x–4)(x+1)
5. Factorize:
(i) x^{3}–2x^{2}–x+2
Solution:
Let p(x) = x^{3}–2x^{2}–x+2
Factors of 2 are ±1 and ± 2
Now,
p(x) = x^{3}–2x^{2}–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
(x+1)(x^{2}–3x+2) = (x+1)(x^{2}–x–2x+2)
= (x+1)(x(x−1)−2(x−1))
= (x+1)(x−1)(x2)
(ii) x^{3}–3x^{2}–9x–5
Solution:
Let p(x) = x^{3}–3x^{2}–9x–5
Factors of 5 are ±1 and ±5
By trial method, we find that
p(5) = 0
So, (x5) is factor of p(x)
Now,
p(x) = x^{3}–3x^{2}–9x–5
p(5) = (5)^{3}–3(5)^{2}–9(5)–5
= 125−75−45−5
= 0
Therefore, (x5) is the factor of p(x)
Now, Dividend = Divisor × Quotient + Remainder
(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)
(iii) x^{3}+13x^{2}+32x+20
Solution:
Let p(x) = x^{3}+13x^{2}+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)= x^{3}+13x^{2}+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
(x+1)(x^{2}+12x+20) = (x+1)(x^{2}+2x+10x+20)
= (x−5)x(x+2)+10(x+2)
= (x−5)(x+2)(x+10)
(iv) 2y^{3}+y^{2}–2y–1
Solution:
Let p(y) = 2y^{3}+y^{2}–2y–1
Factors = 2×(−1)= 2 are ±1 and ±2
By trial method, we find that
p(1) = 0
So, (y1) is factor of p(y)
Now,
p(y) = 2y^{3}+y^{2}–2y–1
p(1) = 2(1)^{3}+(1)^{2}–2(1)–1
= 2+1−2
= 0
Therefore, (y1) is the factor of p(y)
Now, Dividend = Divisor × Quotient + Remainder
(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)
Check out the NCERT Solutions of all the exercises of Polynomials:
Exercise 2.1. NCERT Solutions: Polynomials
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