Polynomials – Exercise 8.1
- Find the degree of the following polynomials.
(i) x2 – 9x + 20
(ii) 2x + 3 + 6x2
(iii) x3 + 2x2 – 5x – 6
(iv) x3 + 17x – 21 – x2
(v) √3x3 + 19x + 14
- If f(x) = 2x3 + 3x2 – 11x + 6, find.
(i) f(0)
(ii)f(1)
(iii)f(-1)
(iv) f(2)
(v)f(-3)
- Find the values of the following polynomials.
(i) g(x) = 7x2 + 2x + 14 when x = 1
(ii) p(x) = -x3 + x2 – 6x + 5 when x = 2
(iii) p(x) = 2x2 + 1/4 x + 13, when x = -1
(iv) p(x) = 2x4 – 3x3 – 3x2 + 6x – 2 , when x = -2
- Verify whether the indicated numbers are zeroes of the polynomials in each of the following cases.
(i) f(x) = 3x + 1, x = – 1/3
(ii) p(x) = x2 – 4, x = 2 , x = -2
(iii) g(x) = 5x – 8, x = 4/5
(iv) p(x) = 3x3 – 5x2 – 11x – 3, x = 3, x = -1 and x = –1/3
- Find the zeroes of the following quadratic polynomials and verify
(i)x2 + 4x + 4
(ii)x2 – 2x – 15
(iii) 4a2 – 49
(iv) 2a2 – 2√2a + 1
- If x = 1 is a zero of the polynomial f(x) = x3 – 2x2 + 4x + k, find the value of k.
- For what value of k, -4 is a zero of the polynomial x2 – x – (2k + 2)?
Polynomials – Exercise 8.1 – Solution:
- Find the degree of the following polynomials.
(i) x2 – 9x + 20
Solution:
p(x) = x2 – 9x + 20 is a polynomial in x of degree 2
(ii) 2x + 3 + 6x2
Solution:
p(x) = 6x2 + 2x + 3 is polynomial in x is degree 2
(iii) x3 + 2x2 – 5x – 6
Solution:
p(x) = x3 + 2x2 – 5x – 6 is a polynomial in x is degree 3
(iv) x3 + 17x – 21 – x2
Solution:
p(x) = x3 – x2 + 17x – 21 is a polynomial in x is degree 3
(v) √3x3 + 19x + 14
Solution:
p(x) = √3x3 + 19x + 14 is a polynomial in x is degree 3
- If f(x) = 2x3 + 3x2 – 11x + 6, find.
(i) f(0)
(ii)f(1)
(iii)f(-1)
(iv) f(2)
(v)f(-3)
Solution:
f(x) = 2x3 + 3x2 – 11x + 6
(i) f(0) = 2(0)3 + 3(0)2 – 11(0) + 6
= 0 + 0 + 0 + 6
= 6
(ii)f(1) = 2(1)3 + 3(1)2 – 11(1) + 6
= 2 + 3 – 11 + 6
= 0
(iii)f(-1) = 2(-1)3 + 3(-1)2 – 11(-1) + 6
= – 2 + 3 + 11 + 6
= 18
(iv) f(2) = 2(2)3 + 3(2)2 – 11(2) + 6
= 2 x 8 + 3 x 4 – 22 + 6
= 16 + 12 – 22 + 6
= 12
(v)f(-3) = 2(-3)3 + 3(-3)2 – 11(-3) + 6
= 2(-27) + 3×9 + 33 + 6
= 12
- Find the values of the following polynomials.
(i) g(x) = 7x2 + 2x + 14 when x = 1
Solution:
g(x) = 7x2 + 2x + 14
if x = 1, then g(1) = 7(1)2 + 2(1) + 14
= 7 + 2 + 14
= 23
(ii) p(x) = -x3 + x2 – 6x + 5 when x = 2
Solution:
p(x) = -x3 + x2 – 6x + 5
When x = 2, p(2) = -(2)3 + (2)2 – 6(2) + 5
= – 8 + 4 – 12 + 5
= – 11
(iii) p(x) = 2x2 + 1/4 x + 13, when x = -1
Solution:
p(x) = 2x2 + 1/4 x + 13
When x = -1, p(-1) = 2(-1)2 + 1/4 (-1) + 13
= 2 – 1/4 + 13
=59/4
= 14.75
(iv) p(x) = 2x4 – 3x3 – 3x2 + 6x – 2 , when x = -2
Solution:
p(x) = 2x4 – 3x3 – 3x2 + 6x – 2
When x = -2, p(-2) = 2(-2)4 – 3(-2)3 – 3(-2)2 + 6(-2) – 2
= 2(16) – 3(-8) – 3(4) – 12 – 2
=32 + 24 – 12 – 12 – 2
= 30
- Verify whether the indicated numbers are zeroes of the polynomials in each of the following cases.
(i) f(x) = 3x + 1, x = – 1/3
Solution:
Given, f(x) = 3x + 1, we have to find out whether at x = ‑1/3 is the zero of the polynomial f(x) = 3x + 1.
If x = – 1/3, f(- 1/3) = 3(- 1/3) + 1
= -1 + 1
= 0
Therefore, x = –1/3 is the zero of the polynomial f(x) = 3x + 1.
(ii) p(x) = x2 – 4, x = 2 , x = -2
Solution:
Given, p(x) = x2 – 4, we have to find out whether at x = 2 and x = -2 is the zero of the polynomial p(x) = x2 – 4.
If x = 2, p(2) = (2)2 – 4
= 4 – 4
= 0
Therefore, x = 2 is the zero of the polynomial p(x) = x2 – 4.
If x = -2, p(-2) = (-2)2 – 4
= 4 – 4
= 0
Therefore, x = -2 is the zero of the polynomial p(x) = x2 – 4.
Thus, x = 2 and x = -2 both are zero of the polynomial p(x) = x2 – 4.
(iii) g(x) = 5x – 8, x = 4/5
Solution:
Given, g(x) = 5x – 8, we have to find out whether at x = 4/5 is the zero of the polynomial g(x) = 5x – 8.
If x = 4/5, g(4/5) = 5(4/5) – 8
= 4 – 8
= -4
Therefore, x = 4/5 is not a zero of the polynomial g(x) = 5x – 8
(iv) p(x) = 3x3 – 5x2 – 11x – 3, x = 3, x = -1 and x = –1/3
Solution:
Given, p(x) = 3x3 – 5x2 – 11x – 3, we have to find out whether at x = 3, x = -1 and x = –1/3 is the zero of the polynomial p(x) = 3x3 – 5x2 – 11x – 3.
For x = 3, p(3) = 3(3)3 – 5(3)2 – 11(3) – 3
= 3×27 – 5×9 – 33 – 3
= 81 – 45 – 33 – 3
=0
Therefore, x = 3 is the zero of the polynomial p(x) = 3x3 – 5x2 – 11x – 3.
For x = -1, p(-1) = 3(-1)3 – 5(-1)2 – 11(-1) – 3
= 3(-1) – 5(1) – 11(-1) – 3
= -3 – 5 + 11 – 3
= 0
Therefore, x = -1 is the zero of the polynomial p(x) = 3x3 – 5x2 – 11x – 3.
For x = –1/3, p(-1/3) = 3(-1/3)3 – 5(-1/3)2 – 11(-1/3) – 3
= – 3x1/27 – 5x1/9 + 11(1/3) – 3
= –1/9 – 5/9 + 11/3 – 3
= 0
Therefore, x = –1/3 is the zero of the polynomial p(x) = 3x3 – 5x2 – 11x – 3.
- Find the zeroes of the following quadratic polynomials and verify
(i)x2 + 4x + 4
Solution:
Given, f(x) = x2 + 4x + 4, we have to find the zero of the polynomial.
f(x) = x2 + 4x + 4 = 0
x2 + 2x + 2x + 4 = 0
x(x + 2) +2(x + 2) = 0
(x + 2)(x + 2) = 0
(x + 2)2 = 0
x + 2 = 0
x = -2
Therefore x = -2 is the zero of the polynomial f(x) = x2 + 4x + 4
Verify:
For x = -2, f(-2) = (-2)2 + 4(-2) + 4
= 4 – 8 + 4
= 8 – 8
= 0.
Hence, x = -2 is the zero of the polynomial f(x) = x2 + 4x + 4.
(ii)x2 – 2x – 15
Solution:
Given, f(x) = x2 – 2x – 15, we have to find the zero of the polynomial.
f(x) = x2 – 2x – 15 = 0
x2 + 3x – 5x – 15 = 0
x(x + 3) – 5(x + 3) = 0
(x – 5)(x + 3) = 0
x + 3 = 0 or x – 5 = 0
x = -3 or x = 5
Therefore x = -3 and x = 5 is the zero of the polynomial f(x) = x2 – 2x – 15
Verify:
For x = -3, f(-3) = (-3)2 – 2(-3) – 15
= 9 + 6 – 15
= 15 – 15
= 0.
Hence, x = -3 is the zero of the polynomial f(x) = x2 – 2x – 15
For x = 5, f(5) = (5)2 – 2(5) – 15
= 25 – 10 – 15
= 25 – 25
= 0.
Hence, x = 5 is the zero of the polynomial f(x) = x2 – 2x – 15
(iii) 4a2 – 49
Solution:
Given, f(a) = 4a2 – 49, we have to find the zero of the polynomial.
f(a) = 4a2 – 49 = 0
4a2 – 49 = 0
4a2 = 49
(2a)2 = 72
2a = 7
a = 7/2
Therefore a = 7/2 is the zero of the polynomial f(7/2) = 4a2 – 49
Verify:
For x = 7/2 , f(7/2) = 4(7/2)2 – 49
= 4(49/4) – 49
= 49 – 49
= 0.
Hence, a = 7/2 is the zero of the polynomial f(a) = 4a2 – 49
(iv) 2a2 – 2√2a + 1
Solution:
Given, f(a) = 2a2 – 2√2a + 1, we have to find the zero of the polynomial.
f(a) = 2a2 – 2√2a + 1 = 0
2a2 – 2√2a + 1 = 0
(√2a)2 – 2(√2a)(1) + 12 = 0
a2 – 2ab + b2 = (a – b)2
⇒ (√2a – 1)2 = 0
⇒ (√2a – 1) = 0
⇒√2a – 1 = 0
⇒√2a = 1
⇒a = 1/√2
Verify:
For a = 1/√2,
f(a) = 2a2 – 2√2a + 1
= 2(1/√2)2 – 2√2(1/√2) + 1
= 2(1/2) – 2 + 1
= 1 – 2 + 1
= 0
Hence, a = 1/√2 is the zero of the polynomial f(a) = 2a2 – 2 √2a + 1
- If x = 1 is a zero of the polynomial f(x) = x3 – 2x2 + 4x + k, find the value of k.
Solution:
Given, if x = 1 is a zero of the polynomial f(x) = x3 – 2x2 + 4x + k. Then we have to find the value of k,
f(1) = (1)3 – 2(1)2 + 4(1) + k = 0
1 – 2 + 4 + k = 0
3 + k = 0
k = -3
- For what value of k, -4 is a zero of the polynomial x2 – x – (2k + 2)?
Solution:
Given, x = – 4 is the zero of the polynomial f(x) = x2 – x – (2k + 2)
f(-4) = (-4)2 – (-4) – (2k + 2) = 0
⇒ 16 + 4 – (2k + 2) = 0
⇒ 20 – 2(k + 1) = 0
⇒2(10 – (k+1)) = 0
10 – k – 1 = 0
-k + 9 = 0
k = 9
Next exercise – Polynomials – Exercise 8.2 – Class X
