Exercise 2.3 NCERT Solutions - Polynomials (Old NCERT)
Q.1. Find the remainder when x3+3x2+3x+1 is divided by
(i) x+1
Solution:
x+1= 0
⇒ x = -1
∴ Remainder:
p(-1) = (-1)3+3(-1)2+3(-1)+1
= -1+3-3+1
= 0
(ii) x-1/2
Solution:
x-1/2 = 0
⇒ x = 1/2
∴ Remainder:
p(1/2) = (1/2)3+3(1/2)2+3(1/2)+1
= (1/8)+(3/4)+(3/2)+1
= 27/8
(iii) x
Solution:
x = 0
∴ Remainder:
p(0) = (0)3+3(0)2+3(0)+1
= 1
(iv) x+π
Solution:
x+π = 0
⇒ x = -π
∴ Remainder:
p(0) = (-π)3 +3(-π)2+3(-π)+1
= -π3+3π2-3π+1
(v) 5+2x
Solution:
5+2x=0
⇒ 2x = -5
⇒ x = -5/2
∴ Remainder:
(-5/2)3+3(-5/2)2+3(-5/2)+1 = (-125/8)+(75/4)-(15/2)+1
= -27/8
Q.2. Find the remainder when x3-ax2+6x-a is divided by x-a.
Solution:
Let p(x) = x3-ax2+6x-a
x-a = 0
∴ x = a
Remainder:
p(a) = (a)3-a(a2)+6(a)-a = a3-a3+6a-a = 5a
Q.3. Check whether 7+3x is a factor of 3x3+7x.
Solution:
7+3x = 0
⇒ 3x = -7
⇒ x = -7/3
∴ Remainder:
3(-7/3)3+7(-7/3) = -(343/9)+(-49/3)
= (-343-(49)3)/9
= (-343-147)/9
= -490/9 ≠ 0
∴ 7+3x is not a factor of 3x3+7x
Check out the NCERT Solutions of all the exercises of Polynomials:
Exercise 2.1. NCERT Solutions: Polynomials
FAQs on Exercise 2.3 NCERT Solutions - Polynomials (Old NCERT)
| 1. What are polynomials in mathematics? |  |
| 2. How do you classify polynomials based on the number of terms? | |
| 3. What is the degree of a polynomial? | |
| 4. How can polynomials be used in real-life applications? | |
| 5. Can polynomials be graphed on a coordinate plane? | |
