- Find the remainder when x3 + 3x2 + 3x + 1 is divided by
(i) x + 1
(ii) x – 1/2
(iii) x
(iv) x + π
(v) 5 + 2x
Solution:
(i) x + 1
p(x) = x3 + 3x2 + 3x + 1 and the zero of x + 1 is – 1
So, p(1) = (-1)3 + 3(-1)2 + 3(-1) + 1
= -1 + 3 – 3 + 1
= 0
So, by the remainder theorem, remainder obtained on dividing p(x) by x + 1 is 0.
So, p(x) is the multiple of x + 1
(ii) x – 1/2
p(x) = x3 + 3x2 + 3x + 1 and zero of x – 1/2 is 1/2
p(1/2) = (1/2)3 + 3(1/2)2 + 3(1/2) + 1
= 1/8 + 3*1/4 + 3*1/2 + 1
= 1/8 + 3/4 + 3/2 + 1
= 1+6+12+8/8 = 27/8
So, by the remainder theorem, remainder obtained on dividing p(x) by x – 1/2 is 27/8.
(iii) x
p(x) = x3 + 3x2 + 3x + 1 and zero of x is 0
p(0) = (0)3 + 3(0)2 + 3(0) + 1
= 0 + 3*0 + 3*0 + 1
= 1
So, by the remainder theorem, remainder obtained on dividing p(x) by x is 1.
(iv) x + π
p(x) = x3 + 3x2 + 3x + 1 and zero of x + π is -π
p(-π) = (-π)3 + 3(-π)2 + 3(-π) + 1
= -π3 + 3π2 – 3π + 1
So, by the remainder theorem, remainder obtained on dividing p(x) by x + π is = -π3 + 3π2 – 3π + 1.
(v) 5 + 2x
p(x) = x3 + 3x2 + 3x + 1 and zero of 5 + 2x is –5/2
p(-5/2) = (-5/2)3 + 3(-5/2)2 + 3(-5/2) + 1
= –125/8 +3*25/4 – 3*5/2 + 1
= -125+150-60+8/8
= ‑27/8
So, by the remainder theorem, remainder obtained on dividing p(x) by 5 + 2x is –27/8.
- Find the remainder when x3 – ax2 + 6x – a is divided by x – a
Solution:
p(x) = x3 – ax2 + 6x – a and zero of x – a is a
p(1) = a3 – a(a)2 + 6(a) – a
= a3 – a3 + 6a – a
= 5a
So, by the remainder theorem, remainder obtained on dividing p(x) by x – a is 5a.
- Check whether 7 + 3x is a factor of 3x3 + 7x
Solution:
7 + 3x = 0
7 = – 3x
x = –7/3
Zero of 7 + 3x is –7/3
p(-7/3) = 3x3 + 7x
= 3(-7/3)3 + 7(-7/3)
= 3(-343/27) – 7(7/3)
= – 343/9 – 49/3
= –490/9
So, remainder = – 490/9 which is different from 0.
Therefore, (3x + 7) is not a factor of the polynomial 3x3 + 7x
2 thoughts on “Polynomial – Exercise 2.3 – Class IX”
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