KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.3

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Karnataka State Syllabus Class 9 Maths Chapter 4 Polynomials Ex 4.3

Karnataka Board Class 9 Maths Chapter 4 Polynomials Ex 4.3

Question 1.
Find the remainder when x³ + 3x² + 3x + 1 is divided by
i) x + 1
ii) x−12
iii) x
iv) x + π
v) 5 + 2x
Answer:
i) p(x) = x³ + 3x² + 3x + 1 g(x) = x – 1
Let x – 1 = 0, then
x = 1.
As per Remainder theorem, r(x) = p(x) = p(a)
p(x) = x³ + 3x² + 3x + 1
p(1) = (1)³ + 3(1)² + 3(1) + 1
= 1 + 3(1) + 3(1) + 1
= 1 + 3 + 3 + 1
P(1) = 8
∴ r(x) = p(x) = 8
∴ Remainder is 8.

ii) p(x) = x³ + 3x² + 3x + 1 g(x) = x−12
If x−12=0 then x=12
p(x) = x³ + 3x² + 3x + 1

∴ r(x) = p(x) = p(a) = 158
∴ Remainder is 158

iii) p(x) = x³ + 3x² + 3x + 1
g(x) = x
If x = 0, then
p(x) = x³ + 3x² + 3x + 1
p(0) = (0)³ + 3(0)² + 3(0) + 1
= 0 + 3(0) + 3(0) + 1
= 0 + 0 + 0 + 1
p(0) = 0
Remainder r(x) = 1.

iv) p(x) = x³ + 3x² + 3x + 1
g(x) = x + π
If x + π = 0, then x = -π
p(x) = x³ + 3x² + 3x + 1
p(-π) = (-π)³ + 3(-π)² + 3(-π) + 1
p(-πt) = -π²³ – 3π² – 3π + 1
r(x) = -π³ – 3v² – 3π+1

v) p(x) = x³ + 3x² + 3x + 1
g(x) = 5 + 2x
If 5 + 2x = 0, then 2x = -5
x=−52
p(x) = x³ + 3x² + 3x + 1

Question 2.
Find the remainder when x³ – ax² + 6x – a is divided by x – a.
Answer:
p(x) = x³ – ax² + 6x – a
If g(x) = x – a, then r(x) = ?
Let x – a = 0, then x = a
p(x) = x³ – ax² + 6x – a
∴ p(a) = (a)³ – a(a)² + 6(a) – a
= a³ – a³ + 6a – a
∴ p(a) = 5a
∴ r(x) = p(a) = 5a.

Question 3.
Check whether 7 + 3x is a factor of 3x³ + 7x.
Answer:
p(x) = 3x³ + 7x
Let g(x) = 7 + 3x = 0. then
If 7 + 3x = 0, then 3x = -7
x=−73
If p(x) is divided by p(z), remainder r(x) – 0, then g(x) is a factor.
p(x) = 3x³ + 7x

Here, r(x)=−5909. This is not equal to Zero.
Hence 7 + 3x is not a factor of p(x).

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