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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 x^{3} + 3x^{2} + 3x + 1 is divided by

i) x + 1

ii) x−12

iii) x

iv) x + π

v) 5 + 2x

Answer:

i) p(x) = x^{3} + 3x^{2} + 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) = x^{3} + 3x^{2} + 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) = x^{3} + 3x^{2} + 3x + 1 g(x) = x−12

If x−12=0 then x=12

p(x) = x^{3} + 3x^{2} + 3x + 1

∴ r(x) = p(x) = p(a) = 158

∴ Remainder is 158

iii) p(x) = x^{3} + 3x^{2} + 3x + 1

g(x) = x

If x = 0, then

p(x) = x^{3} + 3x^{2} + 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) = x^{3} + 3x^{2} + 3x + 1

g(x) = x + π

If x + π = 0, then x = -π

p(x) = x^{3} + 3x^{2} + 3x + 1

p(-π) = (-π)^{3} + 3(-π)^{2} + 3(-π) + 1

p(-πt) = -π^{23} – 3π^{2} – 3π + 1

r(x) = -π^{3} – 3v^{2} – 3π+1

v) p(x) = x^{3} + 3x^{2} + 3x + 1

g(x) = 5 + 2x

If 5 + 2x = 0, then 2x = -5

x=−52

p(x) = x^{3} + 3x^{2} + 3x + 1

Question 2.

Find the remainder when x^{3} – ax^{2} + 6x – a is divided by x – a.

Answer:

p(x) = x^{3} – ax^{2} + 6x – a

If g(x) = x – a, then r(x) = ?

Let x – a = 0, then x = a

p(x) = x^{3} – ax^{2} + 6x – a

∴ p(a) = (a)^{3} – a(a)^{2} + 6(a) – a

= a^{3} – a^{3} + 6a – a

∴ p(a) = 5a

∴ r(x) = p(a) = 5a.

Question 3.

Check whether 7 + 3x is a factor of 3x^{3} + 7x.

Answer:

p(x) = 3x^{3} + 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) = 3x^{3} + 7x

Here, r(x)=−5909. This is not equal to Zero.

Hence 7 + 3x is not a factor of p(x).

**All Chapter KSEEB Solutions For Class 9 Maths**

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