PSEB Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3
PSEB Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3
PSEB 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.3
Question 1.
Find the remainder when x3 + 3x2 + 3x + 1 is divided by
Answer:
The remainder theorem states that when polynomial p (x) of degree greater than or equal to 1 is divided by linear polynomial x – a, the remainder is p (a).
Here. p(x) = x3 + 3x2 + 3x + 1
(i) x + 1
Answer:
Divisor g (x) = x + 1.
Comparing x + 1 with zero, we get x = – 1.
Then, remainder
= p(- 1)
= (- 1)3 + 3(- 1)2 + 3(- 1) + 1
= – 1 + 3 – 3 + 1
= 0
(iii) x
Answer:
Divisor g (x) = x.
x = 0 gives x = 0.
Then, remainder = p (0)
= (0)3 + 3(0)2 + 3(0) + 1
= 0 + 0 + 0 + 1
= 1
(iv) x + π
Answer:
Divisor g (x) = x + π.
x + π = 0 gives x = – π.
Then, remainder
= p(- π)
= (- π)3 + 3(- π)2 + 3(- π) + 1
= – π3 + 3π2 – 3π + 1
Question 2.
Find the remainder when x3 – ax2 + 6x – a is divided by x – a.
Answer:
Here, p (x) = x3 – ax2 + 6x – a and divisor
g (x) = x – a.
x – a = 0 gives x = a.
Then, remainder = p (a)
= (a)3 – a(a)2 + 6(a) – a
= a3 – a3 + 6a – a
= 5a
Question 3.
Check whether 7 + 3x is a factor of 3x3 + 7x.
Answer:
Here, p (x) = 3x3 + 7x and divisor g (x) = 7 + 3x.
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