PSEB 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.3

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 2 Polynomials Ex 2.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths 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

(ii) x – \(\frac{1}{2}\)
Answer:
Divisor g (x) = x – \(\frac{1}{2}\)
x – \(\frac{1}{2}\) = 0 gives x = \(\frac{1}{2}\)
Then, remainder
= p\(\left(\frac{1}{2}\right)\)
= \(\left(\frac{1}{2}\right)^{3}+3\left(\frac{1}{2}\right)^{2}+3\left(\frac{1}{2}\right)+1\)
= \(\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+1\)
= \(\frac{27}{8}\)

(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

(v) 5 + 2x
Answer:
Divisor g(x) = 5 + 2x.
5 + 2x = 0 gives x = – \(\frac{5}{2}\)
Then, remainder
= P\(\left(-\frac{5}{2}\right)\)
= \(\left(-\frac{5}{2}\right)^{3}+3\left(-\frac{5}{2}\right)^{2}+3\left(-\frac{5}{2}\right)+1\)
= \(-\frac{125}{8}+\frac{75}{4}-\frac{15}{2}+1\)
= \(\frac{-125+150-60+8}{8}\)
= \(-\frac{27}{8}\)

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.
7 + 3x = 0 gives x = –\(\frac{7}{3}\).
Then, remainder = p\(\left(-\frac{7}{3}\right)\)
= \(3\left(-\frac{7}{3}\right)^{3}+7\left(-\frac{7}{3}\right)\)
= \(-\frac{343}{9}-\frac{49}{3}\)
= \(\frac{-343-147}{9}\)
= – \(\frac{490}{9}\) ≠ 0
Since the remainder is not zero when
p (x) = 3x³ + 7x is divided by 7 + 3x, it is clear that 7 + 3x is not a factor of 3x³ + 7x.