Jharkhand Board JAC Class 9 Maths Solutions Chapter 2 Polynomials Ex 2.3 Textbook Exercise Questions and Answers.
JAC Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.3
Page – 40
Question 1.
Find the remainder when x3 + 3x2 + 3x + 1 is divided by
(i) x + 1
(ii) x – \(\frac{1}{2}\)
(iii) x
(iv) x + π
(v) 5 + 2x
Answer:
(i) x + 1
Let p(x) = x3 + 3x2 + 3x + 1 be divided by x + 1
By remainder theorem, remainder = P(-1)
= (-1)3 + 3 (-1)2 + 3(-1) + 1
= -1 + 3 – 3 + 1
= 0
Therefore, the remainder is 0.
(ii) x – \(\frac{1}{2}\)
By remainder theorem,
Remainder = p(\(\frac{1}{2}\))
= (\(\frac{1}{2}\))3 + 3(\(\frac{1}{2}\))2 + 3(\(\frac{1}{2}\)) + 1
= \(\frac{1}{8}\) + \(\frac{3}{4}\) + \(\frac{3}{2}\) + 1
= \(\frac{27}{8}\)
Therefore, the remainder is \(\frac{27}{8}\).
(iii) x
By remainder theorem,
Remainder = p (0)
= 03 + 3(0)2 + 3(0) + 1
= 1
Therefore, the remainder is 1.
(iv) x + π
By remainder theorem,
Remainder = p (-π)
= (-π)3 + 3 (-π)2 + 3 (-π) + 1
= -π3 + 3π2 – 3π + 1
Therefore, the remainder is -π3 + 3π2 – 3π + 1.
(v) 5 + 2x
By remainder theorem,
Remainder = p(\(– \frac{5}{2}\))
= \(\left(-\frac{5}{2}\right)^3+3\left(-\frac{5}{2}\right)^2+3\left(-\frac{5}{2}\right)+1\)
= \(\frac{-125+150-60+8}{8}\)
= \(– \frac{27}{8}\)
Therefore, the remainder is \(– \frac{27}{8}\).
Question 2.
Find the remainder when x3 – ax2 + 6x – a is divided by x – a.
Answer:
Lef f(x) = x3 – ax2 + 6x – a be divided by x – a
By remainder theorem,
Remainder = f(a)
= a3 – a(a)2 + 6(a) – a
= a3 – a3 + 6a – a
= 5a
Therefore, remainder obtained is 5a when x3 – ax2 + 6x – a is divided by x – a.
Question 3.
Check whether 7+3x is a factor of 3x3 + 7x.
Ans. Let f(x) = 3x2 + 7x be divided by 7 + 3x
As 7 + 3x = 0 gives
x = \(– \frac{7}{3}\)
By Remainder theorem,
Remainder = f(\(-\frac{7}{3}\))
= \(3\left(-\frac{7}{3}\right)^3+7\left(-\frac{7}{3}\right)\)
= \(-\frac{343}{9}-\frac{49}{3}\)
= \(\frac{-343-147}{9}\)
= \(-\frac{490}{9}\) ≠ 0
As remainder is not zero, so 7 + 3x is not a factor of 3x3 + 7x.