JAC Class 9 Maths Solutions Chapter 2 Polynomials Ex 2.3

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

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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
(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.
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.