Polynomial – Exercise 2.3 – Class IX

1. Find the remainder when x³ + 3x² + 3x + 1 is divided by

(i) x + 1

(ii) x – ¹/₂

(iii) x

(iv) x + π

(v) 5 + 2x

Solution:

(i) x + 1

p(x) = x³ + 3x² + 3x + 1 and the zero of x + 1 is – 1

So, p(1) = (-1)³ + 3(-1)² + 3(-1) + 1

= -1 + 3 – 3 + 1

= 0

So, by the remainder theorem, remainder obtained on dividing p(x) by x + 1 is 0.

So, p(x)  is the multiple of x + 1

 

(ii) x – ¹/₂

p(x) = x³ + 3x² + 3x + 1 and zero of x – ¹/₂ is ¹/₂

p(¹/₂) = (¹/₂)³ + 3(¹/₂)² + 3(¹/₂) + 1

= ¹/₈  + 3*¹/₄ +  3*¹/₂ + 1

= ¹/₈ + ³/₄ + ³/₂ + 1

= ^1+6+12+8/₈ = ²⁷/₈

So, by the remainder theorem, remainder obtained on dividing p(x) by x – ¹/₂ is ²⁷/₈.

 

(iii) x

p(x) = x³ + 3x² + 3x + 1 and zero of x  is 0

p(0) = (0)³ + 3(0)² + 3(0) + 1

= 0  + 3*0 +  3*0 + 1

= 1

So, by the remainder theorem, remainder obtained on dividing p(x) by x  is 1.

 

(iv) x + π

p(x) = x³ + 3x² + 3x + 1 and zero of x + π is -π

p(-π) = (-π)³ + 3(-π)² + 3(-π) + 1

= -π³  + 3π² –  3π + 1

So, by the remainder theorem, remainder obtained on dividing p(x) by x + π  is = -π³  + 3π² –  3π + 1.

 

(v) 5 + 2x

p(x) = x³ + 3x² + 3x + 1 and zero of 5 + 2x  is –⁵/₂

p(-⁵/₂) = (-⁵/₂)³ + 3(-⁵/₂)² + 3(-⁵/₂) + 1

= –¹²⁵/₈ +3*²⁵/₄ – 3*⁵/₂ + 1

= ^{-125+150-60+8}/₈

= ‑²⁷/₈

So, by the remainder theorem, remainder obtained on dividing p(x) by 5 + 2x  is –²⁷/₈.

---

2. Find the remainder when x³ – ax² + 6x – a is divided by x – a

Solution:

p(x) = x³ – ax²  + 6x – a and zero of x –  a is a

p(1) = a³ – a(a)² + 6(a) – a

= a³ – a³ + 6a – a

= 5a

So, by the remainder theorem, remainder obtained on dividing p(x) by x – a  is 5a.

---

3. Check whether 7 + 3x is a factor of 3x³ + 7x

Solution:

7 + 3x = 0

7 = – 3x

x = –⁷/₃

Zero of 7 + 3x is –⁷/₃

p(-⁷/₃) =  3x³ + 7x

= 3(-⁷/₃)³ + 7(-⁷/₃)

= 3(-³⁴³/₂₇) – 7(⁷/₃)

= – ³⁴³/₉ – ⁴⁹/₃

= –⁴⁹⁰/₉

So, remainder = – ⁴⁹⁰/₉ which is different from 0.

Therefore, (3x + 7) is not a factor of the polynomial 3x³ + 7x

---