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#### CLASS 9 MATHS CHAPTER 2 POLYNOMIALS NCERT

#### CHAPTER 2: POLYNOMIALS NCERT

### TEXTBOOKS EXERCISE – 2.3

**1. Find the remainder when x ^{3} + 3x^{2} + 3x + 1 is divided by**

(i) x + 1

(ii) x – 1/2

(iii) x (iv) x + π

(v) 5 + 2x

**Sol. **p(x) = x^{3} + 3x^{2} + 3x + 1

(i) When p(x) is divided by x + 1,

i.e., x + 1 = 0, x = –1 is to be substituted in p(x).

p(–1) = (–1)^{3 }+ 3 (–1)^{2} + 3 (–1) + 1 = – 1 + 3 – 3 + 1 = 0

Remainder = 0.

(ii)
When *p*(*x*) is divided by *x – *1/2 remainder is *p (*1/2
)__ ___ _.

*P( *1/2 ) = (1/2)^{3 } + 3 (1/2 )^{2} + 3 (1/2)+1

= 1/8 + 3/4 + 3/2 +1 = (1+ 6+ 12+ 8) /8

∴ Remainder = 27 /8 =3 3/8

(iii) When p(x) is divided by x, then remainder is p(0).

x = 0, substitute in p(x)

p(0) = 0^{3} + 3 × 0^{2} + 3 × 0 + 1 = 1.

∴ Remainder = 1

(iv) When p(x) is divided by x + π, then, remainder is p(–π).

x = –π to be substituted in p(x)

p(–π) = (–π)^{3} + 3 (–π)^{2} + 3 (–π) + 1.

∴Remainder = –π^{3} + 3π^{2} – 3π + 1

**2. Find the remainder when x**^{3}** – ax**^{2 }**+ 6x – a is divided by x – a. **

**Sol. **p(x) = x^{3} – ax^{2} + 6x – a

When p(x) is divided by x – a, the remainder is p(a).

Substitute x = a in p(x)

p(a) = a^{3} – a^{3} + 6a – a = 5a

**3. Check whether 7 + 3x is a factor of 3×3 + 7x. **

**Sol. **7 + 3x =
0 ⇒ 3x = –7 ⇒ x = −7/ 3

Substitute x = −7/ 3 in p(x) = 3x^{3} + 7x

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