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CLASS 9 MATHS CHAPTER 2 POLYNOMIALS NCERT
CHAPTER 2: POLYNOMIALS NCERT
TEXTBOOKS EXERCISE – 2.3
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
Sol. p(x) = x3 + 3x2 + 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) = 03 + 3 × 02 + 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 = –π3 + 3π2 – 3π + 1
2. Find the remainder when x3 – ax2 + 6x – a is divided by x – a.
Sol. p(x) = x3 – ax2 + 6x – a
When p(x) is divided by x – a, the remainder is p(a).
Substitute x = a in p(x)
p(a) = a3 – a3 + 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) = 3x3 + 7x
Therefore, (3x + 7) is not a factor of the polynomial 3×3 + 7x