source: ncert_exercises
topic: algebra
class_level: class_10
chapter: ch2
exercise: ex2.3
difficulty: intermediate
Class 10 | Ch2: Polynomials | Exercise 2.3
Division Algorithm for Polynomials: Dividend = Divisor × Quotient + Remainder p(x) = g(x) × q(x) + r(x) where degree(r) < degree(g) or r=0.
Q1. Divide the polynomial p(x) by the polynomial g(x) and find quotient and remainder: (i) p(x)=x³-3x²+5x-3, g(x)=x²-2 (ii) p(x)=x⁴-3x²+4x+5, g(x)=x²+1-x (iii) p(x)=x⁴-5x+6, g(x)=2-x² Answer: (i) x³-3x²+5x-3 ÷ (x²-2): Quotient=x-3, Remainder=7x-9. ✅ (ii) x⁴-3x²+4x+5 ÷ (x²-x+1): Quotient=x²+x-3, Remainder=8. ✅ (iii) x⁴-5x+6 ÷ (2-x²)=x⁴-5x+6 ÷ (-x²+2): Quotient=-x²-2, Remainder=-5x+10. ✅
Q2. Check whether first polynomial is a factor of second: (i) t²-3; 2t⁴+3t³-2t²-9t-12 (ii) x²+3x+1; 3x⁴+5x³-7x²+2x+2 (iii) x³-3x+1; x⁵-4x³+x²+3x+1 Answer: (i) Divide: Remainder=0. Yes, t²-3 IS a factor. ✅ (ii) Divide: Remainder=0. Yes, IS a factor. ✅ (iii) Divide: Remainder=2. NOT a factor. ✅
Q3. Obtain all zeroes of 3x⁴+6x³-2x²-10x-5 if two zeroes are √(5/3) and -√(5/3). Answer: Two zeroes = ±√(5/3), so (x-√(5/3))(x+√(5/3))=x²-5/3 is a factor. Divide: 3x⁴+6x³-2x²-10x-5 ÷ (3x²-5) = x²+2x+1 = (x+1)². Remaining zeroes: -1,-1. All zeroes: √(5/3), -√(5/3), -1, -1. ✅
Q4. On dividing x³-3x²+x+2 by polynomial g(x), quotient and remainder are x-2 and -2x+4. Find g(x). Answer: p(x)=g(x)×q(x)+r(x). g(x)=[p(x)-r(x)]/q(x)=(x³-3x²+x+2-(-2x+4))/(x-2)=(x³-3x²+3x-2)/(x-2)=x²-x+1. ✅