Complete Study Material | CBSE Board Exam Preparation
Polynomial: An expression of the form p(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀, where aₙ, aₙ₋₁, ..., a₁, a₀ are real numbers and aₙ ≠ 0.
Degree of Polynomial: The highest power of x in the polynomial.
Types of Polynomials:
- Linear Polynomial: Degree 1 (Example: ax + b)
- Quadratic Polynomial: Degree 2 (Example: ax² + bx + c)
- Cubic Polynomial: Degree 3 (Example: ax³ + bx² + cx + d)
Zero of Polynomial: A real number k is said to be a zero of polynomial p(x) if p(k) = 0.
Solution:
(i) 4x² - 3x + 7
✅ Yes, it is a polynomial. Powers of x are whole numbers (2,1,0).
(ii) y² + √2
✅ Yes, it is a polynomial. Powers of y are whole numbers (2,0).
(iii) 3√t + t√2
❌ No, power of t is 1/2 (not a whole number).
(iv) x + 1/x
❌ No, power of x is -1 (not a whole number).
(v) x¹⁰ + y³ + t⁵⁰
❌ No, it has three variables x, y, t (not in one variable).
Solution:
(i) 2 + x² + x
Coefficient of x² = 1
(ii) 2 - x² + x³
Coefficient of x² = -1
(iii) (π/2)x² + x
Coefficient of x² = π/2
(iv) √2x - 1
Coefficient of x² = 0 (no x² term)
Solution:
Binomial of degree 35: x³⁵ + 1
Monomial of degree 100: x¹⁰⁰
Solution:
(i) 5x³ + 4x² + 7x
Highest power = 3 → Degree = 3
(ii) 4 - y²
Highest power = 2 → Degree = 2
(iii) 5t - √7
Highest power = 1 → Degree = 1
(iv) 3
Highest power = 0 → Degree = 0
Solution:
(i) x² + x → Quadratic (degree 2)
(ii) x - x³ → Cubic (degree 3)
(iii) y + y² + 4 → Quadratic (degree 2)
(iv) 1 + x → Linear (degree 1)
(v) 3t → Linear (degree 1)
(vi) r² → Quadratic (degree 2)
(vii) 7x³ → Cubic (degree 3)
Solution:
p(x) = 5x - 4x² + 3
(i) x = 0: p(0) = 5(0) - 4(0)² + 3 = 0 - 0 + 3 = 3
(ii) x = -1: p(-1) = 5(-1) - 4(1) + 3 = -5 - 4 + 3 = -6
(iii) x = 2: p(2) = 5(2) - 4(4) + 3 = 10 - 16 + 3 = -3
Solution:
(i) p(y) = y² - y + 1
p(0) = 0 - 0 + 1 = 1
p(1) = 1 - 1 + 1 = 1
p(2) = 4 - 2 + 1 = 3
(ii) p(t) = 2 + t + 2t² - t³
p(0) = 2 + 0 + 0 - 0 = 2
p(1) = 2 + 1 + 2 - 1 = 4
p(2) = 2 + 2 + 8 - 8 = 4
(iii) p(x) = x³
p(0) = 0, p(1) = 1, p(2) = 8
(iv) p(x) = (x-1)(x+1)
p(0) = (-1)(1) = -1
p(1) = (0)(2) = 0
p(2) = (1)(3) = 3
Solution:
(i) p(x) = 3x + 1, x = -1/3
p(-1/3) = 3(-1/3) + 1 = -1 + 1 = 0 → Yes, zero
(ii) p(x) = 5x - π, x = 4/5
p(4/5) = 5(4/5) - π = 4 - π ≠ 0 → No
(iii) p(x) = x² - 1, x = 1, -1
p(1) = 1 - 1 = 0 ✓, p(-1) = 1 - 1 = 0 ✓ → Yes
(iv) p(x) = (x+1)(x-2), x = -1, 2
p(-1) = 0, p(2) = 0 → Yes
Solution:
(i) x² - 2x - 8
= x² - 4x + 2x - 8 = x(x-4) + 2(x-4) = (x-4)(x+2)
Zeros: x = 4, x = -2
Sum = 4 + (-2) = 2 = -b/a = 2 ✓
Product = 4 × (-2) = -8 = c/a = -8 ✓
(ii) 4s² - 4s + 1
= (2s)² - 2(2s)(1) + 1² = (2s-1)²
Zero: s = 1/2 (repeated)
Sum = 1 = -b/a = 1 ✓, Product = 1/4 = c/a = 1/4 ✓
(iii) 6x² - 3 - 7x
= 6x² - 7x - 3 = 6x² - 9x + 2x - 3 = 3x(2x-3) + 1(2x-3) = (2x-3)(3x+1)
Zeros: x = 3/2, x = -1/3
Sum = 3/2 + (-1/3) = 7/6 = -b/a = 7/6 ✓
Product = -1/2 = c/a = -3/6 = -1/2 ✓
(iv) 4u² + 8u
= 4u(u + 2)
Zeros: u = 0, u = -2
Sum = -2 = -b/a = -2 ✓, Product = 0 = c/a = 0 ✓
(v) t² - 15
= (t - √15)(t + √15)
Zeros: t = √15, t = -√15
Sum = 0 = -b/a = 0 ✓, Product = -15 = c/a = -15 ✓
(vi) 3x² - x - 4
= 3x² - 4x + 3x - 4 = x(3x-4) + 1(3x-4) = (3x-4)(x+1)
Zeros: x = 4/3, x = -1
Sum = 1/3 = -b/a = 1/3 ✓, Product = -4/3 = c/a = -4/3 ✓
Solution:
(i) p(x) = x³ - 3x² + 5x - 3, g(x) = x² - 2
Quotient = x - 3, Remainder = 7x - 9
(ii) p(x) = x⁴ - 3x² + 4x + 5, g(x) = x² + 1 - x
Quotient = x² + x - 3, Remainder = 8
(iii) p(x) = x⁴ - 5x + 6, g(x) = 2 - x²
Quotient = -x² - 2, Remainder = -5x + 10
Solution:
(i) t² - 3, 2t⁴ + 3t³ - 2t² - 9t - 12
By division, remainder = 0 → Yes, factor
(ii) x² + 3x + 1, 3x⁴ + 5x³ - 7x² + 2x + 2
By division, remainder ≠ 0 → No
(iii) x³ - 3x + 1, x⁵ - 4x³ + x² + 3x + 1
By division, remainder = 0 → Yes, factor
Solution:
Given zeros: √(5/3) and -√(5/3)
Factor: (x² - 5/3)
Dividing polynomial by 3x² - 5, we get x² + 2x + 1 = (x+1)²
Other zeros: x = -1, -1
All zeros: √(5/3), -√(5/3), -1, -1
Solution:
p(x) = g(x) × q(x) + r(x)
x³ - 3x² + x + 2 = g(x) × (x-2) + (-2x+4)
x³ - 3x² + x + 2 + 2x - 4 = g(x) × (x-2)
x³ - 3x² + 3x - 2 = g(x) × (x-2)
g(x) = (x³ - 3x² + 3x - 2) ÷ (x-2) = x² - x + 1
Solution:
Example 1: p(x) = x³ + x² + x + 1, g(x) = x² - 1
q(x) = x + 1, r(x) = 2x + 2
Example 2: p(x) = x⁴ + 2x² + 1, g(x) = x² + 1
q(x) = x² + 1, r(x) = 0
Solution:
(i) p(x) = 2x³ + x² - 5x + 2, zeros: 1/2, 1, -2
p(1/2) = 2(1/8)+1/4-5/2+2 = 0 ✓
p(1) = 2+1-5+2 = 0 ✓
p(-2) = -16+4+10+2 = 0 ✓
Sum = 1/2+1-2 = -1/2 = -b/a = -1/2 ✓
Sum of products = -2 + (-1) + 1/2 = -5/2 = c/a = -5/2 ✓
Product = -1 = -d/a = -1 ✓
Solution:
(i) Sum = 2, Sum of products = -7, Product = -14
Polynomial = x³ - 2x² - 7x + 14
(ii) Sum = 5/2, Sum of products = -11/2, Product = -30
Polynomial = 2x³ - 5x² - 11x + 60
(iii) Sum = -1/2, Sum of products = -1/2, Product = 1/6
Polynomial = 6x³ + 3x² + 3x + 1
Solution:
Sum = (a-b) + a + (a+b) = 3a = 3 → a = 1
Product = (a-b)(a)(a+b) = a(a² - b²) = 1(1 - b²) = -1
1 - b² = -1 → b² = 2 → b = ±√2
a = 1, b = √2 or -√2
Solution:
Zeros: 2+√3, 2-√3
Factor: [x - (2+√3)][x - (2-√3)] = x² - 4x + 1
Dividing polynomial by x² - 4x + 1 gives x² - 2x - 35 = (x-7)(x+5)
Other zeros: 7 and -5
Solution:
After division, remainder = (2k-9)x + (8-5k-k²)
Given remainder = x + a
2k - 9 = 1 → 2k = 10 → k = 5
8 - 5(5) - 25 = 8 - 25 - 25 = -42 → a = -42
Solution:
x² + 7x + 10 = (x + 2)(x + 5)
Zeros: α = -2, β = -5
Sum = -2 + (-5) = -7 = -b/a ✓
Product = (-2)(-5) = 10 = c/a ✓
For Quadratic Polynomial ax² + bx + c:
Sum of zeros (α + β) = -b/a
Product of zeros (αβ) = c/a
Quadratic polynomial = x² - (α+β)x + αβ
For Cubic Polynomial ax³ + bx² + cx + d:
Sum of zeros = -b/a
Sum of product of zeros taken two at a time = c/a
Product of zeros = -d/a
✅ Polynomial: Expression with non-negative integer powers
✅ Degree: Highest power of variable
✅ Zero of Polynomial: Solution of p(x) = 0
✅ Division Algorithm: p(x) = g(x) × q(x) + r(x)
✅ Relationship between zeros and coefficients