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NAME – Ashish Bansal
CLASS- X - D
On the basis of degree
A real number α is a zero of a polynomial
f(x), if f(α) = 0.
e.g. f(x) = x³ - 6x² +11x -6
f(2) = 2³ -6 X 2² +11 X 2 – 6
= 0 .
Hence 2 is a zero of f(x).
 The number of curves tells the degree.
 The number of time it cut the x-axis that
are its zeros.
 Let α, β and γ be the zeroes of the polynomial ax³ + bx² + cx + d
 Then, sum of zeroes(α+β+γ) = -b = -(coefficient of x²)
a coefficient of x³
αβ + βγ + αγ = c = coefficient of x
a coefficient of x³
Product of zeroes (αβγ) = -d = -(constant term)
a coefficient of x³
I) Find the zeroes of the polynomial x² + 7x + 12and verify the relation between
the zeroes and its coefficients.
f(x) = x² + 7x + 12
= x² + 4x + 3x + 12
=x(x +4) + 3(x + 4)
=(x + 4)(x + 3)
Therefore,zeroes of f(x) =x + 4 = 0, x +3 = 0 [ f(x) = 0]
x = -4, x = -3
Hence zeroes of f(x) are α = -4 and β = -3.
2) Find a quadratic polynomial whose
zeroes are 4, 1.
sum of zeroes,α + β = 4 +1 = 5 = -b/a
product of zeroes, αβ = 4 x 1 = 4 = c/a
therefore, a = 1, b = -4, c =1
as, polynomial = ax² + bx +c
= 1(x)² + { -4(x)} + 1
= x² - 4x + 1
Polynomials 120726050407-phpapp01 (1)
Polynomials 120726050407-phpapp01 (1)
Polynomials 120726050407-phpapp01 (1)

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Polynomials 120726050407-phpapp01 (1)

  • 1. NAME – Ashish Bansal CLASS- X - D
  • 2.
  • 3.
  • 4. On the basis of degree
  • 5.
  • 6. A real number α is a zero of a polynomial f(x), if f(α) = 0. e.g. f(x) = x³ - 6x² +11x -6 f(2) = 2³ -6 X 2² +11 X 2 – 6 = 0 . Hence 2 is a zero of f(x).
  • 7.  The number of curves tells the degree.  The number of time it cut the x-axis that are its zeros.
  • 8.
  • 9.  Let α, β and γ be the zeroes of the polynomial ax³ + bx² + cx + d  Then, sum of zeroes(α+β+γ) = -b = -(coefficient of x²) a coefficient of x³ αβ + βγ + αγ = c = coefficient of x a coefficient of x³ Product of zeroes (αβγ) = -d = -(constant term) a coefficient of x³
  • 10. I) Find the zeroes of the polynomial x² + 7x + 12and verify the relation between the zeroes and its coefficients. f(x) = x² + 7x + 12 = x² + 4x + 3x + 12 =x(x +4) + 3(x + 4) =(x + 4)(x + 3) Therefore,zeroes of f(x) =x + 4 = 0, x +3 = 0 [ f(x) = 0] x = -4, x = -3 Hence zeroes of f(x) are α = -4 and β = -3.
  • 11.
  • 12. 2) Find a quadratic polynomial whose zeroes are 4, 1. sum of zeroes,α + β = 4 +1 = 5 = -b/a product of zeroes, αβ = 4 x 1 = 4 = c/a therefore, a = 1, b = -4, c =1 as, polynomial = ax² + bx +c = 1(x)² + { -4(x)} + 1 = x² - 4x + 1