Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case

x^{3} – 4x^{2} + 5x – 2; 2, 1, 1

#### Solution

x^{3} – 4x^{2} + 5x – 2; 2, 1, 1

p(x) = x^{3 }− 4x^{2 }+ 5x − 2 .... (1)

Zeroes for this polynomial are 2,1,1

Substitute x=2 in equation (1)

p(2) = 2^{3 }− 4 × 2^{2 }+ 5 × 2 − 2

= 8 − 16 + 10 − 2 = 0

Substitute x=1 in equation (1)

p(1) = x^{3} − 4x^{2} + 5x − 2

= 1^{3 }− 4(1)^{2 }+ 5(1) − 2

= 1 − 4 + 5 − 2 = 0

Therefore, 2,1,1 are the zeroes of the given polynomial.

Comparing the given polynomial with ax^{3 }+ bx^{2 }+ cx + d we obtain,

a = 1, b = −4, c = 5, d = −2

Let us assume α = 2, β = 1, γ = 1

Sum of the roots = α + β + γ = 2 + 1 + 1 = 4 = `- (-4)/1 (-"b")/"a"`

Multiplication of two zeroes taking two at a time = αβ + βγ + αγ = (2)(1) + (1)(1) + (2)(1) = 5 = `5/1 = "c"/"a"`

Product of the roots = αβγ = 2 × 1 × 1 = 2 = `−(-2)/1="d"/"a"`