Question

Find a cubic polynomial whose zeros are 1 /2, 1 and - 3

Answer 1

Let the zeroes be a, b and c.

Then the polynomial is given by,

x^3 - (a + b + c)x^2 + (ab + bc + ac)x - (abc) = 0

Here, a = 1/2

b = 1 and c = -3

So using the relation between zeroes of a polynomial,

a + b + c = 1/2 + 1 - 3

= -3/2

ab + bc + ac = 1/2 × 1 + 1 × (-3) + 1/2 × (-3)

= (-4)

abc = (-3/2)

So, the required polynomial is,

x^3 - (-3/2)x^2 + (-4)x - (-3/2) = 0

x^3 + 3/2x^2 -4x +3/2 =0

2x^3 + 3x^2 - 8x + 3 = 0

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Answer 2

Answer:

Step-by-step explanation:

Let the zeroes be a, b and c.

Then the polynomial is given by,it is in form of

x^3 - (a + b + c)x^2 + (ab + bc + ac)x - (abc) = 0

Here, a = 1/2 ,b = 1 and c = -3

So using the relation between zeroes of a polynomial,

a + b + c = 1/2 + 1 - 3

              = -3/2

ab + bc + ac = 1/2 × 1 + 1 × (-3) + 1/2 × (-3)

                     = (-4)  

              abc = (-3/2)

So, the required polynomial is,

                        x^3 - (-3/2)x^2 + (-4)x - (-3/2) = 0

                       x^3 + 3/2x^2 -4x +3/2 =0

                        2x^3 + 3x^2 - 8x + 3 = 0

hence it is required equaton

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