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  • Given that the zeroes of the cubic polynomial x^3 – 6x^2 + 3x + 10 are of the form a, a + b, a + 2b for some real numbers a and b, find the values of a and b as well as the zeroes of the given polynomial.

Given that the zeroes of the cubic polynomial x^3 -6x^2 + 3x + 10 are of the form a, a + b, a + 2b for some real numbers a and b, find the values of a and b as well as the zeroes of the given polynomial.

Answers (1)

Answer. [5, 2, –1]         

Solution. Here the given cubic polynomial is x^3 -6x^2 + 3x + 10

A = 1, B = –6, C = 3, D = 10

Given that, \alpha = a, \beta = a + b, \gamma = a + 2b

We know that, \alpha +\beta +\gamma =\frac{-B }{A}

a + a + b + a + 2b =\frac{-(-6)}{1}

3a + 3b = 6

3(a + b) = 6

a + b = 2                                  …..(1)

\alpha \beta +\beta \gamma +\alpha = \frac{C}{A}

a(a + b) + (a + b) (a + 2b) + (a + 2b) (a) =\frac{3}{1}

a(2) + 2(a + 2b) + a^2 + 2ab = 3     (Q  a + b = 2)

2a + 2a + 4b + a^2 + 2ab = 3

4a + 4b + a^2 + 2ab = 3

4a + 4(2 - a) + a^2 + 2a(2 - a) = 3               (Q  a + b = 2)

4a + 8 - 4a + a^2 + 4a - 2a^2 = 3

a^2 + 4a - 2a^2 + 8 - 3 = 0

-a^2 + 4a + 5 = 0

a^2 - 4a - 5 = 0

a^2 - 5a + a - 5 = 0

a(a - 5)+1 (a - 5) = 0

(a -5) ( a + 1) = 0

a = 5, –1

Put a = 5 in (1)                                    put a = –1 in (1)

5 + b = 2                                              –1 + b = 2

b = –3                                                  b = 3

Hence, value of a = 5, b = –3 and a = –1, b = 3

put a = 5, b = –3 and we get zeroes

a = 5

a + b = 5 – 3 = 2

a + 2b = 5 + 2(–3) = –1

Hence, zeroes are 5, 2, –1.

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