Question

If the zeroes of the polynomial f(x)=x^3-3x^2-6x+8 are of the form a-b , a , a+b. Find all the zeroes

Answer 1

Here is your answer :-

Your equation is x³ - 3x² - 6x +8

so, a = 1 , b = - 3 , c = - 6 , d = 8


sum of zeroes = -b/a

a -b +a+a +b = 3

3a = 3

a = 1

Now , product of zeroes = -d/a

(1-b) × 1 × (1+b) = -8

1 - b² = -8

-b² = -9

b² = 9

b = 3

So , all zeroes are -2 , 1 and 4

Answer 2

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