Question

Find the zeros of the polynomial f(x)=x3-12x2+39x-28,if it is given that the zeros are in A. P

Answer 1

ANSWER :

Zeroes of the Polynomials are :

• If α and β are the zeroes of the Quadratic polynomial ax³ + bx + c then ,

       α + β  =  - b/a

       αβ      =    c/a

    Sum of Zeroes  = α + β = -b/a =  - ( Coefficient of x ) /  

    Coefficient of x²

    Product of Zeroes = αβ = c/a = Constant term / Coefficient

    of x²

• If α , β and γ are the zeroes of the Cubic Polynomial ax³ + bx³ + cx + d, then :

    α + β + γ       =   - b / a,

   αβ + βγ + γα  =  c/a,

   and      αβγ    =  - d/a

→

Let , α = a-d , β = a and γ = a + d be the zeroes of the polynomial.

f ( x ) = x³ - 12x² + 39x - 28.

Therefore , α + β + γ = - [ - 12 / 1 ] = 12

And, αβγ = - [ - 28 / 1 ] = 28.

( a - d ) + a + ( a + d ) = 12

And ( a - d ) a ( a + d ) = 28.

3a = 12 and a ( a² - d² ) = 28

a = 4 and 4 ( 16 - d² ) = 28

a = 4 and 16 - d² =  7

a =  4 and d² = 9

a= 4 and d = ± 3

∴ α = a-d = 4 - 3 = 1

β = 4  (value of  a )

and γ = a + d = 7

Final answers :

α = 7

β = 4

And,

γ = 1

Hence, the zeroes of the given polynomial f ( x ) = x³ - 12x² + 39x - 28 are 1 , 4 and 7.

Answer 2

$\purple{\sf{Answer:-}}$

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Let , α = a-d , β = a and γ = a + d be the zeroes of the polynomial.

p( x ) = x³ - 12x² + 39x - 28.

$\red{\sf{--------------------------------}}$

α + β + γ = - (- 12/1) = 12

α + β + γ = - (- 12/1) = 12αβγ = - (- 28/1) = 28

$\red{\sf{--------------------------------}}$

(a - d) + (a) + (a + d) = 12

And (a - d)×a×(a + d) = 28

3a = 12 and a(a² - d²) = 28

a = 4 and 4 ( 16 - d² ) = 28

a = 4 and 16 - d² =  7

a =  4 and d² = 9

a = 4 and d = ± 3

$\red{\sf{--------------------------------}}$

∴ α = a-d = 4 - 3 = 1

∴ β = 4 (value of  a)

∴ γ = a + d = 7

$\red{\sf{--------------------------------}}$

α = 7, β = 4 & γ = 1

$\red{\sf{--------------------------------}}$