Question

find the zeroes of the polynomial x³-5x²-2x+24,if it given that the product of its two zeroes is 12

Answer 1

The sum of the zeroes of a cubic polynomial is -b/a, where b is the coefficient of x² and a is the coefficient of x³.
let α, β, gamma be the zeroes of the polynomial.
then α+β+gamma = 5 ----(1)
product of the zeroes of cubic polynomial is -d/a, where d is the constant term.
therefore, αβgamma = -24  ---- (2)
given that αβ = 12 ----(3)
using (3) in (2) we get 12gamma = -24
⇒ gamma = -2
 also from (1) α+β+(-2) = 5
⇒ α+β = 7 ---- (4)
(α+β)² = (α-β)² + 4αβ
7² = (α-β)² + 4 × 12
49 = (α-β)² + 48
(α-β)² = 1
⇒ α-β = 1 ----- (5)
solving (4) and (5) we get α = 4 and β = 3
therefore 4, 3 and -2 are the zeroes of the given polynomial.

Answer 2

Hi mate,

x³-5x²-2x+24

compare with ax³+bx²+cx+d

α+β+y=-b/a

=5

αβy=-d/a

=-24

12y= -24

(as given the product of two. zeros ie αβ=12)

y= -2.

α+β+y=5

α+β-2=5

α+β =7. .................(1)

(α+β)²=7²

(α-β)²+4αβ=49

(α-β)²+4*12=49

(α-β)²+48= 49

(α-β)² =1

α-β = √1

α-β=1. ....................(2)

subtracting (2) From (1)

α+β-(α-β)=7-1

α+β-α+β=6

2β =6

β=3

putting the value of β in the (2) eq

α-3=1

α=4

α=4,β=3,y=-2

product of α and β = 4 * 3 = 12

I hope it helps you.