Question

find the zeros of the polynomial f(x)=x3-5x2-2x+24 if it is GIven that product of its two zeros is 12

Answer 1

Let the three Zeros of the equation are a,b and c
given , ab=12

sum of zeros
a+b+c = 5. eq1
sum of product of zeros
ab+bc+ca = -2 eq2
product of zeros
abc=-24 eq 3
using eq 3 and given data
(ab)c=-24
(12)c=-24
c=-24/12
c=-2
one zero of this equation is-2
using eq1 we have
a+b=7 eq4
using eq 4 and ab= 12
we have 3,4 is the correct answer.

so all the three zeros of this equation is
-2,3 and 4.

Answer 2

==> In the question

f(x) = x³ - 5x² - 2x + 24

And we have given that the product of it's two zeroes is 12 .

So, let the zeroes of the given cubic polynomial be φ , β and γ .

From the given condition we have,

φβ = 12 ..................(1)

and also we have ,

φ + β + γ = coefficient of x²/coefficient of x³ = 5 ......................(2)

φβγ = - constant term/ coefficient of x³ = -24 ...........................(3)

Substituting the value of φβ in equation no. (3) we get,

φβγ = -24

12γ = -24

γ = -24/12

γ = -2 ......................(4)

Putting the value of γ = -2 in equation no. (2) we get

φ + β + γ = 5

φ + β + (-2) = 5

φ + β = 5 +2

φ + β = 7 ...................(5)

Now, squaring on both sides we will get ,

(φ + β)² = (7)²

We know the expression of quadratic polynomial [ (φ + β)² = (φ - β)²+ 4φβ) ]

∴ ( φ - β )² + 4*12 = 49 [∵ φβ = 12 ]

(φ - β)² + 48 = 49

( φ - β)² = 49 - 48

(φ - β)² = 1

∴ φ - β = 1 .................(6)

Now, we will add the equation no. (5) and equation no. (6) we get,

φ + β = 7

φ - β = 1

----------------

2φ = 8

φ = 8/2

φ = 4

Putting φ = 4 in equation no. (5) we get,

φ - β = 1

4 - β = 1

- β = 1 - 4

-β = - 3 .......(multiplying by -1 on both side)

β = 3

∴ The zeroes are φ, β, γ = 4, 3, -2