Question

find the zeroes of the polynomial f(x)=x^3 - 5x^3 - 2x + 24 , if it is given that the product of it's two zeroes is 12.

Answer 1

f(x)=x3−5x2−2x+24

Letαβ=12

α+β+γ=−(−5)=5

αβ+βγ+αγ=−21=−2

αβγ=−241=−24

αβ=12⟹12γ=−24=γ=−2

α+β+γ=5

α+β−2=5⟹α+β=7

(α−β)2=(α+β)2−4αβ=72−4×12⟹α−β=±1

α+β=7andα−β=1

(or)α+β=7andα−β=1

On solving we get:

α=4,β=3(or)α=3,β=4

Hence the zeroes of the polynominal are 3, 4 and -2

Answer 2

Answer:

Let α,β and y be the zeros of polynomial f(x) such that ab=12

We have, α+β+y=

a

−b

=

1

−(−5)

=5

αβ+βy+yα=

a

c

=

1

−2

=−2 and αβy=

a

−d

=

1

−24

=−24

Putting αβ=12 in αβy=−24, we get

12y=−24 ⇒ y=−

12

24

=−2

Now ,α+β+y=5 ⇒ α+β−2=5

⇒ α+β=7 ⇒ α=7−β

∴αβ=12

⇒(7−β)β=12 ⇒

7β−β

2

=12

⇒β

2

−7β+12=0 ⇒ β

2

−3β−4β+12=0

⇒β=4 or β=3

∴α=3 or α=4