Question

find the zero polynamial of xcube​

Answer 1

Answer:

x^3-4x=x (x^2-4)

x=0 & x=2,-2

the zeroes are 0,-2,2

Answer 2

Answer:

Let α,β,γ be the zeroes of polynomial f(x) = x3 - 5x2 - 2x + 24 such that αβ = 12.

α+β+γ=−(−51)=5 ................ (i)

αβ+βγ+γα=−21=−2

and, αβγ=−241=−24

Putting, αβ=12 in αβγ=−24, we get

12γ=−24

⇒γ=−2412=−2

Putting γ=−2 in eq.(i), we get

α+β−2=5

⇒α+β=7

Now, (α−β)2=(α+β)2−4αβ

⇒(α−β)2=72−4×12

⇒(α−β)2=1

⇒α−β=±1

Thus, we have

α+β=7  and α−β=1 or, α+β=7 and α−β=−1

CASE I: When α+β=7 and α−β=1

Solving α+β=7 and α−β=1 , we get

α=4 and β=3  

CASE II: When α+β=7 and α−β=−1

Solving α+β=7 and α−β=−1 , we get

α=3 and β=4 .

Hence, the zeros of the given polynomial are 3, 4 and - 2 or 4,3 , -2.