Question

find the zeros of the polynomial f(x)= x^3-5x^2-16x+80, if its two zeros are equal in magnitude but opposite in sign

Answer 1

Let α, β, Δ be the roots of given polynomial x^3-5x^2-16x+80.

Given 2 of its roots are equal but of opposite signs

Let α,β be equal roots

such that β = -α

From this polynomial,

Sum of the roots ⇒ α+β+Δ = -coefficient of x²/ coefficient of x³

⇒ α-α+Δ = -(-5)/1

Δ = 5 --------- One of the root (1)

Product of roots ⇒ αβΔ = -constant/coefficient of x³

⇒ α(-α)Δ = -80/1

⇒ -α²Δ = -80

⇒ α²Δ = 80

From(1), Substituting Δ=5

⇒ α²×5 = 80

⇒ α² = 80/5

⇒ α² = 16

⇒ α = √16 = +-4

⇒ α = +4 or -4 {where -4 = -α = β}

Therefore, the three roots of given polynomial x^3-5x^2-16x+80 α,β,Δ are 4,-4,5