Solving polynomials algebraicly using roots (cubic)
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Solving polynomials algebraicly using roots (cubic)

[From: ] [author: ] [Date: 11-05-31] [Hit: ]
β = -2,These roots do indeed each satisfy the equation x^3-3x^2-4x+12=0, so this is the correct answer.The roots are ± 2 and 3.......
x^3-3x^2-4x+12=0 given that the sum of two of its roots are zero: solve this equation.

let roots be α β γ δ

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The roots of a polynomial are associated with its factors:
x^3 - 3x^2 - 4x + 12 = (x - α)(x - β)(x - γ)

From the rules of polynomial multiplication, you know that the product of the roots must be 12.
αβγ = -12
Meanwhile you are given that two roots sum to zero.
α = -β
putting these together:
α^2 γ = 12

Let's start by looking for integer roots.
12 = 2^2 * 3, so the first combination I would test is α = 2, β = -2, γ = 3
These roots do indeed each satisfy the equation x^3-3x^2-4x+12=0, so this is the correct answer.

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Let's just factor it: x^2(x - 3) - 4(x - 3) = 0
(x^2 - 4)(x - 3) = 0
(x + 2)(x - 2)(x - 3) = 0

The roots are ± 2 and 3.
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keywords: Solving,polynomials,using,roots,algebraicly,cubic,Solving polynomials algebraicly using roots (cubic)
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