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You should notice that the polynomial has a complex zero, hence the polynomial will have as zero the conjugate of the complex root.
Hence, if `x = 2+i` is a root for the polynomial, then `x = 2-i` is also a root.
You need to remember that you may write the polynomial in factored form if the roots are given.
The problem provides the information that the polynomial is of third order, hence you need to know three roots to write the factored form.
The problem provides two roots and one of the roots, the complex root provides the third root, hence the polynomial is:
`f(x) = (x-x_1)(x-x_2)(x-x_3)`
`f(x) = (x-4)(x-(2+i))(x-(2-i))`
`f(x) = (x-4)(x-2-i)(x-2+i)`
You should substitute the product `(x-2-i)(x-2+i)` by the difference of squares `(x-2)^2 - i^2` .
Expanding the square and substituting -1 for`i^2` yields:
`f(x) = (x-4)(x^2 - 2x + 4 - (-1))`
`f(x) = (x-4)(x^2 - 2x + 5)`
Opening the brackets yields:
`f(x) = x^3 - 2x^2 + 5x - 4x^2 + 8x - 20`
Collecting like terms yields:
`f(x) = x^3 - 6x^2+ 13x - 20`
Hence, evaluating the third order polynomial under the given conditions yields `f(x) = x^3 - 6x^2+ 13x - 20` .
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