How do you find a third degree polynomial given roots $3$ $3$ and $2 - i$ $2-i$ ?

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How do you find a third degree polynomial given roots $3$ $3$ and $2 - i$ $2-i$ ?

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Answer 1 · 2018-06-18T17:13:45

We are given roots

$x_{1} = 3$ $x_1=3$
$x_{2} = 2 - i$ $x_2=2-i$

The complex conjugate root theorem states that, if $P$ $P$ is a polynomial in one variable and $z = a + b i$ $z=a+bi$ is a root of the polynomial, then $¯ z = a - b i$ $bar z=a-bi$ , the conjugate of $z$ $z$ , is also a root of $P$ $P$ .

As such, the roots are

$x_{1} = 3$ $x_1=3$
$x_{2} = 2 - i$ $x_2=2-i$
$x_{3} = 2 - (- i) = 2 + i$ $x_3=2-(-i)=2+i$

From Vieta's formulas, we know that the polynomial $P$ $P$ can be written as:

$P_{a} (x) = a (x - x_{1}) (x - x_{2}) (x - x_{3})$ $P_a(x)=a(x-x_1)(x-x_2)(x-x_3)$

Where $a$ $a$ is a constant.

$:. P_a(x) = a(x-3)(x-2+i)(x-2-i)$

$:. P_a(x) =a(x^3-7x^2+17x-15)$

One such polynomial would be

$P_1(x) = x^3-7x^2+17x-15$

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How do you find a third degree polynomial given roots $3$ $3$ and $2 - i$ $2-i$ ?

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