How do you find the polynomial function with roots $2 + 3 \sqrt{3}$ $2 + 3sqrt3$ ?

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How do you find the polynomial function with roots $2 + 3 \sqrt{3}$ $2 + 3sqrt3$ ?

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Answer 1 · 2017-08-01T14:26:33

$f (x) = x^{2} - 4 x - 23$ $f(x)=x^2-4x-23$

Explanation:

If a function has the root $a + \sqrt{b}$ $a+sqrtb$ , it must also have the root $a - \sqrt{b}$ $a-sqrtb$ .

So, this function actually has the roots $2 + 3 \sqrt{3}$ $2+3sqrt3$ and $2 - 3 \sqrt{3}$ $2-3sqrt3$ . Since these are roots, when $x$ $x$ takes on these values, $f (x) = 0$ $f(x)=0$ . Thus, each factor of the polynomial will be $x -$ $x -$ the root.

$f (x) = (x - (2 + 3 \sqrt{3})) (x - (2 - 3 \sqrt{3}))$ $f(x)=(x-(2+3sqrt3))(x-(2-3sqrt3))$

$f (x) = (x - 2 - 3 \sqrt{3}) (x - 2 + 3 \sqrt{3})$ $f(x)=(x-2-3sqrt3)(x-2+3sqrt3)$

$f (x) = {(x - 2)}^{2} + (- 3 \sqrt{3}) (3 \sqrt{3})$ $f(x)=(x-2)^2+(-3sqrt3)(3sqrt3)$

$f (x) = x^{2} - 4 x + 4 - 27$ $f(x)=x^2-4x+4-27$

$f (x) = x^{2} - 4 x - 23$ $f(x)=x^2-4x-23$

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How do you find the polynomial function with roots $2 + 3 \sqrt{3}$ $2 + 3sqrt3$ ?

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