How do you find a polynomial function f with real coefficients of the indicated degree that possesses the given zeros: degree 2; 4+3i?

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How do you find a polynomial function f with real coefficients of the indicated degree that possesses the given zeros: degree 2; 4+3i?

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Answer 1 · 2017-07-02T16:56:01

$f (x) = x^{2} - 8 x + 25$ $f(x) = x^2 - 8x + 25$

Explanation:

We know that the function is a degree-2 polynomial, so the function has two solutions, one of which is given as $4 + 3 i$ $4 + 3i$ .

Know that if one imaginary species is a solution to a polynomial equation, its conjugate is also a solution.

Therefore, the other solution to this equation is the conjugate of $4 + 3 i$ $4 + 3i$ , which is $4 - 3 i$ $color(blue)(4-3i$ .

Now that we know the solutions to the polynomial equation, let's derive a function, by setting them equal to zero like so:

$(x - (4 + 3 i)) (x - (4 - 3 i)) = 0$ $(x - (4 +3i))(x - (4 -3i)) = 0$

which is the same as

$(x - 4 - 3 i) (x - 4 + 3 i) = 0$ $(x - 4 - 3i)(x - 4 + 3i) = 0$

Factoring out the polynomial gives us

$x^{2} - 4 x + 3 x i - 4 x + 16 - 12 i - 3 x i + 12 i - 9 i^{2} = 0$ $x^2 - 4x + 3x i - 4x + 16 - 12i - 3x i + 12i - 9i^2 = 0$

Combining like terms..

$x^{2} - 8 x + 16 - 9 i^{2}$ $x^2 - 8x + 16 - 9i^2$

$i^{2} = - 1$ $i^2 = -1$ :

$x^{2} - 8 x + 25$ $x^2 - 8x + 25$

The equation of the function with zeros $4 \pm 3 i$ $4 +- 3i$ is thus

$f (x) = x^{2} - 8 x + 25$ $color(blue)(f(x) = x^2 - 8x + 25$

Answer 2 · 2017-07-02T22:13:27

$x^{2} - 8 x + 25 = 0$ $x^2-8x+25= 0$

Explanation:

Complex roots of real valued polynomials always appear in conjugate pairs, so if one root of $f (x) = 0$ $f(x)=0$ is:

$α = 4 + 3 i$ $alpha = 4+3i$

Then another root is:

$β = 4 - 3 i$ $beta = 4-3i$

Se we are looking for a polynomial of degree $2$ $2$ , ie a quadratic with roots $α$ $alpha$ and $β$ $beta$ .

So then:

Sum $= α + β$ $=alpha+beta$
$= (4 + 3 i) + (4 - 3 i)$ $" " = (4+3i) + (4-3i)$
$= 8$ $" " = 8$

Product $= α β$ $=alphabeta$
$= (4 + 3 i) (4 - 3 i)$ $" " = (4+3i)(4-3i)$
$= 16 - 9 i^{2}$ $" " = 16-9i^2$
$= 25$ $" " = 25$

And so the required equation is:

$x^{2} - (sum of roots) x + (product of roots) = 0$ $x^2-("sum of roots")x + ("product of roots") = 0$
$:. x^2-8x+25= 0$

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How do you find a polynomial function f with real coefficients of the indicated degree that possesses the given zeros: degree 2; 4+3i?

2 Answers

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