How do you write a polynomial function of least degree and leading coefficient 1 when the zeros are -2, -3, i, -i?

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How do you write a polynomial function of least degree and leading coefficient 1 when the zeros are -2, -3, i, -i?

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Answer 1 · 2017-08-13T17:30:04

$P (x) = (x + 2) (x + 3) (x^{2} + 1)$ $P(x) = (x+2)(x+3)(x^2+1)$
$= x^{4} + 5 x^{3} + 7 x^{2} + 5 x + 6$ $" " = x^4+5x^3+7x^2+5x+6$

Explanation:

Suppose the polynomial is $P (x)$ $P(x)$

By the factor theorem, if $x = a$ $x=a$ is root of $P (x) = 0$ $P(x)=0$ , then $x = a$ $x=a$ is a factor of $P (x)$ $P(x)$

We have the following roots of $P (x) = 0$ $P(x)=0$

$x = - 2, - 3, i, - i$ $x=-2,-3,i,-i$

Hence, the following are factors of $P (x)$ $P(x)$

$(x + 2), (x + 3), (x - i), (x + i)$ $(x+2), (x+3), (x-i), (x+i)$

Hencde, we can write the polynoimal of least degree as the product of these factors (any higher degree polynomial would have additional roots)

$P (x) = A (x + 2) (x + 3) (x - i) (x + i)$ $P(x) = A(x+2)(x+3)(x-i)(x+i)$

We want our polynomial to have leading coefficient $1 \Rightarrow A = 1$ $1=>A=1$ , and also to have real coefficients, so let us multiply out the complex factors (as they are conjugates we will get a real product)

$(x - i) (x + i) = x^{2} + i x - i x - i^{2} = x^{2} + 1$ $(x-i)(x+i) = x^2 + ix - ix -i^2 = x^2+1$

Thus we have:

$P (x) = (x + 2) (x + 3) (x^{2} + 1)$ $P(x) = (x+2)(x+3)(x^2+1)$
$= x^{4} + 5 x^{3} + 7 x^{2} + 5 x + 6$ $" " = x^4+5x^3+7x^2+5x+6$

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How do you write a polynomial function of least degree and leading coefficient 1 when the zeros are -2, -3, i, -i?

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