How do you write a polynomial with zeros 2, -2, -6i and leading coefficient 1?

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How do you write a polynomial with zeros 2, -2, -6i and leading coefficient 1?

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Answer 1 · 2017-05-10T23:35:20

Identify the four (not three) zeroes and write out the zeroes in factored form. Answer: $x^{4} + 32 x^{2} - 144 = 0$ $x^4+32x^2-144=0$

Explanation:

Original question: Find polynomial with zeroes $2, - 2, - 6 i$ $2, -2, -6i$ and leading coefficient of $1$ $1$

We know that each complex zero must have a pair that is also a complex zero, which is its conjugate. Therefore, if $- 6 i$ $-6i$ is a zero, then $6 i$ $6i$ is a zero too.

We also know that zeroes of a polynomial come in the form $(x - a) (x - b) (x - c) (x - d) = 0$ $(x-a)(x-b)(x-c)(x-d)=0$ where $a, b, c, d$ $a,b,c,d$ are zeroes, therefore we can write our polynomial in this same form:
$(x - 2) (x + 2) (x + 6 i) (x - 6 i) = 0$ $(x-2)(x+2)(x+6i)(x-6i)=0$

Now, we can write out the expanded form by using the fact that $(a + b) (a - b) = a^{2} - b^{2}$ $(a+b)(a-b)=a^2-b^2$ and expand the resulting product of binomials:
Note that ${(6 i)}^{2} = 36 (- 1) = - 36$ $(6i)^2=36(-1)=-36$
$(x^{2} - 4) (x^{2} + 36) = 0$ $(x^2-4)(x^2+36)=0$
$x^{2} \cdot x^{2} + x^{2} \cdot 36 - 4 \cdot x^{2} - 4 \cdot 36 = 0$ $x^2*x^2+x^2*36-4*x^2-4*36=0$
$x^{4} + 36 x^{2} - 4 x^{2} - 144 = 0$ $x^4+36x^2-4x^2-144=0$

Combining like-terms, we get:
$x^{4} + 32 x^{2} - 144 = 0$ $x^4+32x^2-144=0$

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How do you write a polynomial with zeros 2, -2, -6i and leading coefficient 1?

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