Respuesta :
Answer:
[tex](x-6i)(x+6i)(x-3)(x+3)[/tex]
Step-by-step explanation:
If 6i is a zero then -6i is a zero.
In general, if a+bi is a zero then a-bi is a zero (if the polynomial has real coefficients which this one does: 1,27,-324).
Let's test it to see:
Check [tex]x=6i[/tex]
[tex]P(6i)=(6i)^4+27(6i)^2-324\\
P(6i)=6^4(i^4)+27(6)^2(i^2)-324\\
P(6i)=1296(1)+27(6^2)(-1)-324\\
P(6i)=1296-27(36)-324\\
P(6i)=1296-972-324\\
P(6i)=1296-1296\\
P(6i)=0\\[/tex]
Check [tex]x=-6i[/tex]
[tex]P(-6i)=(-6i)^4+27(-6i)^2-324\\
P(-6i)=(6i)^4+27(6i)^2-324\\
P(-6i)=P(6i)\\
P(-6i)=0\\[/tex]
So yep they both give us 0 when we plug it in.
If x=6i is a zero then x-6i is a factor by factor theorem.
If x=-6i is a zero then x+6i is a factor by factor theorem.
What is (x-6i)(x+6i)?
Let's use the multiply conjugates formula: [tex](u-v)(u+v)=u^2-v^2[/tex].
[tex](x-6i)(x+6i)=x^2-36i^2=x^2-36(-1)=x^2+36[/tex]
Now we know [tex](x^2+36)[/tex] is a factor of [tex]x^4+27x^2-324[/tex].
We can use long division or we could try to find two numbers that multiply to be -324 and add up to be 27 since this is a quadratic in terms of [tex]x^2[/tex] with leading coefficient of 1.
Well we already know we are looking for number times 36 that would give us -324.
So -324=-9(36) and 27=-9+36
So the factored form in terms of real numbers is:
[tex](x^2+36)(x^2-9)[/tex]
We already know the first factor can be factored as (x+6i)(x-6i).
The other can factored as (x-3)(x+3) since (-3)(3)=-9 and -3+3=0.
So the complete factored form is
[tex](x-6i)(x+6i)(x-3)(x+3)[/tex].
