how to write a polynomial function with the given zeros: 5, -11, 2-5i

Question

anonymous · Answer

The factors must be (x-5)(x+11)(x-(2+5i))

Multiply them out to get your polynomial.

anonymous · Answer

it says to put it in standard form

anonymous · Answer

multiply and then you'll put in standard form

anonymous · Answer

ok so i dont understand how you get the (x(2+5i))

anonymous · Answer

Since you know that there must be a root (in other words, a zero) at the value of 2+5i, then you think about it like a normal root.

If there's a root at 11, you write (x-11), right?
So think about it the same way. If there is a zero at 2 + 51, then you must do (x - (2+5i).

If you plug in 2 + 5i for x in that case, you get (2+5i) - (2+5i), which is zero. Does that make sense to you?

anonymous · Answer

ok yes.. 
so what do you do after  (x-5)(x+11)(x-(2+5i))

anonymous · Answer

because it says to have real coefficients

anonymous · Answer

You have to multiply it out. So do your FOIL method stuff.

anonymous · Answer

even with the 5i

anonymous · Answer

can someone help me symplify this?

anonymous · Answer

does it say you have to have real coefficients or find the real coefficients?

dumbcow · Answer

imaginary roots always come in pairs, so if "2-5i" is root then so is "2+5i"
to obtain quadratic factor which may be easier than multiplying everything out since you can eliminate the "i" term
$$x = 2 \pm 5i$$
$$x-2 = \pm5i$$
$$(x-2)^{2} = -25$$
$$(x-2)^{2}+25 = 0$$
$$x^{2} -4x +29 = 0$$
thus polynomial is:
$$\rightarrow (x-3)(x+11)(x^{2} -4x+29)$$
multiply
$$= (x^{2}+8x-33)(x^{2}-4x+29)$$
$$=x^{4}+4x^{3}-36x^{2}+364x-957$$

anonymous · Answer

thanks! it was kinda late though