This is a third degree polynomial because we are given three roots to multiply together to get it. Even though we only see "2 + i" the conjugate rule tells us that 2 - i MUST also be a root. Thus, the 3 roots are x = -4, x = 2 + i, x = 2 - i.

Setting those up as factors looks like this (keep in mind that the standard form for the imaginary unit in factor form is ALWAYS "x -"):

If x = -4, then the factor is (x + 4)

If x = 2 + i, then the factor is (x - (2 + i)) which simplifies to (x - 2 - i)

If x = 2 - i, then the factor is (x - (2 - i)) which simplifies to (x - 2 + i)

Now we can FOIL all three of those together, starting with the 2 imaginary factors first (it's just easier that way!):

(x - 2 - i)(x - 2 + i) = $x^2-2x+ix-2x+4-2i-ix+2i-i^2$

Combining like terms and canceling out the things that cancel out leaves us with:

$x^2-4x+4-i^2$

Remembr that $i^2=-1$ , so we can rewrite that as

$x^2-4x+4-(-1)$ and

$x^2-4x+4+1=x^2-4x+5$

That's the product of the 2 imaginary factors. Now we need to FOIL in the real factor:

$(x+4)(x^2-4x+5)$

That product is

$x^3-4x^2+5x+4x^2-16x+20$

which simplifies down to

$x^3-11x+20$

And there you go!

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3 years ago