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Mathematics 10 Online
OpenStudy (anonymous):

Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form. 5, -3, and -1 + 2i

OpenStudy (anonymous):

helpp

OpenStudy (anonymous):

If \(-1+2i\) is a root, then so is \(-1-2i\).

OpenStudy (anonymous):

yeah! i dont know how to multiply (x-(-1+2i))(x-(-1-2i))

OpenStudy (anonymous):

Hold on give me a second...

OpenStudy (anonymous):

Look at the quadratic formula \[ x = \frac{-b\pm \sqrt{b^2-4ac}}{2a} \]Supposing that \(a=1\) then clearly \(b=2\) and \(b^2-4ac = -16\)

OpenStudy (anonymous):

I'm trying to reverse engineer the answer here.

OpenStudy (anonymous):

\[ 4-4c = -16 \implies 1-c=-4\implies c=-5 \]

OpenStudy (anonymous):

Check to see if \(x^2+2x-5\) has the roots we want.

OpenStudy (anonymous):

Woops, \(c=5\) And checking \(x^2+2x+5\) gives us the correct roots of \(-1-2i,-1+2i\)

OpenStudy (anonymous):

yeahh

OpenStudy (anonymous):

\[ \begin{split} (x-(-1+2i))(x-(-1-2i)) &= x^2 - (-1-2i)x - (-1+2i)x +(-1+2i) (-1-2i)\\ &= x^2 +x+2ix +x-2ix + 1-4i^2 \\ &= x^2+2x + 1-4(-1) \\ &= x^2+2x+5 \end{split} \]Multiplying them works well too.

OpenStudy (anonymous):

So in general, when you have an imaginary number, say it is \(a+bi\) Then it should turn out to be: \[ x^2-2ax+a^2+b^2 \]

OpenStudy (anonymous):

hmmm f(x) = x4 - 4x3 + 10x2 + 20x + 75 f(x) = x4 - 14x2 - 40x - 75 f(x) = x4 - 4x3 - 10x2 - 20x - 75 f(x) = x4 + 10x2 - 40x - 75 these are my answer choice though...@wio

OpenStudy (anonymous):

@wio

OpenStudy (anonymous):

\[ (x-5)(x+3)(x^2+2x+5) \]

OpenStudy (anonymous):

oohhh okay!!

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