Question

Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form.

2, -4, and 1 + 3i

Answer 1

note if \(z=a+bi\) is a zero of a polynomial with real coefficients then \(\bar z=a-bi\) must be zero as well. in this case \(f(1+3i)=0\) means that \(f(1-3i)=0\) as well. we therefore have a minimum of four distinct zeros: \(-4,2,1-3i,1+3i\) -- can you construct such a polynomial?

Answer 2

complex roots always occur in pairs
roots are \[2,-4,1+3\iota,1-3\iota \]
required polynomial \[=\left( x-2 \right)\left( x+4 \right)\left( x-1-3\iota \right)\left( x-1+3\iota \right)\]
simplify it.

Answer 3

must be A* zero as well... sorry

Answer 4

how im not sure how to do imaginary

Answer 5

well @griffin5210 if \(k\) is zero then we need a factor like \(x-k\) right? so we know our function will look like:$$f(x)=(x-(-4))(x-2)(x-(1-3i))(x-(1+3i))$$

Answer 6

we can simplify \(x-(-4)=x+4\). in a similar sense we can simplify \((x-(1-3i))(x-(1+3i))\) -- foil it out:$$(x-(1-3i))(x-(1+3i))=x^2-(1-3i)x-(1+3i)x+(1-3i)(1+3i)$$

Answer 7

if we combine like terms we have:$$-(1-3i)x-(1+3i)x=(-1+3i-1-3i)x=-2x$$and if we foil the last product we get:$$(1-3i)(1+3i)=1-3i+3i-(3i)^2=1-(-9)=1+9=10$$

Answer 8

An easier way to do the complex pair.
Just pick one of the complex roots.
x = 1 + 3i
x - 1 = 3i
(x-1)^2 = (3i)^2 = -9
x^2 - 2x + 1 = -9
x^2 - 2x + 10 = 0

Answer 9

so \((x-(1-3i))(x-(1+3i))=x^2-2x+10\)

Answer 10

now we try to finish it off:$$f(x)=(x+4)(x-2)(x^2-2x+10)$$we know \((x+4)(x-2)=x^2+4x-2x-8=x^2+2x-8\) so:$$f(x)=(x^2+2x-8)(x^2-2x+10)$$now just distribute:$$\begin{align*}f(x)&=x^2(x^2-2x+10)+2x(x^2-2x+1)-8(x^2-2x+10)\\&=x^4-2x^3+10x^2+2x^3-4x^2+2x-8x^2+16x-80\\&=x^4-2x^2+18x-80\end{align*}$$

Answer 11

\[\left\{ \left( x-1 \right)+3\iota \right\}\left\{ \left( x-1 \right)-3\iota \right\}=\left( x-1 \right)^{2}-\left( 3\iota \right)^{2}\]
\[=x^{2}-2x+1+9=x ^{2}-2x+10\]

Answer 12

oops there's something wrong there in my work

Answer 13

f(x) = x4 - 2x2 + 36x - 80

f(x) = x4 - 3x3 + 6x2 - 18x + 80

f(x) = x4 - 9x2 + 36x - 80

f(x) = x4 - 3x3 - 6x2 + 18x - 80

these are the choices

Answer 14

fixed it:$$\begin{align*}f(x)&=x^2(x^2-2x+10)+2x(x^2-2x+1\color{red}0)-8(x^2-2x+10)\\&=x^4-2x^3+10x^2+2x^3-4x^2+2\color{red}0x-8x^2+16x-80\\&=x^4-2x^2+\color{red}{36}x-80\end{align*}$$