Question

Contruct a polynomial withe degree 5, has only real coefficients, has zeros of -1,2,5,3+2i, and the y-intercept is (0,3)

Answer 1

\[(x+1)(x-2)(x-5)(x-(3+2i))(x-(3-2i))\]

Answer 2

you have no other choice but to multiply this mess out
the hardest part is the last two factors, but that is not too bad
since the zeros are \(3+2i\) and \(3-2i\) the quadratic from the last two factors is
\[x^2-6x+14\]

Answer 3

once you multiply out
\[a(x+1)(x-2)(x-5)(x^2-6x+14)\] then set \(x=0\) and the result equal 3, solve for \(a\)

Answer 4

damn damn typo sorry

Answer 5

\[a(x+1)(x-2)(x-5)(x^2-6x+13)\] is correct.

Answer 6

if the zeros are \(a+bi\) and \(a-bi\) the quadratic is \(x^2-2ax+a^2+b^2\)

Answer 7

so I just have to find a?

Answer 8

yes