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)

Question

anonymous · Answer

$$(x+1)(x-2)(x-5)(x-(3+2i))(x-(3-2i))$$

anonymous · Answer

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 
$$x^2-6x+14$$

anonymous · Answer

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 $a$

anonymous · Answer

damn damn typo sorry

anonymous · Answer

$$a(x+1)(x-2)(x-5)(x^2-6x+13)$$ is correct.

anonymous · Answer

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

anonymous · Answer

so I just have to find a?

anonymous · Answer

yes