algebra
posted by Vikky .
Find a quadratic function with zeros 7i and 7i

(x7i)(x+7i) = x^2+49

Use the strategy to solve the higher degree polynomial equation and find the roots of the equation.
6x3 + 14x2  3x  7 = 0 
6 x^3  3 x + 14 x^2  7 = 0
3 x (2 x^2  1) + 7 (2 x^21) = 0
(3x+7)( 2x^21) = 0
does that help? 
If f(x) is a polynomial function of degree 3 and has zeros of
1 and 4i, then find the third zero. 
complex zeros come in conjugate pairs. Just as in the first problem above. So, what do you think?

@ Steve. i still don't get it. should i take the "i" out of the 4? because if i do something like this, (x)(x+4i) i will have 4ix, of which its confusing.

Vikky, what Steve is telling you that complex roots, as well as irrational roots, always come in conjugate pairs.
e.g. if one root is 5  7i, then there has to be a matching 5 + 7i
e.g. if one root is (4 + 3√5)/2 , then there has to be a (4  3√5)/2
So, if you know one root is 4i , then there has to be a +4i
Furthermore, since 1 is also a root, we have 3 factors,
(x1)(x + 4i)(x  4i)
= (x1)(x^2  16i^2)
= (x1)(x^2 + 16)
= x^3  x^2 + 16x  16
equation:
x^3  x^2 + 16x 16 = 0 
Now i get it. Thanks a lot

find the zeros and multiplicity of the equation, Range and interval q is positive.
q(x)=x^2(x+1)^2(x2)