The equation 4x^3+ax^2+bx+c=0 has solutions of x=-4, x=5-2i, and x=5+21. Find the values of a, b, and c.

Question

parthkohli · Answer

Do you know what Vieta's Formulas are?

anonymous · Answer

no

parthkohli · Answer

OK, never mind that. 

We know that if $k$ is a root, then $x-k$ is a factor, and a polynomial is a product of its factors. In this case, the factors are:

$x+4$
$x - 5 + 2i$
$x-5 - 2i$

So,$$4x^3 + ax^2+bx + c = 4(x+4)(x-5+2i)(x-5-2i)$$

anonymous · Answer

thank you!

anonymous · Answer

can you help me with another problem?

anonymous · Answer

Find a fifth degree polynomial function that has a bounce point at 5, a pass-through point at 8, additional roots at 8-3i and 8+3i, and a y-intercept at the point (0,7).

cwrw238 · Answer

lol  - great sounding terms! I guess a bounce point is a local maximum.  But what's a pass through point at 8???

anonymous · Answer

pass through as in the multiplicity is odd