How do you write a polynomial function given the real zeroes -2,-2,3,-4i and coefficient 1?

Question

1598 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-05T03:07:17

$x^{5} + x^{4} + 8 x^{3} + 4 x^{2} - 128 x - 192$ $x^5+x^4+8 x^3+4 x^2-128 x-192$

The major trick with this problem is remembering that complex roots always come in pairs.

Thus, along with the root of $- 4 i$ $-4i$ , the polynomial will also have a root of $4 i$ $4i$ .

The polynomial can be written as:

${(x + 2)}^{2} (x - 3) (x + 4 i) (x - 4 i)$ $(x+2)^2(x-3)(x+4i)(x-4i)$

$= (x^{2} + 4 x + 4) (x - 3) (x^{2} + 16)$ $=(x^2+4x+4)(x-3)(x^2+16)$

When distributed completely, this gives

$x^{5} + x^{4} + 8 x^{3} + 4 x^{2} - 128 x - 192$ $x^5+x^4+8 x^3+4 x^2-128 x-192$

graph{x^5+x^4+8 x^3+4 x^2-128 x-192 [-10, 10, -500, 301.6]}

As you can see, the graph has an odd degree $(5)$ $(5)$ , has a root of $- 2$ $-2$ with multiplicity $2$ $2$ and a root at $3$ $3$ .