Question

find a third-degree polynomial equation with rational coefficants that has roots -4 and 2+i

please explain

Answer 1

it wants you to find a glorified fraction: that is equal to zero when you use those values

Answer 2

read rational in the wrong place.... its just a poly

Answer 3

(a3/b3)x^3 + (a2/b2)x^2 + (a1/b1)x + (a0/b0)

Answer 4

the 2 + i root means that there is a bend in the graph along the x=2 line, but that it is either above or below the graph ....

Answer 5

f'(2) = 0 would be a part of the equation i think

Answer 6

might be wrong tho ...

Answer 7

(x-1)^2 + 3 is a quadratic with roots at:

1 +- i sqrt(3) ; which means that its bend, its vertex, is along the x=1 line and that it is either above or below it.... so I am assuming that is a good analogy to this

Answer 8

the cubic formula tho is alot more complicated ..

Answer 9

(x+4)(x+4)(x - (2+i)) or (x+4)(x - (2+i))(x - (2+i)) seem to be the possibilities; lets try to find the products of these...

Answer 10

(x-2-i)(x-2-i) =

x^2 -2x -ix
-2x -ix +3 + 4i
--------------------

(x^2 -4x -2ix +4i +3)
(x+4)
-------------------
x^3 -4x^2 -2ix^2 +4ix +3x
+4x^2 -8ix -16x +16i +12
-----------------------------------
x^3 -2ix^2 -4ix -13x +16i +12

kinda gotta wonder abt my technique lol

Answer 11

(x+4)^2 = x^2 +8x +16

(x^2 +8x +16)
(x-2-i)
--------------
x^3 +8x^2 +16x
-2x^2 -16x -32
-ix^2 -8ix -16i
--------------------------------
x^3 -6x^2 -32 -ix^2 -8ix -16i


(x^3 -6x^2 -32) - i( x^2 +8x +6)

Answer 12

(x^3 -13x +12) - 2i( x^2 +2x -8)