
Andy C. answered 11/21/17
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(x-2)(x-2)(x-2)(x+4)(kx+C)
(x^2 - 4x + 4)(x^2 + 2x - 8)(kx + C)
x^4 - 4x^3 + 4x^2
+ 2x^3 - 8x^2 + 8x
-8x^2 + 32x - 32 times (kx + C)
(x^4 - 2x^3 - 12x^2 + 40x - 32)(kx + C)
kx^5 - 2kx^4 - 12kx^3 + 40kx^2 - 32kx
+Cx^4 - 2Cx^3 - 12Cx^2 +40C x - 32C
kx^5 + (C - 2k)x^4 - (2C + 12k)x^3 + (40k - 12C)x^2 + (40C - 32k)x - 32C
Since the leading coefficient is k=3.
3x^5 + (c-6)x^4 - (2c + 36)x^3 + (120 - 12C)x^2 + (40C - 96)x - 32C
We now need either the other zero, the y-intercept, or some
other initial condition in order to find the coefficient C
checking:
For x=2: 3*32 + (c-6)*16 - (2c +36)*8 + (120 - 12C)*4 + (40C - 96)*2 - 32c
96 + 16c - 96 - 16c - 288 + 480 - 48c + 80c - 192 - 32c
yes, they all cancel out
For x= -4:
3*(-4)^5 + (c-6)*(-4)^4 - (2c +36)*(-4)^3 + (120 - 12c)*16 + (40C - 96)(-4) - 32C =
-3072 + 256c - 1536 + 128c + 2304 +1920 - 192c - 160 c + 384 - 32c
yes they cancel out
(x^2 - 4x + 4)(x^2 + 2x - 8)(kx + C)
x^4 - 4x^3 + 4x^2
+ 2x^3 - 8x^2 + 8x
-8x^2 + 32x - 32 times (kx + C)
(x^4 - 2x^3 - 12x^2 + 40x - 32)(kx + C)
kx^5 - 2kx^4 - 12kx^3 + 40kx^2 - 32kx
+Cx^4 - 2Cx^3 - 12Cx^2 +40C x - 32C
kx^5 + (C - 2k)x^4 - (2C + 12k)x^3 + (40k - 12C)x^2 + (40C - 32k)x - 32C
Since the leading coefficient is k=3.
3x^5 + (c-6)x^4 - (2c + 36)x^3 + (120 - 12C)x^2 + (40C - 96)x - 32C
We now need either the other zero, the y-intercept, or some
other initial condition in order to find the coefficient C
checking:
For x=2: 3*32 + (c-6)*16 - (2c +36)*8 + (120 - 12C)*4 + (40C - 96)*2 - 32c
96 + 16c - 96 - 16c - 288 + 480 - 48c + 80c - 192 - 32c
yes, they all cancel out
For x= -4:
3*(-4)^5 + (c-6)*(-4)^4 - (2c +36)*(-4)^3 + (120 - 12c)*16 + (40C - 96)(-4) - 32C =
-3072 + 256c - 1536 + 128c + 2304 +1920 - 192c - 160 c + 384 - 32c
yes they cancel out