Degree 4; zeros -2,0,2,4
2006-11-04 14:47:29 · 6 answers · asked by im_all_about_spades2003 1 in Science & Mathematics ➔ Mathematics
f(x) = x(x+2)(x-2)(x-4)
Multiply it out if you need it in general form.
There could also be a constant in front of the x.
2006-11-04 15:10:19 · answer #1 · answered by just♪wondering 7 · 0⤊ 0⤋
Multiply the linear components featuring the roots.
That is
(x+2) for the -2 root
x for the zero root
(x-2) for the 2 root
(x-4) for the 4 root.
You can do that, can't you?
2006-11-04 22:53:00 · answer #2 · answered by Vincent G 7 · 0⤊ 0⤋
Degree 4, so we the highest exponent is 4.
(x+2)(x-0)(x-2)(x-4) = 0
x(x^2-4)(x-4) = 0
x(x^3 - 4x^2 - 4x + 16) = 0
x^4 - 4x^3 - 4x^2 + 16x = 0
Thanks!!
2006-11-04 22:53:34 · answer #3 · answered by iggry 2 · 0⤊ 0⤋
Set it up (x+2)(x-0)(x-2)(x-4) and multiply through....
(x^2 + 2x)(x^2-6x+8) =
x^4-4x^3-4x^2+16x
2006-11-04 23:06:32 · answer #4 · answered by Action 4 · 0⤊ 0⤋
x*(x-2)*(x+2)*(x-4)
do the math.
2006-11-04 22:59:49 · answer #5 · answered by Dr. J. 6 · 0⤊ 0⤋
(x+2)(x)(x-2)(x-4)
x(x^2-4)(x-4)
(x^3-4x)(x-4)
x^4-4x^3-4x^2+16x
2006-11-04 22:52:08 · answer #6 · answered by Greg G 5 · 0⤊ 0⤋