f (x) = (x^2 - 16)^2
Find all of the zeros of the polynomial function and state the multiplicity of each.
A. 4 with a multiplicity of 2
B. 4 with a multiplicity of 4
C. -2 with multiplicity 2 and with multiplicity 2
D. -4 with mulitplicity 2 and 4 with mulitplicity 2
2007-03-30 12:16:39 · 5 answers · asked by Lennett O 1 in Science & Mathematics ➔ Mathematics
f(x)= ( (x-4)(x+4) )^2 = (x-4)^2 (x+4)^2 ... so D
2007-03-30 12:19:49 · answer #1 · answered by hustolemyname 6 · 1⤊ 0⤋
D; if you expand this you have:
f(x) = (x^2-16)(x^2-16).
By difference of squares one x^2-16 has zeroes of +/-4, and there are 2 of them ,so multiplicity of 2.
2007-03-30 19:20:36 · answer #2 · answered by J Z 4 · 1⤊ 0⤋
It is D. There has to be four solutions since this is a 4th degree equation. So x^2=16 --> x = +/- 4. So x = 4,4,-4,-4.
2007-03-30 19:24:04 · answer #3 · answered by ironduke8159 7 · 1⤊ 0⤋
You + Open Math book = The answer
2007-03-30 19:23:56 · answer #4 · answered by Anonymous · 0⤊ 1⤋
Hi. I really dont know wat u mean!!! its very very confusing!!!
2007-03-30 19:22:57 · answer #5 · answered by ? 2 · 0⤊ 3⤋