simplify (x^2+16)(x^2-16=?

Question

explain

Answer 1

well there is no answer to the exact question because of the two variables so: ?=y
(x^2+16)(x^2-16)=y
first distribute out the x^2
x^2 * x^2 = x^4
then
x^2 * -16 = -16x^2
then distribute out the +16
16 * x^2 = 16x^2
then
16 * -16 = -256
so
x^4+16x^2-16x^2-256=y simplifies down to
x^4 - 256 = y
^---Simplified Version



Good Luck
Tyler

Answer 2

It really depends on what you mean by "simplify". "Simplify" as in "expand and collect like terms", or "simplify" as in "factor completely"? If you're referring to the first one, then:

(x^2 + 16)(x^2 - 16) = (x^2)^2 + 16x^2 - 16x^2 - (16)^2 (using FOIL)
= x^4 - 256

If you were referring to the second one (which I am more inclined to assume), then:

(x^2 + 16)(x^2 - 16) = (x^2 + 16)(x + 4)(x - 4),

on factoring x^2 - 16 (i.e. since it is a difference of two squares,
then we may apply a^2 - b^2 = (a + b)(a - b); the result follows by setting a = x and b = 4).

Ultimately, your choice of the appropriate meaning for your use of the word "simplify" in the original problem would depend on where in your high school algebra syllabus you already are.

Answer 3

i could be way off but i used the FOIL technique(first, outside, inside, last). Multiply x^2 & x^2 together to get x^4. Then multiply x^2 & -16 to get -16x^2. Next multiply 16 & x^2 to get 16x^2. Finally multiply 16 & -16 to
get -256. The two 16x^2's cancel each other out to leave you with x^4 - 256. Sorry if that doesn't work ...i tried.

Answer 4

well, you have (x^2+16)(x-4)(x+4) = 0
x^2 + 16 can never be 0 because the square of any real number is a positive number
thus, your answer is x = 4 and x = - 4

Answer 5

(x^2+16)(x^2-16)= (x^4 - 256)

or ...
(x^2+16) (x^2-16)=
(x^2+16) (x + 4) (x - 4)

{{{ hint general form: a^2 - b^2 = (a - b) (a + b ) }}}
for "real numbers" a^2 + b^2 is "irreducible", meaning it
cannnot be factored down any further, hence
"(x^2+16) " is as far down as you can take this factor
if you restrict yourself to "real numbers"


to "solve" this for "real numbers,

x = +4
x = -4
and that's all for "reals"
since (x-4) = 0 reduces to x = 4
since (x+4) = 0 reduces to x = -4


to "solve" this for "complex" numbers,
x = +4
x = -4
x = 0 + 4i
x = 0 - 4i

(where "i" = sqrt(-1) )


part of my difficulty here is understanding what you mnean by "simplify" ...

Answer 6

Where is the end bracket? If it is after 16 then It is nothing but X^4-256.

Answer 7

Let's solve the quadratic equation:
x^2+16+x^2-16 = 0
First we need to put the equation in the form:
ax^2 + bx + c = 0
For this, we arrange the terms from the highest exponent to the lowest:
2x^2 = 0
We have:
a=2 b=0 c=0
We can now apply the quadratic formula:
x = ( -b ± √(b²-4ac) ) / 2a
We find one solution:
x=0

Answer 8

(a+b)(a-b) = a^2 - b^2

Here a = x^2 and b = 16
so (x^2+16)(x^2-16) = x^2^2 - 16^2
x^4 = 256 and
x = +4 or -4

Answer 9

always remember a formula to solve such equations..

(a+b)(a-b) = a^2 - b^2

now what u have asked is (X^2 + 16) (x^2 - 16)
it will become ((x^2)^2) - 16^2
ie., x^4 - 256 .....(as ((a^b)^b) is a^(b*b) that is, a^2b)
now x = +4, -4 ( if x^4 - 256 = 0)

Answer 10

(x^2+16)(x^2-16)
=(x^2)^2-(16)^2
=x^4-256

it is a simple formula:a^2-b^2=(a-b)(a+b)