Help on Simple Math Problem!!!?

2008-12-17 12:05 pm
Okay, x (exponent 4) -16
i know you can factor it: (x (exponent 2) -4) (x(exponent 2) + 4)
But can you simplify it further and how do u check your work?

回答 (6)

2008-12-17 12:15 pm
✔ 最佳答案
You know that (x^4 - 16) simplifies to (x^2 - 4)(x^2 + 4) because it is the difference of squares.

Notice how (x^2 - 4) is also the difference of squares. That can be simplified to (x - 2)(x + 2)

Your final answer is (x^2 + 4)(x - 2)(x + 2)

You can check by doing FOIL.
(x - 2)(x + 2) = x^2 - 4 + 2x - 2x = (x^2 - 4)
2008-12-17 8:19 pm
Well, you have done half of it, Now do the same again with (x² - 4), which gives you

(x - 2) (x + 2) (x² + 4)

And for a simple check, give x a numerical value - one is usually good enough.

x^4 - 15 = -15 for x = 1

And (x - 2) (x + 2) (x² + 4) = (-1) (3) (5) = -15

Same answer, so it looks good. Do it with some higher numbers if you want to be extra-sure.
2008-12-17 10:05 pm
a^2 - b^2 = (a + b)(a - b)

x^4 - 16
= (x^2)^2 - 4^2
= (x^2 + 4)(x^2 - 4)
= (x^2 + 4)(x^2 - 2^2)
= (x^2 + 4)(x + 2)(x - 2)
2008-12-17 8:16 pm
x^4 - 16 = (x^2 + 4) ( x^2 -4) = (x^2 + 4) ( x -2) (x + 2)

The best ways is to work back from your answer. The easiest way which does not guarantee they are the same is to test a point or two. For example x = 1
1 - 16 = (1+4)(1-2)(1+2)
-15 = (5)(-1)(3) = -15
2008-12-17 8:16 pm
x^4-16=
x^4-2^4=
(x^2+4)(x^2-4)=
(x^2+4)(x-2)(x+2)
On real numbers

Over complex space continue like:
(x^2+4)(x-2)(x+2)=
(x^2-(2i)^2)(x-2)(x+2)=
(x-2i)(x+2i)(x+2)(x-2)
2008-12-17 8:14 pm
x^4 - 16 = (x^2 + 4)(x^2 - 2) = (x^2 + 4)(x + 2)(x - 2)

I assure that the above factoring is correct.


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