Help me in my math problems.?

Question

1) Is this polynomial fully factored?
x^2- 16 = (x+4)(x-4)
Why?

2) Can you factor this polynomial
x^4 - 16 = (x^2 + 4)(x^2 - 4)

Please for number 2 question include your solution and some explanation.

Answer 1

1) Yes x^2 - 16 = (x - 4)(x + 4) is fully factored. Why? There's simply nothing to factor out and it's very obvious.

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

See Theorem 1 in the following link for generalization:

http://www.geocities.com/sem26k

It says that x^n - y^n is always divisible by (x - y). If n is odd then x^n + y^n is divisible by x + y and if n is even then x^n - y^n is divisible by x + y.

And have a nice day!

Answer 2

1) This is fully factored since there are no terms left with a squared term. True, 4 is a perfect square, but x is not known to be a perfect square, so you can not factor the (x - 4).

2) This can be further factored, since the 2nd parentheses contains a perfect square : (x² - 4) = (x + 2)(x - 2), so the factors of this are (x² + 4)(x + 2)(x - 2).

In higher mathematics you can get all 4 factors of this as (x + 2i)(x - 2i)(x + 2)(x - 2) -- but you must use imaginary numbers (which involve √-1) to factor the SUM of two squares.

Answer 3

1) Yes b/c you cannot break it down any further -- meaning, you can't pull anything out of x AND 4 that will simplify it.

2) That is the correct factor. There is a possibility to break it down further b/c you could pull something out of x^2 AND 4, in other words, you COULD pull out an x from x^2 and a 2 from 4: (x^2+4) = (x - 2)(x - 2) OR (x+2)(x+2) -- however neither of these equal (x^2 + 4) when FOILd out.

However, looking at (x^2-4) there are possibilities, again by pulling out an X and a 2:

(x^2-4) = (x+2)(x-2). So, to further simplify the answer, you would have:

(x^2+4)[(x+2)(x-2)]

Answer 4

1. It is fully factored. Unless you take (x-4) = (sqrt(x)+2)(sqrt(x)-2). But the convention is to reduce to the lowest integer power of x so the polynomial is fully factored

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

The above is now fully factored

Answer 5

1. yes because it can no longer be factored. try thinking if there's any possible way you can factor a linear function without radicals.

2. x^4-16 = (x^2+4)(x^2-4)
= (x^2+4)(x+2)(x-2)
again the two factors (x+2) and (x-2) are linear. (x^2+4) cannot be factored anymore because if you use the quadratic formula:
x= [-b +- sqrt (b^2 - 4ac)] / 2a
a= 1, b=0, c=4
x = [0+- sqrt (0 - (4)(1)(4)] / (2)(1)
x = (+- sqrt -16) / 2
the square root of a negative number does not exist (imaginary)
=>>> thus, it can no longer be factored

Answer 6

Okay the first polynomial is completely factored.


the second is not because you can break down the x^2-4 into (x+2) (x-2) (it is a difference of perfect squares.)

so the final answer for the second will be....

(x^2 +4)(x+2)(x-2)

Answer 7

This would normally be very, very easy for me, but the summer has drained me of all knowledge. Coulda done it in a second two months ago. oh well, sorry. Anyway, don't ask for answers online, do the work yourself so yiou actually learn it.

Answer 8

thats hard