2018年4月7日 · x^4+64 = (x^2-4x+8)(x^2+4x+8) Given: x^4+64 Note that this is positive and therefore non-zero for any real value of x. So it has no linear factors with real coefficients. We can find quadratic factors.

2016年9月1日 · log_4 64 = x (this follows from the basic definition of log)

2014年9月17日 · (x^2+4x+16)/(x-4) Replace all x's with the limit being approached (4)... ((4)^2+4(4)+16)/((4)-4) Combine terms... 48/0 The limit approaches infinity since division by 0 is undefined, but division by 0 also approaches infinity.

2015年2月2日 · The answer is: x^3+64=(x+4)(x^2-4x+16). It's easy if we remember the formula of sum of cubes, that says: x^3+y^3=(x+y)(x^2-xy+y^2).

x^4-64 I take the 4th root of x and get x, and the 3rd root of -64 and get -4, which can be simplified further to 2. My answer then is x-2. On the other hand I looked at factoring x^4-64 by factoring like so, (x^2+8)(x^2+8), then (x^2+8)(x+4)(x-2)but this would give me a middle term. Please help. Answer by Earlsdon(6294) (Show Source):

You can put this solution on YOUR website! Evaluate log4 64----Means "the power of 4 that gives you 64". Ans: 4

2015年4月15日 · If you write #x^6-64 = (x^2)^3 - 4^3# you can factor it into: #(x+2)(x-2)(x^4 +4x^2+16)# If you write: #x^6-64 = (x^3)^2 - 8^2#, then you factor as #(x^3+8)(x^3-8)# which can be further factored as: #(x+2)(x^2-2x+4)(x-2)(x^2+2x+4)# The second answer is factored into irreducibles (over #RR#).

2015年10月27日 · Either using polynomial long division or synthetic division we can discover: #color(white)("XXX")(x^4-256) = (x-4)(x^3+4x^2+16x+64)#

2016年4月15日 · As soon as you see a negative value to have its square root calculated you know that complex numbers are involves.

(x-4)^2=64 x = 4 ± 8 |Taking the square root of both sides of the Equation x = 12 x = -4 are the roots of ...