2016年1月19日 · Use some standard identities to find: 64x^6-1=(2x-1)(4x^2+2x+1)(2x+1)(4x^2-2x+1) The difference of squares identity can be written: a^2-b^2=(a-b)(a+b) The difference ...

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#).

2016年8月20日 · #x^6-64i = 0# Add #64i# to both sides to get: #x^6 = 64i# #color(white)(x^6) = 2^6(0 + 1i)# #color(white)(x^6) = 2^6(cos (pi/2) + i sin (pi/2))# From de Moivre's theorem, we have: #(cos theta + i sin theta)^n = cos n theta + i sin n theta# Hence principal root: #x_1 = root(6)(64i) = 2(cos (pi/12) + i sin (pi/12))#

2018年1月14日 · First, we can use these rules for exponents to eliminate the outer exponent: #a = a^color(red)(1)# and #(x^color(red)(a))^color(blue)(b) = x^(color(red)(a) xx color ...

2016年1月20日 · x=+-2,-1+-sqrt3i,1+-sqrt3i Knowing the following factoring techniques is imperative: Difference of squares: a^2-b^2=(a+b)(a-b) Sum of cubes: a^3+b^3=(a+b)(a^2-ab+b^2) Difference of cubes: a^3-b^3=(a-b)(a^2+ab+b^2) x^6-64=0 Apply difference of squares: x^6=(x^3)^2,64=8^2. color(red)((x^3+8))color(green)((x^3-8))=0 Use both sum & difference of cubes: x^3=(x)^3,8=2^3. color(red)((x+2)(x^2-2x+4 ...

2015年2月4日 · You can use the definition of logarithm: log_a(b)=x -> a^x=b In your case you get: 2^x=64 which can be written as: 2^x=2^6 (because 2^6=64) and finally (same base, 2 and comparing the exponents): x=6 Precalculus

2018年4月24日 · The solutions are S={2i,-2i,-sqrt3+i, -sqrt3-i, sqrt3-i,sqrt3+i} We need a^3+b^3=(a+b)(a^2-ab+b^2) Therefore, x^6+64=(x^2)^3+(2^2)^3=(x^2+2^2)(x^4-4x^2+16) {(x^2+4=0 ...

We have that x^6 + 16x^3 + 64=(x^3)^2+2*8*x^3+(8)^2=(x^3+8)^2 But x^3+8 can be factored further as follows x^3+8=x^3+2^3=(x+2)*(x^2-2x+4) Finally we get x^6+16x^3+64=(x+2)^2*(x^2-2x+4)^2

If f(x)= log(2,x) and g(x)= 2x^2 + 14, determine the value of (f o g) (5)? The "o" in the (f o g) is a symbol that I was not able to do, and I don't think it represents multiplication...or does it? Here is how I thought it would be done: f(g(5)) = log (base 2)(2(5^2) + 14) = log (2, 64) = 2^x =64 x = 6 Is this right? -----fog(x) means f[g(x)]----

2017年9月16日 · sqrt(64x^6) => sqrt(8^2x^6) => 8x^3