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 of cubes identity can be written: a^3-b^3 = (a-b)(a^2+ab+b^2) The sum of cubes identity can be written: a^3+b^3=(a+b)(a^2-ab+b^2) We find: 64x^6-1 =(8x^3)^2-1^2 =(8x^3-1)(8x^3+1 ...

2016年1月2日 · Use some standard identities to find: x^6-1=(x-1)(x^2+x+1)(x+1)(x^2-x+1) Use the difference of squares identity: a^2-b^2 = (a-b)(a+b) the difference of cubes identity ...

2017年2月16日 · x^3 (x^6)^(1/2) may also be written as " "x^(6xx1/2) = x^(6/2) = x^3 ~~~~~ By the way (x^6)^(1/2) is another way of writing sqrt(x^6)

Given: #(x^3-7x-6)/(x+1)# Using place keepers of no value. Example: #0x^2# #color(white)("ddddd.ddddd.d")x^3+0x^2-7x-6#

2015年2月19日 · we can rewrite the function into a form which is easily integrated Think of the function like this (x+6)/((1)(x+1))=A/1+B/(x+1) We are doing a partial fraction decomposition. Multiply the expression above by (1)(x+1) x+6=A(x+1)+B(1) x+6=Ax +A +B Equating coefficients we get A=1 and A+B=6 Since A=1 we can conclude that B=5 Therefore we can rewrite as follows int(x+6)/(x+1)dx=int1+5/(x+1)dx ...

2016年11月7日 · How do you use the Binomial Theorem to expand # (1/X + X)^6#? Precalculus The Binomial Theorem The Binomial Theorem. 1 Answer

f^'(x) = 5/(2 sqrt (5 x+6)) f(x) = (5 x+6)^(1/2) f^'(x) = 1/2(5 x+6)^(1/2-1)* 5 or f^'(x) = 1/2(5 x+6)^(-1/2)* 5 or f^'(x) = 5/(2 sqrt (5 x+6)) [Ans]

What is the function that is a reflection of #f(x) = 6(1/3)^x#? Precalculus. 1 Answer

2018年3月11日 · (x^3-1)^2(x^3+1) x^9-x^6-x^3-1 x^6(x^3-1)-1(x^3-1) (x^3-1)(x^6-1) (x^3-1)[(x^3)^2-(1^2)] (x^3-1)(x^3-1)(x^3+1) (x^3-1)^2(x^3+1)

2016年8月30日 · How do you evaluate the limit #cos((x^5+1)/(x^6+x^5+100))# as x approaches #-oo#? Calculus Limits Determining Limits Algebraically. 1 Answer