If #abs(x) >= 4# then #(x^2-16) >= 0#, #(x-3)^2 > 0# and #9x^2 > 0#, so #f(x) > 0#. So the only Real roots of #f(x) = 0# are in the range #-4 < x < 4# . The rational root theorem tells us that …

2015年4月12日 · Set this expression to 0 to determine a root; then use synthetic division to determine the second factor If x^3 - 16 =0 then x^3 = 16 and x = 2root(3)(2) So (x-2root(3)(2)) …

2018年6月27日 · Given: #x^4+2x^3-8x-16# First observation is that #2xx8=16# so we have a possible connection there.

2017年2月3日 · First, we can use this rule of exponents to rewrite this expression: #x^(color(red)(a) xx color(blue)(b)) = (x^color(red)(a))^color(blue)(b)#

2018年4月7日 · Here, #I=intx^3sqrt(16-x^2)dx=intx^2sqrt(16-x^2)*xdx# Let, #x=4sinu=>x^2=16sin^2u=>2xdx=32sinucosudu# i.e. #xdx=16sinucosudu#

2015年5月22日 · In general a^3+b^3= (a+b)(a^2-ab+b^2) 16x^3+54 =2((2x)^3+(3)^3) =2(2x+3)(4x^2-6x+9) How do you find the two numbers by using the factoring method, if one …

2018年1月28日 · Try #x=4 tan theta#.Then #x = 4 sec^2theta d theta# So #int x^3/sqrt{x^2+16} dx = int {64 tan^3 theta times 4 sec^2theta}/{4sectheta}d theta #

2017年2月4日 · #intx^3/(sqrt(16-x^2))dx# If we look at the denominator, it looks like a rearrangement of Pythagoras' theorem, #a^2+b^2=c^2 rarrc^2-b^2=a^2# So the length of the …

Question 345875: if 16 is added to one-third of a number, the result is 3 times the number. What is the number? Does the proper equation start with: 16 + x/3 = 3x? Answer by rfer(16322) …

2015年9月13日 · 1/(3(16-x^3))+C Choose t=16-x^3, then dt=-3x^2dx => x^2dx=-dt/3 intx^2/(16-x^3)^2dx=int1/t^2(-dt/3)=-1/3intt^(-2)dt= =-1/3 t^(-1)/-1+C=1/3 1/t+C=1/(3(16-x^3))+C