How do you factor x^6-1 completely?
Here is the complete factorization. First you use the difference of perfect squares and then you use the difference and sum of perfect cubes formulas. x^6-1 (x^3-1)(x^3+1) (x-1)(x^2+x+1)(x+1)(x^2-x+1) which cannot be factored furthermore. But you can set each one of the factors to zero and solve using the quadratic formula for the two quadratics.
How can x+1 become a factor of x^n+1?
Let’s assume [math]x^n+1 = (x+1)(F(n))[/math]Then [math]F(n) = \frac{x^n + 1}{x+1}[/math]Now note that[math]F(n) = x^{n-1} + \frac{-x^{n-1} + 1}{x+1}[/math]Simplify last term to get,[math]F(n) = x^{n-1} - x^{n-2} + F(n-2)[/math]Using above expression recursively, we get[math]F(n) = x^{n-1} - x^{n-2} + x^{n-3} - x^{n-4} + x^{n-5} - x^{n-6} + \ldots + F(m),[/math]where [math]m = 1[/math] if [math]n[/math] is odd and [math]m = 0[/math] if [math]n[/math] is evenNow [math]F(0) = 1/(1+x)[/math] = can’t be factorized any furtherand [math]F(1) = 1 = x^0[/math].Thus [math]x^n+1[/math] can be factorized for odd values of [math]n[/math] as,[math](1+x)(1 - x + x^2 - x^3 + \ldots + x^{n-1})[/math]
How does -1 x -1 become +1?
Good Question! We can show other multiplications with the help of number line but not this one. It goes like this -Long long time ago , when Brahmgupta defined zero (0) , he wrote one property of zero as -a × 0 = 0(Wondering what this has to do with your question?)Let us put a = -1 (Please read as Negative one. Minus one is an Operation , Negative one is a number)(-1) × 0 = 0(-1) × (-1 + 1) = 0Distributing it -[ (-1) × (-1) ] + [(-1) × 1] = 0Let us denote [ (-1) × (-1) ] as K. Also , We know that (-1) × 1 = 1. Hence , putting that into the equation -K + (-1) = 0Therefore , K has to be 1 to satisfy the equation.That is why (-1) × (-1) = 1 .Hope this helped!
If f(x) is a polynomial such that f(1) =1, f(2) =2, f(3) =3 and f(4) =16. Find the value of f(5)?
Cannot be determined, unless assumptions taken.Any polynomial of order [math]n[/math] requires you to have [math]n+1[/math] data sets/information.A linear polynomial/line [math]y=ax+b[/math] requires two points to be defined completely.A quadratic polynomial/parabola [math]y=ax^2+bx+c[/math] requires three points to be defined completely.So, a 4th order polynomial will require 5 data sets.If you consider [math]f(x)[/math] in the above question as a 3rd order polynomial (assumption), then the question can be solved.In general, one may assume [math]f(x)[/math] as a 4th order polynomial and to solve it, assume the coefficient of highest order/leading coefficient as 1. There have been answers to this question in which people have solved it in rote ways. But, I would like to highlight the beauty of the problem by considering a 3 degree polynomial.Notice one similarity in the question.The value of the function (independent variable) is equal to the dependent variable for [math]x=1,2,3[/math].Now, since [math]f(x) = x[/math] is satisfied by [math]x=1,2,3[/math]and [math]f(x)[/math] is a 3rd order polynomial (assumption).Define a function-[math]g(x) = f(x) - x[/math]The function [math]g(x)[/math][math] = [/math][math]0[/math] at [math]x=1,2,3[/math]which means they are the roots of the function.Hence,[math]g(x) = f(x) - x = a(x-1)(x-2)(x-3)[/math][math]g(4) = f(4) - 4 = a*3![/math][math]a=2[/math][math]g(5)= f(5) - 5 = 2*4*3*2[/math][math]f(5)= 53[/math]This approach would have been beautiful if [math]f(4)[/math] had been [math]4[/math] then,[math]g(x) = f(x) - x = (x-1)(x-2)(x-3)(x-4)[/math][math]f(5)=5+4!=29[/math]A similar question to this can be:If [math]f(x)[/math] is a 4th order polynomial with leading coefficient as 1 and [math]f(-1) = 1, f(1) = 1, f(2) = 4, f(3) = 9[/math][math].[/math] Find [math]f(5)[/math][math]?[/math]Off you go.
The coefficient of x^2 in the binomial expansion of (1+x/2)^n, where n is a +ive integer, is 7?
The coefficient of x^2 in the binomial expansion is 7 The term of a binomial expansion is nCr*q(n-r)*p^r In the given binomial expansion q = 1 and p = x/2 If r = 2, then only x^2 will be there in the term The term containing x^2 in the binomial expansion is nC2*1^(n-2)*(x/2)^2 = nC2*1*x^2/4 = nC2*x^2/4 The coefficient of x^2 is nC2/4 According to the question nC2/4 = 7 nC2 = 28 Since nCr = n! / (r!*(n-r)!) n!/2!*(n-2)! = 28 n! = n*(n-1)*(n-2)! Therefore n*(n-1)*(n-2)! / 2*1*(n-2)! = 28 n*(n-1) = 56 n^2-n-56 = 0 (n-8)(n+7) = 0 n = +8 or -7 Since n is a positive integer n shall be taken = 8 >>>>>>>>>>>>>>>> b) If r= 4, then only x^4 will be there in the term The term containing x^4 in the given binomial expansion is 8C4*1^(8-4)*(x/2)^4 = 70*1*x^4 / 16 = 70/16 x^4 = 35/8 x^4 Therefore the coefficient of x^4 is 35/8
D/dx ((1-x^2)^(1/2))/x?
d/dx ((1-x^2)^(1/2)) / x in another words y = (1 - x² ) ^ 0.5 ------------------- x dy/dx = {using quotient rule} y = u/x u = (1 - x² ) ^ 0.5 v = x du/dx = -2x(0.5) (1 - x² ) ^ -0.5 = -x(1 - x² ) ^ -0.5 dv/dx = 1 dy/dx = [v.du/dx - u.dv/dx] / v² dy/dx = { [x].[-x(1 - x² ) ^ -0.5] - [(1 - x² ) ^ 0.5].[1] } / x² dy/dx = { [ [-x²]/[(1 - x² ) ^ -0.5] ] - [(1 - x² ) ^ 0.5]} / x² dy/dx = [(1 - x² ) ^ -0.5] { [-x²] - (1 - x² ) } / x² dy/dx = [(1 - x² ) ^ -0.5] { -x² - 1 + x² ) } / x² dy/dx = [(1 - x² ) ^ -0.5] { - 1 } / x² dy/dx = - [(1 - x² ) ^ -0.5] / x² dy/dx = - 1 / { x² [(1 - x² ) ^ 0.5] }
Integrate x/(x^2+1)?
Integral { x/(x^2+1) dx Use u subsitution u = x^2+1 , du= 2x dx , 1/2 du=xdx Plug in u for x. Integral { du/u 1/2 , bring out the fraction 1/2 to front Integral 1/2 {1/u du Now integrate 1/2 { 1/u du = 1/2 * ln |u| Plug u back in the equation Integral {x/x^2+1 dx = 1/2 ln|x^2+1| + C ----derivative of ln x = 1/x , when integrating you are solving for the antiderivative, therefore integral {1/u = ln |u| Cheers
What is the simplified form for [math]x^2\cdot x^{\frac{1}{2}}[/math]? How do I add fractional exponents?
Several of these answers are plain wrong.Use the rule:[math] x^a\cdot x^b = x^{a+b} [/math]and recall that [math] x^{1/2} = \sqrt{x} [/math] and[math] (x^a)^b = x^{a\cdot b} [/math].Combining those rules, you get[math] x^{5/2} = (x^{1/2})^5 = \sqrt{x}^5 [/math].For some numbers this is straightforward to calculate yourself - other times you'll need a calculator. For instance:[math] \sqrt{4}^5 = 2^5 = 32 [/math].You can verify that that's the same as plugging 4 into your original expression:[math] 4^2\cdot 4^{1/2} = 16\cdot 2 = 32 [/math].I hope that helps.
Let f be the function defined by f x ={ root (x+1) for 0 As written, the function you have described is not continuous at x = 3 because it is not defined at x = 3. The limits from the left and right at x = 3 are equal, but there is still a point discontinuity there. Continuity requires lim f(3) = f(3), but f(3) is undefined. For the same reason, g(x) is not differentiable at x = 3 because g(3) is undefined. But let's assume g(3) is meant to be defined by one or the other equation. Then, in order for g(x) to be differentiable, the limits of g(3) and of g'(3) must be the same from the left and right. g(x) = k*sqrt(x+1) for 0
How can I solve integral of e^(x^(1/2)) from 1-4 using both substitution and integration by parts?
Well, since no one responded and I had some time iI worked this one out as well. Was actually much shorter than I thought. ∫e^(x^1/2) dx Start by letting z = x^(1/2) ∫e^z dx So dz/dx = (1/2) x^(-1/2) and as such makes. dx = 2 x^(1/2) dz So far this gives us ∫2 x^(1/2) e^z dz 2∫ x^(1/2) e^z dz Bad, we got x's and z's and dz. Anyway to relate x to z? Yep z = x^(1/2) The his makes the integral. 2 ∫ z e^z dz And from here, Integrate by parts u = z du = dz dv = e^z v = e^z (note i dropped the 2 for now) ∫ z e^z dz = z e^z - ∫e^z dz ∫ z e^z dz = z e^z - e^z Bring the 2 back 2 ∫ z e^z dz = 2[z e^z - e^z] 2 ∫ z e^z dz = 2e^z[z - 1] And you can change your limits here and evaluate or put it in terms of x in terms of z it's 1 to 2 if you put it back in terms of x and go 1 to 4 ∫ e^√x dx = 2e^√x)[√x - 1] either way you should get 2e^2