(a) Factor x^6 - 64 = (x^2)^3 - 64 as a difference of two cubes. then factor one of the factors as a difference of squares.\n(b) factor x^6 - 64 = (x^3)^2 - 64 as a difference of two squares, then factor the two factors: one as a sum of cubes and one as a difference of cubes\n(c) compare the factors you get from factorizations in parts (a) and (b), and disregard the factors in each that are obviously equal. write an equation that equates the factors in each expression that are different.\n(d) generalize the equation you wrote in part (c) with an arbitrary integer a instead of 2 (ie write the corresponding factors in the factorization of x^6 - a^6)\n\nthanks for your help!!
136946
This is the best I could come up with:\n\n(a)  (x^2-4)(x^4+4x^2+16)\n      (x-2)(x+2)(x^4+4x^2+16)\n\n(b)  (x^3-8)(x^3+8)\n     (x-2)(x^2+2x+4)(x+2)(x^2-2x+4)\n\n(c)  If you take out the factors that are the same, what's left should be equal hence:\n     x^4+4x^2+16=\n(x^2+2x+4)(x^2-2x+4)\n\n(d)  x^(a^2)+2ax^a+4a^2=\n(x^a+ax+2a)(x^a-ax+2a)
534862
1. When you square root both sides you get x - 3 = +4 or - 4\nNow add 3 to bot sides and get x = 7 or -1\n\n3 and 4 are similar.  (x-3)^2 means (x-3)(x-3) which is x^2 - 6x + 9\n\nSo x^2 - 6x + 9 = 16\n\nSubtract 16 from both sides to get x^2 - 6x - 7 = 0\n\nTo use the zero product rule, factor by FOIL:(x-7)(x+1)=0\n\nAnd to use quadratic formula, you can see\n that a= 1, b= -6, and c= -7
595739
Write it as:\n2x^2-4x+3x-6\n=2x[x-2]+3[x-2]\n=[2x+3][x-2]\nSimple middle-term factorisation!!!
363551
a.xa+xb\nb.21x^3+24x^2y\nc.3x(8x+5)+2(8x+5)\n=24x^2+15x+16x+10\n=24x^2+31x+10\nd.4x(2x+1)-6(2x+1)\n=8x^2+4x-12x-6\n=8x^2-8x-6\ne.8x(5x^2+7x+3)+2(5x^2+7x+3)\n=40x^3+56x^2+24x+10x^2+14x+6\n=40x^3+66x^2+38x+6\nf.x^2-1\ng.x(x^2+2x+4)-2(x^2+2x+4)\n=x^3+2x^2+4x-2x^2-4x-8\n=x^3-8
652710
2(x^2+3x-40) <<<<<The answer\n\n\nfor the guy below me...that wouldn't be factoring.
362336
You can factor b^4 + 2b^2 - 24 to:\n(b^2 + 6)(b^ - 4) = 0\nTreat these as two separate equations, just for convenience's sake:\nb^2 - 4 = (b + 2)(b - 2) = 0\nb = 2, -2\nb^2 + 6 = 0\nb^2 = -6\nb = +/-(sqrt)(-6)\nb = +/-(sqrt)6 X i\n\nThus, the answer is the second of the four you listed (-2,2,-i*6,i*6)
426622
Okay, the goal here is to get values of x that fit the equation.  These are not like easy linear ones, they have x^2 in them.  We can still solve them.  \n\nx^2=16\nWhat is the reverse sign of the square?  It's the square root.  Square root both sides and get\nx=sq(16)\nSo x=+/-4\nBoth work because (-4)^2 becomes 16 as well.  Those are your answers.\n\n\n2x(x+3)=0\nThe goal here is to make this equation equal zero.  Do you see how there are two "factors?"  Things that are multiplied together?  Well, to equal zero, one of them must equal zero.  Either one can equal zero, so you have different solutions that work.\n\nTake 2x:\n2x=0\nx=0\n\nNow x+3:\nx+3=0\nx=-3\nSo\nx=0,-3\n\n\nHere we have to "factor" this equation:  Find two things that, when multiplied together, equal this.  y^2 and 8y both have y, so factor out the y and get:\ny(y+8) \nWe can do as we did before, and get\ny=0\nand \ny+8=0, so\ny=-8\nOur answers are\ny=0,-8\n\nThe next two are harder:\n\nFirst bring the 28 over to the other side:\n\nx^2+3x-28=0  \n\nThe factored form of this actually has two factors, but none with just x.  To factor this, we need two factors of -28.  When one of them is multiplied by x and added to the other one multiplied by x, they equal 3x.  Now, a negative times a positive is a negative so we are looking for one positive and one negative factor of 28.  Let's try these two:\n\n7 and -4.\n7(-4)=-28\n7x+(-4x) = 7x-4x=3x!\n\n-4 and 7 are our factors, so:\n(x+7)(x-4) is our factored form.  Solve the components:\nx+7=0\nx=-7\n\nx-4=0\nx=4\n\n x=-7,4 are our two answers.\n\n\n\n2x^2+9x-5\n\n\nThe next one is similar, but more complicated.  We have a 2x^2.  No matter.  We need to find two factors of -5, and one of them will be multiplied by 2x instead of x.  The only factors of -5 are +/-1, and +/-5.  It looks like the 5 will be positive since 9x is fairly big and positive.  Let's try -1 and 5:\n2x(5)=10x\nx(-1)=-x\n10x+(-x)= 10x-x=9x!\n\nSo 2x will be multiplied by 5, and x by -1.  In factored form, this is:\n\n (2x-1)(x+5)\n\nBecause you use FOIL (first outside, inside last) so the 2x is not multiplied by -1, but by 5.  Now solve these components.\n\n2x-1=0\n2x=1\nx=1/2\n\nx+5=0\nx=-5\nSo\n x=1/2, -5 are our two answers. \n\nAll done.  Hope this helps.
412246
Monomial factor is 3x^4  The factorization is\n\n3x^4 * (2x^3 + 5x^2 - 3x - 3)\n\nI cannot see how 2x+1 is a factor of 2x^3 + 5x^2 - 3x - 3; polynomial division does not result in integral coefficient and also leaves a remainder.
350162
Go for it... ask some questions and I'll try and help.  However, be forewarned, I like to give people the *method* to solve a problem, so they can do it later on their own.  You will rarely find me just giving out answers.  Hope that is okay.\n\nIn your example:\n64 - 40ab\n\nThe best you can do is factor out a common factor from 64 and 40.  Both of these numbers are divisible by 8 (8 x 8 and 8 x 5)... that means you get:\n\n8(8 - 5ab)\n\nSince 8 and 5 share no further common factors, you can't factor them any further.  Also, a and b are only part of the second number, so you can't factor those...  the answer is:\n8(8 - 5ab)
178383
Seems like I have been doing most of your homework, SMILE4LISA *wink.  But I do not mind.  I noticed a couple of errors in your question, so I have taken the liberty of fixing them myself.\na) Should read: factor a^6 + 1 = (a^2)^3 + 1 as a sum of two cubes.\nANSWER: Using the formula, a^3 + b^3 = (a + b)(a^2 - ab + b^2), I can write a^6 + 1 = (a^2)^3 + 1^3.  Hence, a^6 + 1 = (a^2)^3 + 1^3 = (a^2 + 1)((a^2)^2 - a^2 + 1)\nb) Simple plug n chug: \n2^6 + 1 = (2^2)^2 + 1^3 = ((2^2) + 1)((2^2)^2 - 2^2 + 1), both check out to be 65 when simplified.\n2^12 + 1 = (4^2)^3 + 1^3 = ((4^2) + 1)((4^2)^2 - 4^2 + 1), both check out to be 4097.\nc)It should read: show that a number of the form 2^n + 1 (n > 0) cannot be prime if n is a multiple of 3 (ie if n = 3k for some integer k)\nANSWER: 2^n + 1 = 2^3k + 1 = (2^k)^3 + 1^3 = (2^k + 1)(2^2k - 2^k + 1) and this implies that it is divisible by the factors of (2^k + 1) and (2^2k - 2^k + 1); primes can only be divisible by 1 and itself.
549817
First of all, solve both equations. So the first would be..\n\n3(2x+1)>5-2x \n     6x+3>5-2x (add 2x to both sides)\n         8x>2 (take 3 from both sides)\n           x>0.25 (divide both sides by 8)\n\nThe second would be..\n\n2x^2-7x+3>0\n4x^2-14x+6>0 (multiply both sides by 2)\n(2x-6)(2x-1)>0 (factorisation)\n \nReplacing > with =..\n\n2x-6=0,      2x-1=0\nx=3,                 x=0.5\n\nSo to satisfy both criteria, x must be greater than 0.25, greater than 0.5 and greater than 3. \n\nSo the answer is that x > 3.
548495
at the first 4 you use a^3-b^3=(a-b)(a^2+ab+b^2)\n and a^3+b^3=(a+b)(a^2-ab+b^2)\n\nfor the next 2 :\nthe first 2 factors and the second 2 factors have something in comon\n\nthe next ones use the above formulas plus one more \na^2-b^2=(a-b)(a+b)
658238
You know the first term is x^2 so you can write:\n(x     )(x     )\nYou know that the second term is -4x:\nIf factors are of the form (x+A)(x+B) then  Ax+Bx = -4x\nYou also know that the last term is -5, so A*B=-5\n\nJust find the values of A and B that satisfy these equations:\nAx+Bx = -4x\nA*B = -5\n\nA = -5 and B = 1\n\nNOW that we have the factors\n(x-5)(x+1)=0\nTo get 0 either section of the left hand side can be 0, so solve both cases:\nCASE 1:\nx-5 =0\nx = 5\nCASE 2:\nx+1=0\nx=-1\n\nANSWER x=5 or x=-1
90163
1) \nxy - x + 4y - 4\nx(y-1) + 4(y-1)\n(x+4)(y-1)\n\n2)note: x^3 means x cubed; x^2 means x squared\nx^3 -6x^2 + 2x - 12\nx^2(x-6) + 2(x-6)\n(x^2 + 2)(x-6)\n\n3)\n6tx + r - 3t - 2xr\n3t(2x - 1) - r(2x - 1)\n(3t - r)(2x - 1)\n\n4)Are you sure it wasn't -x^2 - 2x - xy - 2y?\n-x^2 - 2x - xy - 2y\n-x(x + 2) - y(x + 2)\n(-x - y)(x + 2)\n\nTo "factor by grouping," you must split the original polynomial that you are given into two different binomials by "grouping" them together.  This is done with perenthesis in the problems above.  Then, you factor out the "greatest common factor" from each of the binomials.  After factoring, the two quantities still within the perenthesis should be the same.  You could call these quantities the "common binomial."  Lastly, you factor out the common binomial to get the final answer.
47386
(x+3)(x^2-x+4)/(x+3) common factor\n\nmultiply by x+3\n\nx^2-x+4
362098
The equation to start with is : x^3 = 27\nor, x^3 = 3^3\nor, x^3 - 3^3 = 0\n\nThis is a difference of two cubes,\nso it can be factored :\n\n(x - 3)(x^2 + 3x + 9) = 0\n\nOne solution is seen to be x = 3.\n\nThe other solutions are obtained by\napplying the quadratic formula to :\n\nx^2 + 3x + 9 = 0\n\nThis gives :\n\nx = {-3 ± sqrt[3^2 - 4(1)(9)]} / (2 * 1)\n\n= [-3 ± sqrt(-27)] / 2\n\n= [-3 ± i*3*sqrt(3)] / 2\n\n= (-3 / 2) ± [3*sqrt(3) / 2] * i
559180
take the equation x^2 - 7x + 12 and do the following  (x ) (x) now...see that there is a - and a + in the equation....you need to do this ( x -  ) ( x - ) 'cause you put the minus in the first factor in the second factor you take the - and the + and multiply them....So you get (x- ) (x -) the other 2 terms for each equation comes from the 7x and the 12.  What you do is to find 2 numbers that multiplied gives you 12 and added give you 7....The numbers are 3 and 4 ...Then put them into your factors (x-3)(x-4).\n\nFor the other 2 equations you do the same process\n\n(x^2 - 8x + 7) = (x-  ) (x-  ). In this one, you have to find 2 numbers that multiplied give you 7 and added give you 8. They are 1 and 7. Then (x^2 - 8x + 7) = (x- 1 ) (x- 7 ).\n\nw^2 - 8w + 15 = (w - ) (w- ). As you saw in one and two, three is the same....so the factors are 3 and 5. Then, w^2 - 8w + 15 = (w -3 ) (w-5 )\n\nFor the solve...\n1. x^2 - 6x = 0\n2. x^2 - 2x - 3 = 0\n\nTHe 1 is like this...you have to complete your quadratic equation then you do the following x^2 - 6x = 0 --> x^2 - 6x + 9 =0+ 9...The 9 in the right side was added in order to keep the two sides equal...IN the left side you got a perfect quadratic equation, hence it will be (x-3)^2= 9....Now you pass the 9 of right to the left ...then it will be [(x-3)^2]-9=0\n\n\nThe other equation x^2 - 2x - 3 = 0....This one is quite simple, it is the same as the part 1...The difference is that when you find the factors, they will be 2 numbers that multiplied give you 3 but substracted gives you 2.....The #s are 3 and 1....Here is the thing....when you break the equation into factors, you will have the following (x- ) (x+ )...Since you have a -2 that means that it is the -3 and the +1...Hence your factors will be the following (x-3)(x+1)