factorize x^2+x+1=0!!\n\ni wanna know how far u fellas out there can go!
120153
(x+ [1+i*sqrt(3)]/2) * (x+[1-i*sqrt(3)]/2) = 0
228335
I think u are talkin abt this method....\n\n\nif u hav a cubic like  x^3 + 2x^2 - 2x - 1 = 0\n\nto find out whether (x-1) is a factor....\n\nfirst write the coeff of the eq in this way n write the factor ( here it iz 1 since x-1=0 or x=1) on the left hand side Like this\n\n{one}  1     2     -2     -1\n--------  0     1      3      1\n_______________________________\n{sum}  1     3      1      0\n\n\nHere what i hav done is write the 0 under the coeff of x^3\nnow add that coeff n the zero\nu get 1...\nnow multiply this one wid the number written on the left hand side(here it is also one)\nwite what u get under the coeff of x^2....\nas above add the new no and the coeff of x^2\nin a similar way multiply one and the sum u got n write it under the coeff of x...\nadd the no with coeff of x n then write it below....\nfinally u multiply the sum wid one and write it under the constant term.... n then add...\nif u get 0 as the sum then whatever no u hav taken in da LHS is on of the root.... here we get 0 in the last case... so (x-1) is a fctor.....\n\nnow this eq can b written as \n\n    (x-1) * (1*x^2 + 3 * x + 1) = 0\nhere 1 , 3 and 1 of the quadratic are the sum u got in the above method......
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)
244069
u need to factor:\n\n(x^3-9x)=0\nx(x^2-9)=0\nx(x-3)(x+3)=0\nx=0//x-3=0//x+3=0\n      +3 +3  -3 -3\n     x=3    x=-3  \n\nso ur final answer is x= 3, -3\n\nhope this made sense!
363968
I worked this problem. I did get both sides to zero. I don't even know how to begin explaining it.\n\nFor the sake of typing this with some readability, let (k+1) = b and (k+2) = c.\n\nFirst you subtract (2^c)/(k+3) from both sides to get 0 on one side. Then to get a common denominator, you multiply top and bottom of the first term by (k+3) and top and bottom of the third term by (k+2). You get:\n\n[ (k+3)(2^b) + (k+1)(2^b) - (k+2)(2^c) ] / [ (k+2)(k+3) ] = 0\n\nThen you distribute out the terms:\n\n[ k(2^b) + 3(2^b) + k(2^b) + (2^b) - k(2^c) - 2(2^c) ] / [ (k+2)(k+3) ]\n\nThen you factor out 2^b from the first four terms:\n\n[ (2^b)(k + 3 + k + 1) - k(2^c) - 2(2^c) ] / [ (k+2)(k+3) ]\n\nThen you combine like terms on the left four. On the right two, you reduce the exponent by one and multiply the term by two. Since c is one more than b, this turns (2^c) into (2^b). You can do this since it's an exponent on two:\n\n[ (2k + 4)(2^b) - 2k(2^b) - 4(2^b) ] / [ (k+2)(k+3) ]\n\nYou can factor 2^b out of the right two terms to get:\n\n[ (2k + 4)(2^b) - (2k + 4)(2^b) ] / [ (k+2)(k+3) ] = 0\n\nThe two terms in the numerator cancel out:\n\n0 / [ (k+2)(k+3) ] = 0\n\nAnd:\n\n0 = 0\n\nI must admit that I liked this problem. I don't do math for fun, but every once in a while I like a challenge.\n\nI hope this helped.
439489
c)\n\n= x^3 + x^2 + x - x^2 - x -1\n= x^3 -1
354471
x^4 + 5x^3 - 30 / x^2 - 3\n\nThe important thing is that, especially with division, polynomials need place value so 0 coefficients apply.\n\nRewrite as follows\n\nx^4 + 5x^3  + 0x^2 + 0x - 30 / x^2 + 0x - 3\n\n.................... x² + 5x + 3\n................. ____________________\nx² + 0x - 3 )x^4 + 5x³ + 0x² + 0x - 30\n.................. x^4 + 0x³ - 3x²\n.................  ------------------ .....↓\n........................... 5x³ + 3x² + 0x\n........................... 5x³ + 0x² - 15x\n........................... ------------------ ....↓\n.................................... 3x² + 15x - 30\n..................... ............... 3x² + 0x - 9\n.................................... ------------------\n....................... ................... 15x - 21\n\nSo x^4 + 5x³  - 30 = (x²  - 3)(x² + 5x + 3) + 15x - 21\n(Or quotient is x² + 5x + 3 and the remainder is 15x - 21)
483354
I enjoy algebra, and this is easily my favorite type of problem.  So I hope you, too, one day enjoy them for their inner beauty.\n\nFor now, let's tackle this "factoring problem."  We may wish to use the method of "synthetic division" which is arguably easier than "long division of polynomials."\n\nThink back to grade school where you learned that 6 = 2 * 3.  This simple statement enabled you to 'deconstruct' the number 6 as either "two times three" or "three times two".  Even if you forgot one of the 'factors' you could conjure it up again quickly by asking yourself, "What is six divided by two?" for example.\n\nIn problem (a), you are essentially asking, "What are the coefficients of P(x) if the only factor we know is (x-1)?"\n\nLet's set up synthetic division (see link below), by placing all the coefficients of P(x), and using 1 (the zero implied by (x-1)) as the divisor:\n\n1 ||  a  b  c  d\n___________\n.....a\n\n1 ||  a  b  c  d\n_____a_____\n.....a (a+b)\n\n1 ||  a  b        c       d\n_____a___(a+b)____\n.....a (a+b) (a+b+c)\n\n1 ||  a  b        c                d\n_____a___(a+b)___(a+b+c)_\n.....a (a+b) (a+b+c) (a+b+c+d)\n\nNote that this last term must equal 0 since it is assumed that (x-1) is a factor ... thus, the "remainder" term must equal zero!\n\nThere's the result you were after:  a+b+c+d = 0\n\nHope this helps you tackle the others!
592176
x^3 + 5x^2 - 8 divided by x -1 is:\n\nx^2 + 6x + 8\n\nor...\n(x + 2) (x + 4)\n\ndon't listen to byh. sorry I can't show you HOW to do the problem. I just couldn't figure out how to in type.
89468
The problem factors to\n\n(x - 2)(x^2 + 1)(x^2 + 1) or (x - 2)(x^2 + 1)^2\n\nx = 2, i, -i\n\nthe only real zero is 2.
405147
x^2+x+1=0\nx^2+2(1/2)x+1=0\nX^2+2(1/2)x+(1/2)^2+1-(1/2)^2=0\n(x+1/2)^2+3/4=0\n(x+1/2)^2=-3/4\nx=-1/2+/-√(3/4) i=-1/2+/-√3 * i/2
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!!!
482020
You need to give some directions.  What is the problem asking you to do?\nNote: The person below is incorrect.  (4+3x)^5 does NOT equal\n4^5 + (3x)^5
88938
x^3+4x+1\n\n(3x^4+2x^3+12x^2+11x+2)=\n=(3x+2)*(x^3+4x+1)
492617
wow its been a long time.. i use a calculator to do that now.. but here goes one thing i remember...\n\nlets say we have a polynomial like this\n\nx^2 -3x-10  and we need to factor it..\n\nwe take a look at what factors of -10 we could use that would add up to -3  in this case there are two way to get -10 (-2*5) or (2*-5)  if we add the two sets together one gives us -3 and the other gives us 3 so we can throw out the -2*5 terms.\n\nSo now we lay out the factors.... we need two x's to get x^2 so we need two x's in the first terms (remember FOIL?)\n\nlay the terms out so they give us the correct values\n\n(x+2)(x-5)  now we can use FOIL (firs, outer, inner, last) to check ourselves...\nX*X = x^2\nX*-5 = -5X\n2*x =2x\n2*-5=-10\n\nso we get [x^2] +[-5x] +[2x] +[-10]\nor x^2 -5x+2x-10\ncombine like terms now\n-5x+2x =-3x\n\nwhich gives us\nx^2-3x-10\n\nhope that helps
79133
I don't think (x-1)(x+3) are factors for x^2 + x - 3 = 0\nExpanding that will give x^2+2x-3=0\n\nQuadratic formula?\n\nFactors for x=(-1+sqrt13)/2, (-1-sqrt13)/2\nHence x= 1.30, -2.30\n\nTo solve for inequality of x^2+x-3<0\n\nTest x=2, (2)^2+2-3>0 [This is to test x>1.3)]\nTest x=0, (0)^2+0-3<0 [This is to test -2.3 <x<1.3]\nTest x=-3, (-3)^2-3-3>0 [This is to test x<-2.3]\n   \n\nHence solutions for x^2+x-3<0 are \n\n  -2.30 < x < 1.30 \n\nOR better still (-1-sqrt 13)/2 < x < (-1+sqrt 13)/2\n\nmeans that when x values are between this two numbers, x^2+x-3 is less than zero.
489565
You just factor\n(2x-1)(x-2)=0\nx=2, 1/2