Factor and solve for x?
601402
depends on what you mean\nif (9x)^2 - 81 = 0 then\n(9x)^2 = 81\n81x^2 = 81\nx^2 = 1\nx = 1 or -1\n\nif 9x^2 -81 = 0 then\n9x^2 = 81 \nx^2 = 9\nx = 3 or -3\n\ntosmith
94666
The only way to pull out the largest common factor will be to solve for x squared & x. Is that what you want to do?
394036
Try to solve the eqs with quad rule:\na*x^2 + b*x + c=0\nx=(-b +- Sqrt(b^2-4*a*c))/(2*a)\n\nx^2+2x+5=0 => Get two Complex answers, \nso no it can't be factorized.\n\nx^2+4x+5=0 => Get two Complex answers, \nso no it can't be factorized.\n\nx^2+4x+3=(x+3)*(x+1)=0; \nSo, yes, this can be factorized. \nSolutions are x=-3 and x= -1\n\nx^2+2x-3=(x+3)*(x-1)=0;\nSo, yes, this can be factorized. \nSols are x= -3 and x=1
498372
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......
894442
f(x)=2x^3-3x-1\nf(-1)=-2+3-1=0\nx+1 is a factor\ndividing by x+1\n2  0  -3  -1\n0 -2   2   1\n-------------\n2x^2-2x-1\nso the factors are\n(x+1)(2x^2-2x-1)
1095294
5) find the factors\nsquareroot of x * x * x * x * x * x /  7*7*5*5*2*2\nthe factors can come out of the square root by pairs only\nyou have 3 pairs of x's = x^3\nyou have 1 pair of 7's so it comes out of the square root as a 7\nx^3/7*5*2\nx^3/70 is it's simplified term\n\n__________________\nif u want all the correct answers then that would tarenirator's answers
686655
No prob dude... take a deep breath, grab a cold drink, and concentrate on this... ready? ok....\n\nOften, the simplest way to solve "ax2 + bx + c = 0" for the value of x is to factor the quadratic, set each factor equal to zero, and then solve each factor. But sometimes the quadratic is too messy, or it doesn't factor at all, or you just don't feel like factoring. So, while factoring may not always be successful, the Quadratic Formula can always find the solution. \nThe Quadratic Formula uses the "a", "b", and "c" from "ax2 + bx + c", where "a", "b", and "c" are just numbers. The Formula is derived from the process of completing the square, and is formally stated as:\n\n For ax2 + bx + c = 0, the value of x is given by:\n   \nNote that, for the Formula to work, you must have "(quadratic) = 0". Note also that the "2a" at the bottom of the Formula is underneath everything above, not just the square root. And don't forget that it's a "2a" under there, not just a "2"! And make sure that you are careful not to drop the square root or the "plus/minus" in the middle of your calculations, or I can guarantee that you will forget to "put them back" on your test, and you'll mess yourself up. And remember that "b2" means "the square of ALL of b, including the sign", so don't leave b2 being negative, even if b is negative, because the square of a negative is a positive. In other words, don't be sloppy and don't try to take shortcuts, because it will only hurt you in the long run. Trust me on this!
959197
I have factored it please refer to.\n\nFrom this solution can be found
234504
Multiply 3 times each of the factors in the parenthesis\n3x+12=27\nmove all like factors to each side, rememberr to switch signs if you move a factor to the other side\n\n3x=27-12\n\n3x=15\n divide each factor by 3\nx=5\n\nX is the variable
614345
(x+10)(x-10)\n(x-4)(x²+4x+16)\n(x+w)(x²-wx+w²)\n\nUnfortunately you just have to memorize the formulas.  There is no other way to figure out the factoring of the cubic expressions.
898442
You just factor\n(2x-1)(x-2)=0\nx=2, 1/2
541628
One of the factors will be x-x=0.\nSo, the answer is 0.\nThis is because 0 times any number is 0.
954300
ok\nso its factoring.this is hella easy.10x+40 is 10(x+4)\nso the answer is 10(x+4)there you go
141310
to solve for X, you have to separate X from everything else. in that problem 2X+3X=5, you would combine the 2X and 3X to get 5X. so the equation end up as 5X=5. from there, divide both side by 5. 5X/5=X and 5/5=1 so X=1.
283601
1) First we move all the factors with x to one side, and the ones without x to the other.  Result will be:\n\n4x/3 - 2x + x = 3/2 + 1/5\n\n2)  Then, we factor out an x from the left side and get:\n\nx(4/3 - 2 + 1) = 3/2 + 1/5\n\n3)  Calculate the actual numbers on both sides\n\nx(4/3 - 6/3 + 3/3) = 15/10 + 2/10\nx(1/3) = 17/10\n\n4)  Divide both sides by 1/3 and get a final result of...\n\nx = (17/10)/(1/3)\nx = (10/17)*(3/1)\nx = 30/17\n\n5)  Comform that I PWN n00bs.  And... I'm pretty sure I pwn n00bs.  Therefor, we are done!  ^-^
561923
You need to determine 2 numbers that add to -11 (the x term) and multiply to -60 (the numerical term).  You can do it by formula, or by a little trial-and-error.\n\nThe answers are -15 and 4, so the equation factors:\nx^2 - 11x - 60 = 0\n(x - 15)(x + 4) = 0\n\nIn order for the total to be zero, at least one of the factors must be zero.  So your solutions are x = 15 (which makes the first factor zero) and x = -4 (which makes the second factor zero).
219588
3x^3 + 2x^2 + x - 6\n= 3x^3 + 5x^2 - 3x^2 + 6x - 5x - 6\n= x(3x^2 + 5x + 6) - 1(3x^2 + 5x + 6)\n= (3x^2 + 5x + 6)(x - 1) \n\nThe factors are : (3x^2 + 5x + 6) and (x - 1) \n\nAlternatively, using polynomial division : \n(3x^3 + 2x^2 + x - 6) divided by (x-1) yeilds \n(3x^2 + 5x + 6)