3c^2 - 5cd - 6d^2\n\nI can't seem to find any factors of 6*3 (18) that will equal 5.  Help?
210139
Actually, -18 *does* have something to do with it. Try to find two numbers which multiply to give -18 and which add to give -5. In this case, it doesn't work, so the polynomial doesn't factor.\n\nConsider 8x^2-10x-3. Multiply the 8 and -3 to get\n-24. Now find to numbers which multiply to give -24 and which add to give -10. In this case -12 and 2 work. Now do this:\n\n8x^2-10x-3=\n8x^2-12x+2x-3= (using the -12+2=-10)\n4x(2x-3)+(2x-3)= (factoring 1st two and last two terms)\n(4x+1)(2x-3).\n\nThis gives a better way to factor that simple guessing. It will always work if the quadratic can be factored.
457847
You need a common denominator .  You could reduce the third expression before going on; it is 2/3x\n\nThe CD for 3x and 6x is 6x\n\nMultiply top & bottom of first expression and new third expression by 2; leave the secons as it already has the C.D.  You get\n\n8/6x - 5/6x + 4/6x\n\nNow just add the tops: 8-5+4=7 so it is 7/6x
391750
Since ABC divided by 5 and by 6 are both integers and that 5 and 6 are relatively prime, it follows that ABC is divisible by 30, so C is 0.\n\nBy the division, DE and ED are integers with DE the larger and divisible by 6 and ED the smaller and divisible by 5.  Also we can deduce that DE = (6/5)*ED.\n\nSince ED is divisible by 5, then D is either 0 or 5, but if it were 0, then DE would be a one digit number and less than ED.  So D must be 5.\n\nSo we are looking for a single digit E so that 5E = (6/5)*E5, and that 5E > E5.  Since we only have 1, 2, 3, and 4 as our candidates (0 is right out by a similar one digit number argument as above).  We can check each case to find that E=4.\n\nSo ABC = 6*ED = 6*45 = 270.  Thus the numbers are\n\nA = 2\nB = 7\nC = 0\nD = 5\nE = 4.\n\nCheers!
660632
18r^4s^5t^6\n---------------- =\n-3r^2st^3\n\n-6r^(4-2)s^(5-1)t^(6-3)= -6r^2s^4t^3\n\nhope it's not too late.\n\nGood luck!
435368
x^3 - 6x^2 - 3x + 18 \n= x^2 (x -6) - 3(x-6) \n= (x^2 - 3)(x - 6)
112276
15cd = 3*5*c*d\n30c^2d^2 = 2*3*5*c*c*d*d\n\n15cd+30c^2d^2 = 3*5*c*d(1+2*c*d) = 15cd(1+2cd)
469770
You could factor that equation this way:\n\n9c^2-64\n= (3c-8)(3c+8)
21535
rt11guru is correct up to the point where he says:\n\nX = ((2x2)^15)x((3x3)^6 x 3)x((5x5)^2 x 5)x((7x7)x7)\n\nThe number of possibilities is then given by:\n\nN = (16 * 7 * 3 * 2) - 1 = 671\n\nYou need to subtract 1 to account for the case where the divisor has no prime factors (and so equals 1).  If 1 counts as a solution (perfect square and divisor) the number of solutions is 672.
411003
6z+11=5z+3\n=>6z-5z=3-11\n=>z=--8\n/Therefore, (C) is corret
125381
r is probably 27 and t is 8. Here's my reasoning:\n\n18 * 18 * 18 = r * r * t\n5832/r = r * t\nLets factorize 5832\n5832 = 27 * 27 * 8\nAnd Since 27 > 8, it can be accepted as the solution.\n\nI don't know if this is the correct way of doing it. Anyway, in my opinion worth a shot.
291500
In 6ab + 4a + 3b + 2, we group the first 2 terms and the last 2 terms:\n\n6ab + 4a + 3b + 2 =\n(6ab + 4a) + (3b + 2)\nthen factor out the common factor 2a in (6ab + 4a)\n(Note: to factor out is the same as dividing each term with the common factor)\n\nwe get:\n2a(3b + 2) + (3b + 2)\nNotice that in each term of the above we have a common factor of (3b + 2)\nBy dividing each term by (3b + 2) or factoring out (3b + 2) we have:\n(3b + 2)(2a + 1)
573287
factors\n1=1\n2=2x1\n3=3x1\n4=2x2=4x1\n5=5x1\n6=3x2=6x1\n7=7x1\n8=2x2x2=4x2\n9=3x3=9x1\n10=5x2=10x1\nMultiples\n\n1=1,2,3,4,5,6.......\n2=2,4,6,8,10,\n3=3,6,9,12,15\n4=4,8,12,16\n5=5,10,15,20\n6=6,12,18,24\n7=7,14,21,28\n8=8,16,24,32\n9=9,18,27,36\n10=10,20,30,40
427548
HI!\nFor the first: 5 times six equals 30.\n\nSecond: x to the yth power.\n\nWhen in doubt...don't be afraid to ask! :) :)
130462
You need to factor everything, then cancel common factors.\n4x+8 factors as 4(x+2)\n6x-3x^2 factors as 3(2-x)\n3x^2-4x-4 factors as (3x+2)(x-2)\n9x^2-4 factors as (3x+2)(3x-2)\n\nCancel the (3x+2)'s, and the (x-2) can cancel with the (2-x) leaving a factor of -1.  The factors that are left are:\n[-4(x+2)]/[3(3x-2)].
503129
5a             3b\n------ +   ---------=   \n6b^2        8a^2c\n\n\n5a*8a^2c + 6bc^2*b  \n--------------------------=     (you add the rational numbers)\n8a^2c*6bc^2             \n\n\n\n40a^3c + 18b^2c^2\n------------------------=       (you multiply the expressions)\n48a^2bc^3\n\n\n\n2c(20a^3+9b^2c)\n---------------------=         (you factorize using the LCD=2)\n48a^2bc^3\n\n\n\nthe simplest form:\n\n20a^3+9b^2c\n-----------------\n24a^2bc^2
443809
Its 5=2n-6 \n\nwell done!
654713
No problem, i just finished a calc3 course in college.\n\nYour problem is the same as 125 * (6x^2 + 5x + 7)^(-1/3)\nYou have to use the chain rule so you get\n125 * (-1/3) * (12x + 5) * (6x^2 +5x + 7)^(-4/3)\n\nthen you can simplify how you like from there