Plz help me with this one i have a problem factoring it.
207966
When you have groups of terms like your problem you factor out the things that each group of terms share. \n\nFor your first part: (a^2c^2 + b^2c^2 ) the terms share a c^2 so you factor that out. \n\nyou have c^2 (a^2 + b^2)\n\nThe second terms (-4a^2d^2-4b^2d^2) share 4d^2 so you factor that out. \n\nSo you're left with c^2(a^2+b^2) - 4d^2 (a^2 + b^2). \n\nWhen you've done this part you can actually go farther because if you look at it both sections share (a^2 + b^2). \n\nso you can simplify the whole problem with \n\nc^2 - 4d^2 (a^2 + b^2).
395786
Notice that both terms have a 'factor' of (p+q) \nThe first term is (p+q)*(p+q) and the second is (5)*(p+q)\nSo 'factor out' the (p+q) from each :\n= (p+q) * [ (p+q) + (5) ]\n= (p+q) * (p+q+5)\nis the factorization.
489565
You just factor\n(2x-1)(x-2)=0\nx=2, 1/2
447345
When you retry the problem remember to move only yout z^-2 up don't move the 4 that is a constant in the denominator. \n\nyou should get this . (12x^5z^3/30x^7y^7) * (15x^2z^10/4y^3)\n\nremember to simplify as best you can prior to multiplying.\n\ntry from here and let me know what you get.
519078
T=m1*-9.8+m1*a\nT=m2*a--m2*-9.8\n\nset them equal and solve for a\n\ntrust me, i had a test on this today
413909
1.factor\n(2x+1)(x-4)=0\n2.set it to 0\n2x+1=0\nor\nx-4=0\n3.solve\n{x=-1/2 or x=4} <=answer
441522
Do you mean (2x^2 + 3x + 1/ 2x^2 + x - 3) for your first fraction?  I'm going to assume you do.\n\nFirst, factor each numerator and denominator:\n\n2x^2 + 3x + 1 = (2x+1)(x+1)\n\n2x^2 + x - 3 = (2x-3)(x+1)\n\n4x - 4 = 2(2x-2)\n\nNow let's look out our multiplied fraction, rewritten using our factors we just found:\n\n[ (2x+1)(x+1)2(2x-2) ] / [ (2x-3)(x+1)(2x+1) ]\n\nNotice we can cancel out the (2x+1)'s and the (x+1)'s, to get:\n\n2(2x-2)/2x-3\n\nANSWER:   2(2x-2)/2x-3
615323
Good job!  You're halfway there!\n\nAs you are writing it out on your homework, underline the (4x+1) in each of the two terms.  That helps you see what you are factoring.\n\nFactoring out the (4x+1) from each term, you should get your final answer.\n\n=(5x + 4y)(4x+1)
391848
1. 5m^2 - 15mp - 2mr - 6pr <-- Impossible to factor by grouping. It needs to be either, exactly one of the signs needs to be reversed.\n\n2. 6-3x-2y+xy\n(6 - 3x) - (2y - xy)\n3(2 - x) - y(2 - x)\n(3 - y)(2 - x)
69503
Next time use ^2 instead of typing out "to the 2". Anyway, write it in standard form, so that the variables are in descending order.\n\n14w^3 + 49w^2 - 8w - 28\n\nYou can't factor this. If it was 7w instead of 8w, then you could factor out a 7 by dividing all the coefficients by 7 and pulling it outside the parentheses. Alternatively, you could factor the first two terms separately from the second two:\n\n7w^2(2w + 7) - 4(2w + 7)\n\nNow it's in a great form for factoring. Group them.\n\n(2w+7)(7w^2-4)\n\n---\n\nDo the same thing for the second example.\n\n44t^3 - 4t^2 + 66t - 6\n\nFor this one, you can factor out a 2.\n\n2(22t^3 - 2t^2 + 33t - 3)\n\nOr, again, you can factor the terms separately:\n\n4t^2(11t - 1) + 6(11t - 1)\n\nGroup.\n\n(11t-1)(4t^2+6)\n\n\n\nHope this helped.
33423
i got it!\n\nit's 3(2c+7)(c-4d)(c+d)!!!!\n\n6c^3 + 21c^2 - 18dc^2 - 63cd - 24cd^2 - 84d^2\nfirst we take out factor 3:\n3(2c^3 + 7c^2 - 6dc^2 - 21cd - 8cd^2 - 28d^2)\nthen we take out (2c+7) from each pain inside the parenthesses:\n3[(2c+7)c^2 - (2c+7)3cd - (2c+7)4d^2)]\nwe take (2c+7) out:\n3(2c+7)[c^2 - 3cd - 4d^2]\nwe factor [c^2 - 3cd - 4d^2] trinomially:\nc^2 - 3cd - 4d^2 = (c-4d)(c+d)\nthen we get:\n3(2c+7)(c-4d)(c+d)\n\nam i right??\n\n:) yeah i didn't believe myself...\nbut if u really write down all your factors, it's much more easier to take them out. and it WAS fun :))
239281
start pulling out common factors\n\n2 (x^3 + 27)\n\n2(x^3 + 3^3)
522019
First, check for rational roots. If this doesn't have\nany we are in trouble, but fortunately it does have some.\nIf r/s is a rational root of this equation, then r must\ndivide -35 and s must divide 2.\nBy trial we find that -1 is  a root of this equation,\nand so x + 1 is a factor. But x+1 is also a root\nof the quotient, so x+1 is a double root.\nNow work on the resulting cubic.\nWe  find that x = 5/2 is one of its roots.\nThe final quadratic is\nx² - 4x + 7 = 0\nwhich you can now solve by the quadratic formula.\nHope that helps a bit!
78408
I haven't done factoring for a while but I'm pretty sure this is right.\n\nSo you start with 125q²-n³q²\nIf you take out both the q² 's you have q^4 \n(q x q x q x q)\nSo you are left with q^4(125-n³)\nThe cubed root of 125 is 5\n(5 x 5 x 5)\nand the cubed root of n³ is n\n(n x n x n)\nSo your answer is q^4(5-n)\n\nI think that's right.
130751
are you trying to solve for x?\n\nFirst, find the factors for the equation.  (x+2)(x+1)\n\nCheck it to see if it's right.  x times x equals x squared. 1x plus 2x equals 3x. 2 times 1 equals 2.\n\nMake one of the parenthesis groups equal to zero.  For example, x+2 = 0   To solve this, x must equal -2.\n\nPut -2 back into the original equation and see if it works.\nNow make the parenthesis group equal to zero.  x+1=0  To solve this, x must equal -1.\n\nTherefore, the answer is using either -1 or -2 will make the equation zero.\n\nHope this helps.  ;)
39161
U need to use the quadratic formula.
652710
2(x^2+3x-40) <<<<<The answer\n\n\nfor the guy below me...that wouldn't be factoring.