factor the polynomial: 6ab+4a+3b+2?
610613
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)
966048
(3ab^2)(3ab)^2\n(3ab^2)(9a^2b^2)\n27a^3b^4
259003
Use the quadratic formula.  It works everytime!\nx = (-b plus or minus sprt(b^2 - 4ac))/2\nwhere\nax^2 + bx + c = 0\nSome of these will factor, however not all of them will.
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)
402319
ya it is a^2 +b ^2 +2abcos(c)
917477
where else?\n\nhttp://www.factoring-polynomials.com/
397605
12-13b+b^2=(12-b)-(12b-b^2)\n  = (12-b)-(12-b)b\n  =(12-b)(1-b)\n\nThis might not be easy to see, so here is the general approach:\n  Cb^2 + Db + E = (b-B1)(b-B2)\nwhere B1 = (-D + sqr(D^2-4CE) ) /(2C)\n and B2 = (-D - sqr(...) ) /(2C)\n\nIf (D^2-4CE)<0, you cannot factor it.\n\nsqr is square root. In your case C=1, D=-13, E=12, so you get B1=12 and B2=1
532215
Your question is to factor in the following form y=a(x-p)^2+q for the question y=x^2-8x+8. Okay, let's roll.\n\ny=x^2-8x+8\n = (x-4)^2 -16 +8   [using (a-b)^2 = a^2 -2ab +b^2]\n = (x-4)^2 -8\n\nTherefore, in the form  y=a(x-p)^2+q, a=1, p=4 and q= -8.
914755
your equation must be like x^2-3x-6 and your trying to factor this\n\n2 numbers say a and b must multiply to make -6 and add to make -3 so a*b=-3 and a+b=-6\n but i can't think on any that fits this criteria so you must use the quadratic equation\n\nx= [-b +-squareroot(b^2-4ac)]/2a\nwhere a=1 b=-3 and c=-6\nx= 3+-squareroot(9-4(1)(-6))/2\n\nx=(3+squareroot(33))/2 and x= 3-squareroot(33)/2\n\nin decimals x=4.3723 and x=-1.3723
767438
b1=(-b+ (b^2-4ac)^1/2 )/2a=(-8+(64-48)^1/2)/6=(-8+4)/6\n    =-4/6=-2/3\nb2=(-b- (b^2-4ac)^1/2 )/2a=(-8-4)/6=-12/6=-2
760715
If it is divisible by x - 3, then (x - 3) is a factor of that polynomial, and if x - 3 = 0, x = 3.  \n\nIf we substitute x = 3 to the polynomial, then it must yield 0 (since 3 is a root).\nx³ + 3x² + kx - 24\n\nThus,\n(3)³ + 3(3)² + k(3) - 24 = 0\n\nThen,\n27 + 27 + 3k - 24 = 0\n\nTranspose,\n3k = -30\n\nThus,\nk = -10\n\nThe polynomial then becomes\nx³ + 3x² - 10x - 24 = 0\n\nWe then divide it by x - 3, to factor it out.  The easiest method goes like this\nx³ - 3x² + 6x² -18x + 8x - 24 = 0\n\nNow, we group\n(x³ - 3x²) + (6x² - 18x) + (8x - 24) = 0\n\nWe factor out from each group\nx²(x - 3) + 6x(x - 3) + 8(x - 3) = 0\n\nWe then factor out x - 3\n(x - 3)(x² + 6x + 8) = 0\n\nWe can factor the quadratic trinomial\n(x - 3)(x + 2)(x + 4) = 0\n\nTherefore, your polynomial is also divisible by (x + 2) and (x + 4).\n\n^_^
1144353
4a+4b+4c=2ab+2bc+2ac \n4(a+b+c)=2(ab+bc+ac)\na=2\nb=2\nc=2\n\nor \na=1\nb=2\nc=4
62092
if you mean 3(2ab-x) ... then 6ab-3x
528677
2ab+ 2ab=4ab-ab=3ab   answer is 3ab
447887
You probably mean\np(x) = 2x^4 +5x^3 – 3x^2 -2x -6\nThe answer is e) Both b,d\nbecause when you multiply the factors you should get the original expression and (x+3)(2x^3 –x^2 -2)= 2x^4 +5x^3 – 3x^2 -2x -6\nThis was quickly seen because you need x^4 which is only obtainable by x multiplied by x^3 in your problem :)\nHope this helps
78119
I guess you must factorize them.\nAll the polynomials you wrote are to be factorized by partial factorization.\nI think you know the general rule.\nA(x)P(x)+A(x)Q(x) + B(x)P(x) + B(x)Q(x) = \n---> factorize between the 1°-2° and 3°-4° <---\nA(x)[P(x) + Q(x)] + B(x)[P(x) + Q(x)]\nYou get the same binomial term [P(x) + Q(x)] which is a common factor. Factorize:\n[P(x) + Q(x)]*[A(x)+B(x)]\n\nFor example, in the first:\nA()=a\nB()=b\nP(x)=2x\nQ(y)=3y\nYou get \na(2x+3y)+b(2x+3y) = (2x+3y)(a+b)\n\nIn all the others you must rearrange the terms to use the above steps:\nin xy+32+8x+4y the repeating patterns are 8x + xy and 32 + 4y ....\n\nThink about it and good work
902478
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