Factor completely. ax – ay + x^2 – xy?
88953
=\na(x-y)+x(x-y)\n=(a+x)(x-y)\ndone
780960
I also have difficulty typing this thypes of math problems.  First, simplify the ax^2+bx+c parts in the first half to (x+c)(x+c) looking things first.  So...\n\n[(x-3)(x-1)]} is the first part.  With the bottom of the second half, divide by 3, so that you get 3(x+1).  Now line everything up like so...\n\n(x+1)(x+3)(x-3)\n_____________\n3(x-3)(x-1)(x+1)\n\nNow do the cancellations and you get the right answer...\n(x+3)\n____\n3\n\nSorry for not working to convention, but it was easier to type it that way.
223322
3x*(x^2-8xy+16y^2) Same factor\n3x*(x-4y)^2 Perfect squares.
208740
EDIT: ** Baah. When I typed this answer I was first. Then at the end, I'm like the 3rd. Oh well..**\n\nThere are two ways. Which one is easier depends greatly on your thinking skills.\n\nYou can take the expression x2+7x+12 and assume it should break up in two factors:\n(x+ ? )(x+ ?)\nThen, search for two numbers that add up to the 7, but when multiplied equal 12.\nIt helps if you break down 12 into it's components, 12= 2*2*3. So, obviously, the answers could be:\n\n2 * (2*3) = 2 * 6 (but 2+6 is not =7)\n3 * (2*2) = 3 * 4, voila! 3+4 also = 7, so:\n\n(x+3)(x+4).\n\nThe second way could be just solving the cuadratic formula for second degree equations:\n\nb=(-b +/- SQRT(b2 - 4ac))/2a\n\nWhere ax2 + bx + c, so you'll have: a=1, b=7, c=12.\n\nReplace, then solve twice: Once for the + value of the SQRT, another for the - value of the SQRT. You should get something like:\n\nFirst run:\nx= -3\n\nSecond run:\nx= -4\n\nx=-3 is the same thing as (x+3)=0\nand \nx=-4 is also (x+4)=0\n\nNotice the two factors are precisely (x+3) and (x+4)?\n\nHope this helped..
1172065
if it is a quadratic (ax^2 + bx + c) then put it into the quadratic equation. that automatically will give you 0, 1, or 2 real answers. the equation is: x = [-b +/- (square root of b^2 - 4*a*c)] / 2a\n\nif (b^2 - 4*a*c) is negative, you have no real solutions. if it is zero, you have one real solution. if it is positive, you have 2 real solutions.\n\nthen, if for example you get  x=4 and x= -2, your factors are (x-4) and (x+2)
1156185
You know the first term is x^2 so you can write:\n(x     )(x     )\nYou know that the second term is -4x:\nIf factors are of the form (x+A)(x+B) then  Ax+Bx = -4x\nYou also know that the last term is -5, so A*B=-5\n\nJust find the values of A and B that satisfy these equations:\nAx+Bx = -4x\nA*B = -5\n\nA = -5 and B = 1\n\nNOW that we have the factors\n(x-5)(x+1)=0\nTo get 0 either section of the left hand side can be 0, so solve both cases:\nCASE 1:\nx-5 =0\nx = 5\nCASE 2:\nx+1=0\nx=-1\n\nANSWER x=5 or x=-1
744954
f(x) = ax^2 + bx + c\n\nf(0) = c = 3\n\nf(1) = a + b + c = 7\n--> a + b = 4\n\nf(-3) = 9a -3b +c = 9*f(1) = 63\n9a -3b + c = 63\n9a -3b = 60\n9a - 3(4-a) = 60\n9a -12 +3a = 60\n12a = 72\na = 6\nb = -2\n\nf(x) = 6x^2 -2x +3
565133
We start by factoring the polynomial.  A little trial and error (testing the factors of 2, which is the constant term) reveals that 2 is a root.  So, we know that (x-2) is a factor.\n\nDividing the polynomial by (x-2) using synthetic division, we find the other factor is 6x^2+x-1.  This can be factored as (3x-1) (2x+1).\n\nSo, the equation, in factored form, is\n(x-2)(3x-1)(2x+1)=0\n\nSo, the roots are x=2, x=1/3, and x=-1/2.
323323
This is the best I could come up with:\n\n(a)  (x^2-4)(x^4+4x^2+16)\n      (x-2)(x+2)(x^4+4x^2+16)\n\n(b)  (x^3-8)(x^3+8)\n     (x-2)(x^2+2x+4)(x+2)(x^2-2x+4)\n\n(c)  If you take out the factors that are the same, what's left should be equal hence:\n     x^4+4x^2+16=\n(x^2+2x+4)(x^2-2x+4)\n\n(d)  x^(a^2)+2ax^a+4a^2=\n(x^a+ax+2a)(x^a-ax+2a)
526107
You are only allowed to factor if the equation is equal to zero; your's is equal to 4.  You need to solve "x^3 - 6x^2 +9x - 4 = 0"  using synthetic division.
770101
3ax+4by-2ay-6bx\n=3ax-6bx-2ay+4by\n=3x(a-2b)-2y(a-2b)\n=(3x-2y)(a-2b)\n2ax-2by+ay-4bx\n=2ax-4bx+ay-2by\n=2x(a-2b)+y(a-2b)\n=(2x+y)(a-2b)\nso 3ax+4by-2ay-6bx over2ax-2by+ay-4bx\n=3ax+4by-2ay-6bx /2ax-2by+ay-4bx\n=(3x-2y)(a-2b)/(2x+y)(a-2b)\n= (3x-2y)/(2x+y)
898442
You just factor\n(2x-1)(x-2)=0\nx=2, 1/2
993104
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)
591721
Let x^2=a\nTherefore we get\na^2-8a+15=0\nFactoring\na=5,3\n\nHence\nx=+/-sqrt 3\nx=+/-sqrt 5
978722
You can't, algebraically.  You have two equations and three variables.  Sorry.  You need one more equation, such as a = b.
631080
You can noy factor it out... it makes it easier to figure out when you put it into lowest terms... so it would be \n\n   x^2+7x-25..... so you cant factor it.
1174486
First look at the highest powers of a and b which in this case is "squared" and the product 6ab is present, so factoring will likely involve (a+b)^2\nIn fact 3a^2+6ab+3b^2 = 3(a^2+2ab+b^2) = 3(a+b)^2\nSo your problem reduces to:\n\n3(a+b)^2 +7(a+b) - 6\n\nIf we choose to factor -6 as -2 x 3 then the it is easy to factor, and the solution is:\n\n(3(a+b) -2)((a+b)+3)