explain factoring in Algebra 2?
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"A factor of a number is any integer by which the number can be divided evenly, with no reminder.  The factors of 6, for example,  are\n1,2,3,and 6, . . ."\nFor more details you may wish to consult Encarta encyclopedia--factoring in algebra.
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The instruction is to write a polynomial function with the zeros 0, 1, and 1+i. Please keep in mind that when a zero consists radicals or imaginary numbers, the conjugate of that zero is also a zero. For exmaple, here 1+i is a zero. Then 1-i must also be a zero.\n\nThis problem is like working backwards. Instead of factoring the polynomial, you are asked to do it backwards, starting with the zeros. Since 0, 1, and 1+i and 1-i are zeros, you would expect the last step of the factoring to look something like this:\n\nx(x-1)[x-(1+i)][x-(1-i)]\n\nNow simplify this expression a bit. It becomes:\n\nx^2-x[(x-1-i)(x-1+i)]\n\nSimplify this expression even further and you get:\n\nx^4-3x^3+4x^2-2x
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You can reduce you work by knowing that if your polynomial looks like  Ax^n + ... + B that if you have a factor that looks like (ax + b) that a has to be a divisor of A and b has to be divisor of B. That means that the only possibilitys in your example are\n(x + 1), (x - 1), (-x + 1) and (-x -1)\nSince (-x - 1) = -1(x+1) and (-x + 1) = -(x - 1)\nyou only have to look at two cases (x+1) and (X-1)\n\nTo find out if either of the two is a factor, you can do long division on the polynomial just like you would with regular numbers. If you don't know how to do that look at the link below for an example.\nI can't show you here because it won't format correctly.
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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
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You just factor\n(2x-1)(x-2)=0\nx=2, 1/2
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I'm assuming you're looking for information on factoring polynomials since it's a College Algebra courese.  \nThis website is very helpful, with the steps for factoring laid out, and it also provides websites you can use if you need more information.  Good luck!\n\nhttp://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut7_factor.htm
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Write it as:\n2x^2-4x+3x-6\n=2x[x-2]+3[x-2]\n=[2x+3][x-2]\nSimple middle-term factorisation!!!
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I'm going to assume the placement of parenthesis.\n(x^2 – 3x + 2)/(8x - 8) ÷ (x^2 – 4)/(5x +10)\n\nStep 1: Flip the second fraction, so that you have multiplication instead of division.\n(x^2 – 3x + 2)/(8x - 8) * (5x +10)/(x^2 – 4)\n\nStep 2: Factor.\nx^2 – 3x + 2 = (x - 2)(x - 1)\n8x - 8 = 8(x-1)\n5x +10 = 5(x + 2)\nx^2 – 4 = for a^2 - b^2, the factors are (a + b)(a - b). So: (x + 2)(x -2)\n\n(x - 2)(x - 1) / 8(x-1) * 5(x + 2) / (x + 2)(x -2)\n\nStep 3: Cancel out, multiply\n(x - 2) / 8 * 5 / (x -2)\n5 / 8 (solution!)
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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.
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When you have a quadratic equation Nx² + Mx + P, you can sometimes factor it into (ax+r)(bx+s).  Here's a way to determine a, b, r and s:\n\nWrite down N and M and P, then look at what two numbers a & b when multiplied will equal N, and what two numbers r and s when multiplied will equal P and at the same time when all multiplied together (as + rb) will equal M.  It sounds scarier than it is!  Here's an example with your first equation:\n\nb² + 4b - 12\n\nSo we have our coefficients (N,M,P):  1  4  -12\n\nFor 1 = ab our only choices are a=1, b=1\n\nFor -12 = rs, we have several choices 1,12 ... 2,6 ... 3,4  with one factor being positive and one negative so we can get -12.\n\nWhich two factors have a difference of 4?  Well it's 2 and 6.  And since 4 is positive we want +6 and -2.   So putting it all together we get:\n\nb² + 4b - 12 = (b-2)(b+4)\n\nLet's try one more (and then I have to go to a meeting so I can't help you anymore right now, sorry!) ...\n\n2b² - 13b +15\nOur coefficients are:  2  -13  +15\nFactors of 2:  1,2\nFactors of 15:  1,15 ... 3,5\nHow can we combine factors of 2 and factors of 15 to get -13?\nNotice that -15+2 = -13, so our factors are:\n(2b-15)(b+1) = 2b² - 13b +15\n\nHope that helped a little.  Good luck!\n\n~ ♥ ~
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(x+7)(x+3)\n(x-11)(x+2)\nforget this one.  its AC factoring though\nthis ones AC too
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If you're talking about algebra I, you are probably dealing with quadratic expressions like the following:\n\nx^2 + 7x + 6\n\nThese are a bit like a puzzle.  If the number in front of the x^2 term is 1 (as in the above example) then the puzzle is like this:\n\nLook at the 7 and the 6.  What two numbers multiply to 6 and ad to 7?\n\nConsider all the pairs of numbers that multiply to 6\n\n1,6 or 2,3.  Which pair adds to 7? The 1 and 6.\n\nSo the above equation factors to (x + 6)(x + 1)\n\nAnother example is x^2 - 6x - 16\n\nThe ques ion becomes what two numbers multiply to -16 and add to -6.\n\nThe pairs of factors of -16 are \n+ or - 1 and 16\n+ or - 2 and 8\n+ or - 4 and 4\n\nWhich pair of these factors add to -6?  With a little checking you can see the pair is -8 and +2\n\nSo this equation factors to\n(x - 8)(x + 2)\n\n\nThis even works for equations like the following:\nx^2 - 16\n\nWhat you have to remember is that this can be re-written as\n\nx^2 + 0x -16\n\nSo, what multiplies to -16 and add to zero.  Use the last example and you should see this one for yourself.\n\nGood luck.
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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)].
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Factor a^2 + 8a + 16\na^2 + 8a + 16 = \n(a+4)(a+4) =\n(a+4)^2\nNow \n(a^2)+8a+16-(b^2)\n= (a+4)^2 - b^2\nThis is a difference of squares.\n(a+4+b)(a+4-b)
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The expression was not set equal to anything.\n\nIf you're doing simple factoring in algebra class, the \nsimplest factors would be (b+4) (b+3).\n\nIf the expression was set to equal 0, solve for b, then your answer set would include -4 and -3.\n\nHope this helps!
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With factorisation you turn the equation x^2 + px + q = 0 into two smaller equations in brackets like this:\n\n(x+a)(x+b) = 0\n\nSo that a+b = p\nand a*b = q\n\nEG. Your first example you have p = 4 and q = 3\n\nSo you've got a+b = 4 so a could be 2 and b 2 or a 1 and b 3 etc\n\nand a*b = 3 so a could be 1 and b 3... Yes it matches.\n\n(x+3)(x+1) = 0\n\nThe solution is just x = -a and x = -b\n\n-> x = -1 or x = -3
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Not factorable unless you use imaginary numbers; if you haven't learned about those, then the answer is no.  \n\nA real solution would be similar to the form\n(3x+5)^2 or (3x-5)^2, but when you square these expressions (FOIL), you end up with an extra term +30x or \n-30x.\n\nAnother method is to apply the . . . can't remember the name, but the formula is this:\n\nFor ax^2 + bx + c = 0, solution may be found by \n\nx = (-b +/- sqrt(b^2-4ac))/2a\n\nfor 9x^2 + 25, b = 0, so when you do the square root part, you will be trying to take the square root of a negative number, which doesn't work for most algebra.  The graph of this function is a parabola that never crosses the x axis and as a result, doesn't have any roots.  Therefore, it is not factorable either.