FACTORING.. please help.. its a take home test?
487623
While I have no moral compunctions about math, being that math is a science, I would seriously suggest you double-check my math (especially on the 2nd one):\n\n(x^2+9)(x+1)(x-1)\n\nx(2x^2-3x+4)-6\n\n3x^2(rad^3(x-2))\n\n25(4x+1)(4x-1)\n\n3x^2(x+10)(x-10)\n\nrad^3(4x+7)
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.
883460
Please see the answer above and find out the missing factor in your calculation. You are in the right track. All the best!
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)
402918
I think you wrote 10x*2yz - 22x*3yz.  To factor this oyu must take out like terms by dividing.  You can see that both terms share a 2, y, z, and an x*2.  After the  division you get  2x*2yz( 5 - 11x).  I hope this helps.
899751
If you can do basic math (add, subtract, multiply or divide) you can do it. Unless you are going for a Subject Test for Math, then you need to get a study guide available from the test company or online.\n\nGood Luck
495916
3x^2+9x-84 \nGCF = 3\n3(x^2 + 3x - 28) now we factor the binomial\n\nwe could do this by multiplying the first coefficient (1) by the last coefficient/term (-28). we get -28\nnow we need to select two numbers that multipli to -28 and add to the middle coefficient, 3.\n\nthe two numbers are (7) and (-4)\nwe rewrite the problem with these two terms\n(x^2 + 7x - 4x - 28) i did not write the GCF 3 just yet\nfactor by grouping\n\nx(x + 7) - 4(x + 7)\n\nfactor completely\n\n(x + 7)(x - 4)\n\nfinal answer, don't forget about the 3\n\n3(x + 7)(x - 4)\n\n----------------------------\n4x^2 - 4x - 3   \n1. check for GCF - none\n2. first and last (4)(-3) = -12\n\ntwo numbers that multiply to -12 and add to -4 = 2 & -6\n\n4x^2 + 2x - 6x - 3\n\nfactor by grouping\n\n2x(2x + 1) - 3(2x + 1)\n\n=(2x + 1)(2x - 3)\n\n----------------------------\n3x^2-17x+22 \n\n1. check for GCF - none\n2. first and last   (3)(22) = 66\n3. two numbers are (-6) & (-11)\n4. rewrite\n\n3x^2 - 6x - 11x + 22\n\n5. factor by grouping\n3x(x - 2) - 11(x - 2)\n\n(x - 2)(3x - 11)\n\n----------------------------\nx^2-9 \n\ndifference of squares\n\na^2 - b^2 = (a + b)(a - b)\n\n1. GCF - none\n2. factor\nx^2-9 = (x)^2 - (3)^2 = (x + 3)(x - 3)\n\n---------------------\n\n-3x^2+48?\n\nalso difference of squares\n1. GCF = -3\n\n= -3(x^2 - 16)  \n\n2. factor: x^2 - 16 \nx^2 - 16 = (x)^2 - (4)^2 = (x + 4)(x - 4)\n\n3. final answer = -3(x + 4)(x - 4)
283601
1) First we move all the factors with x to one side, and the ones without x to the other.  Result will be:\n\n4x/3 - 2x + x = 3/2 + 1/5\n\n2)  Then, we factor out an x from the left side and get:\n\nx(4/3 - 2 + 1) = 3/2 + 1/5\n\n3)  Calculate the actual numbers on both sides\n\nx(4/3 - 6/3 + 3/3) = 15/10 + 2/10\nx(1/3) = 17/10\n\n4)  Divide both sides by 1/3 and get a final result of...\n\nx = (17/10)/(1/3)\nx = (10/17)*(3/1)\nx = 30/17\n\n5)  Comform that I PWN n00bs.  And... I'm pretty sure I pwn n00bs.  Therefor, we are done!  ^-^
272590
1/a + 1/b = c  multiply both sides by ab\n\nb + a = abc  get the terms with b on one side\n\na = abc - b factor out a b\n\nb(ac - 1) = a  divide by ac - 1\n\nb = a/(ac - 1)
282749
You will need to place the parentheses at the proper places. To help you out, I have assumed where they will be and solved it. Please confirm if these are correct. If not, rewrite the problem and I will try to solve it.\n\nThe text you find in between the steps is an explantion of what I have done...\n\n(7x/(x^2 + 4x)) divided by (35x^2/ x+4). This is my assumption.\n\nRewriting the problem in mathematical form\n= (7x/(x^2 + 4x))/(35x^2/ x+4)\n\n\nDividing by a fraction is the same as multiplying by its reciprocal\n= (7x/(x^2+4x)))*((x+4)/35x^2)\n\nRewriting after moving out the dommon factor in the denominator of the first fraction. [x^2+4x = x(x+4)]\n= (7x/(x(x+4)))/35x^2/(x+4) .)\n\nCanceling out the common terms in the numerator and denominator. Here it is (x+4)\n= (7x/x)*(1/(35x^2))\n\nCanceling out more terms. Here it is x\n= 7/(35x^2)\n\nCanceling out more terms. Here the number 7\n= 1/5x^2\n\nThis will be the final answer.
1095294
5) find the factors\nsquareroot of x * x * x * x * x * x /  7*7*5*5*2*2\nthe factors can come out of the square root by pairs only\nyou have 3 pairs of x's = x^3\nyou have 1 pair of 7's so it comes out of the square root as a 7\nx^3/7*5*2\nx^3/70 is it's simplified term\n\n__________________\nif u want all the correct answers then that would tarenirator's answers
178199
alright, you need to understand this much more than you need the correct answer. You won't have Yahoo! answers to help you on any tests.\n\nYou need help factoring. A real-life tutor who can talk to you and show you in person is best, but I'm going to try to help you understand. \n\na polynomial is in the form ax^2 + bx + c. This is from the formula of (Z*x + q) * (R*x + p). Don't let all these letters scare you. They are just place holders for numbers. \n\nThe only thing that can get the letter A in Ax^2 +bx +c is the first two parts of the "terms" you need to multiply together. With letters a= z*r. In order to get C, you only need to worry about the last parts in the terms. Using letters c= q*p. B is what makes sure you know what two terms were multiplied together. B = z*p + q*r. In order to "factor" out a term from a polynomial you have to find the set of z, p, q, and r that works. I'll show you now with numbers.\n\nLet's say you have the equation 2x^2 + 8x +8. See, it's in the form of ax^2 +bx + c. a=2, b=8, c=8. NOw you have to figure out the facotrs of 2 and 8 (which are a and c)\nfor a=2 the numbers that are multiplied together to get two(the factors) are 2 and 1. Since you know that only the first two parts of the things you multiply together have to get you a, you can set up something that looks like this: \n(2x + ?) *(x + ?) \n\nNow you need to find what two numbers fit into your equation to get you c=8. The factors of 8 are 1*8 or 2*4. Let's try putting 1*8 into the equation.\n(2x+1)* (x+8) = 2x^2 +17x + 8. That's not right, so lets try to reverse them.\n(2x+8)* (x+1) = 2x^2 +10x + 8. That's not right either, so we know only the factors of 2*4 will work in this equation.\nLet's try them\n\n(2x+2)*(x+4) = 2x^2 +10x +8. It may seem like we've tried each of the factors, but we can still reverse 2*4.\n\n(2x+4) * (x+2) = 2x^2 +8x+8. That's right. Once you get more practice and understand this better, you can "see" certain factors that have to be included just because "factoring" will be automatic. But, just to make sure you should always check you answer like I did when I showed you the example.\n\nNow that you hopefully understand the hard part, perfect squares are easy. All you do is take  the square root of 'a' and take the square root of 'c' and put them into the equation. When it is a PERFECT SQUARE this works because of: ax^2 + bx +c = (z*x+ q)^2. ax^2 = (zx)*(zx) so a = z^2, and the square root of a = z. c = (q)*(q) so c = q^2, and the square root of c= q. I'll use numbers now.\n\n100x^2 + 240x + 144. a = 100, so the quare root of 100 = 10. z=10. Then the square root of 144 is 12, so q=12. That means y=(10x+12)^2. Now, to find b, all you have to do is foil it out.
930308
x=[2+/-rt(4+60)]/2\n=[2+/-8]/2\n=-3 or 5\nso the factors are\n(x+3)(x-5)
175207
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.
87985
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!
1054627
assuming this is (3a+1)/(a^2-1) - 1/(a+1)\n\nfactorise the (a^2-1) to give (a+1)(a-1)\n\nmultiply the second fraction by (a-1)\n\n-->  (3a+1)/[(a+1)(a-1)] - 1(a-1)/[(a+1)(a-1)]\n\nfactorising this gives [(3a+1)-(a-1)]/[(a+1)(a-1)]\n\nwhich equals (2a+2)/[(a+1)(a-1)]\n\n= [2(a+1)]/[(a+1)(a-1)] \n\nthe (a+1)'s cancel out...\n\ntherefore = 2/(a-1)
907035
You have \n100 +4X^2 -16y -40x\n\nFirst let's factor out a 4, since 4 is a factor of each term, that gives\n\n4*(x^2 -10X +25) -16Y\n\nOf course I rearranged the terms so the highest powers of X are\non the left.  I didn't factor 4 out of -16Y because I want the \nquadratic (X^2-10X+25), which I know I can factor.\n\nsince (X^2-10X+25) = (X-5)*(X-5) = (X-5)^2, we have\n\n4*(X-5)^2 -16Y\n\nThe secret is being able to recognize that \n(x^2 -10X +25) = (X-5)^2