PLEASE HELP!!  Factor   2a (squared) - 10a   check by multiplying?? thanks for any help on this!?
831044
2x² - 10a\n\n2a(a - 5)\n\nThe answer is 2a(a - 5)\n\n- - - - - - - -s-
91424
x*SQRT(8+3x)*SQRT(50-4)*2x^2?????\n\nYour statement can be expressed as different algebraic expressions.\n\nExample: The last statement, "2x squared" can be 2x^2 or (2x)^2.\n\nTry to write it in algebraic form with, paranthesis if needed.
193745
1.25 * 10^-3 * 8 * 10^-5\n\n(1.25 * 8) * (10^-3 * 10^-5)\n\n(10^1) * (10^-8)\n\nANSWER : 1.0 * 10^-7
890098
I enjoy algebra, and this is easily my favorite type of problem.  So I hope you, too, one day enjoy them for their inner beauty.\n\nFor now, let's tackle this "factoring problem."  We may wish to use the method of "synthetic division" which is arguably easier than "long division of polynomials."\n\nThink back to grade school where you learned that 6 = 2 * 3.  This simple statement enabled you to 'deconstruct' the number 6 as either "two times three" or "three times two".  Even if you forgot one of the 'factors' you could conjure it up again quickly by asking yourself, "What is six divided by two?" for example.\n\nIn problem (a), you are essentially asking, "What are the coefficients of P(x) if the only factor we know is (x-1)?"\n\nLet's set up synthetic division (see link below), by placing all the coefficients of P(x), and using 1 (the zero implied by (x-1)) as the divisor:\n\n1 ||  a  b  c  d\n___________\n.....a\n\n1 ||  a  b  c  d\n_____a_____\n.....a (a+b)\n\n1 ||  a  b        c       d\n_____a___(a+b)____\n.....a (a+b) (a+b+c)\n\n1 ||  a  b        c                d\n_____a___(a+b)___(a+b+c)_\n.....a (a+b) (a+b+c) (a+b+c+d)\n\nNote that this last term must equal 0 since it is assumed that (x-1) is a factor ... thus, the "remainder" term must equal zero!\n\nThere's the result you were after:  a+b+c+d = 0\n\nHope this helps you tackle the others!
814352
(12x^2 +16x) - (12x-7)- (4x-5)\n12x^2 +16x -12x +7 -4x +5\n\n12x^2+12\n\nthat is what I got....\nI think you might be able to put 12 on the other side of the = sign and turn it into 12x^2 = -12 and then since both sides have 12 in common you can divide both sides by 12 and you will have x^2 = -1 then you can add 1 to both sides and you will have x^2 +1 but I think I'm getting off track by that point....it's been a while since I had to do this stuff.
386814
They are correct.  Observe the order of operations or PEMDAS\n\nParenthesis\nExponents\nMultiplication-Division\nAddition-Subtraction\n\nso first multiply   -89 * 658 =   -58562\nThen add 2   =                    -58560\nThen add 6   =                    -58554\nThen subtract 89 (or add -89) =   -58643
842934
Ok, first you must get the -8 and the x/-2 out of parentheses, so u just multiply them both by one. So then is -8 * x / -2. First, you do -8 * x, which is -8x. The problem then becomes -8x / 2. In this step, just don't mind the negative or the x and just put them back in at the end. So you're really just doing 8 / 2, which is 4. You must put the - and the x back in, so the answer is -4x. Ta Da!
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.
959197
I have factored it please refer to.\n\nFrom this solution can be found
770521
When your first term is x², you can look at the other two coefficients -- in this case -8 and +15.\n\nTo factor, look to see what two numbers when multiplied together will give you your last coefficient (+15) and when added together will give you your middle coefficient (-8).\n\nSo in your case, we see that we need two NEGATIVE numbers because 15 is positive and only two negs or two positives can give you a positive when multiplied.\n\nWe know that the two must be negative, because we need to get a negative 8 when we add them together.\n\nSo, we can see that +15 = -5 * -3\nand -8 = -5 - 3\n\nSo, our factorization is (x-5)(x-3).\n\nNotice we plugged in our -5 and -3 into (x__)(x__).\n\nHope that helped somewhat!\n\n.
967957
There really isn't any way to do this other than guess and check. If you find that you substitute some value in and find f(a) = 0, then (x-a) will be a factor. You can repeat that a few times until you have found some roots, and then do a long division to something which you *can* solve by hand (a quadratic or linear).\nNote also that you only need to check things which are factors of the constant term, and there aren't too many.\nIn the first case, you will find that substituting in -2, -1 and 3 will all give 0, but you may have trouble finding the last root. So you will have to do some long divisions (divide by (x+2)(x+1)(x-3)), and you will end up finding that the -1 was a repeated root - so the answer is (x+2)(x+1)^2(x-3).\n\nSame thing for the second: you will find -1 as one root. You can then do a long division and find -1 is repeated, and again, and again, and again. The actual factorisation is (x+1)^5 - you may have noticed this if you knew much about pascals triangle and recognised the coefficients are an entire row - 1,5,10,10,5,1.
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
841476
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.
774014
2^3+2*2^2-3(2)-10=8+8- 6 -10 =0\nTherefore x-2 is a factor of the given polynomial
781910
not possible to factor easily ... try the quadratic formula.
550823
wow, if you typed in the question correctly, it's going to be tough to solve...maybe even impossible, are you sure that the result isn't supposed to be a^2 +8a -20?  (perhaps you left off the zero at the end of the two??)\n\nif that's the case, then the answer is (a-2)(a+10)=a^2+8a-20\nhere's how to get this answer, and again, i'm making the assumption that 2 should be 20 instead...\n\nyou know from (a+10) ... = a^2... that you're going to need something multiplied by a that gives a^2, so the only possible choice is a\n\nnow you have (a   )(a+10)=a^2 +8a -20 since the 20 has a negative sign before it, you know that something times a positive 10 needs to give a minus 20, so -2 works.  and conveniently when you have (a-2)(a+10) you get a^2 +8a -20.  \n\nuse the FOIL method (first, outer, inner, last)\n\nthe first terms (a and a) multiplied is a^2, then the outer (a and 10) gives 10a  then the inner (-2 and a) gives -2a and the last (-2 and 10) gives -20 \n\na^2 + 10a -2a -20 is the same as a^2 +8a -20.
437832
Homework.\n\nSuppose a1*x+a2*Tx+..+am*T^(m-1)x=0.\n\nApply T^(m-1) to get a1=0.\nApply T^(m-2) to get a2=0, etc.