Everyone explained the first part now I do not understand this section.  Please help me.  Explanations are wonderful.\nMy teacher does not know how to explain these problems.\n\nx^2 + 11x + 30\n\nx^2 - 6x + 8\n\nx^2 + 19xy +48y^2\n\n6x^2 + 18x + 12
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well lets take the first example- ok so the middle number is 11. and the number with no 'x' in it is 30. what makes 30 = 6 x 5 or 3 x 10. so try adding and subtracting those numbers in your mind: 6+5=11 ... 3+10=13 ...6-5=1...5-6=-1...10-3=7...3-10=-7. so which one of those give 11  ->> 6 + 5. \nso all u do is x^2+ 6x + 5x + 30 [split 11 as 6 + 5]\nur next step ; take 'x' common for the first to terms and 5 common for the 2nd two terms, u get\nx(x+6)+ 5(x+6)\ntake the bracket term multiplied by the un-bracket terms\n(x+5)(x+6)\n\nvoila...problem solved
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x^2+y^2=(x+y)^2-2xy\nsubstituting\nx^2+y^2=3^2-2(1)\n            =9-2\n            =7\nx^2+y^2=7\nmy 10 pointsplease
231686
It has been a long time, here are suggestions:\n1. 5x^3 + 11x^2 + 3x + 1  \n2. 4x^3 - 13x^2 - 8x + 7 \n3. 2x^4 - x^3\n4. 21x^2 + 26 x + 8
413661
a.)\nx^2 - 12x\nx^2 - 12x + 36 = 36\n(x - 6)^2 = 36\n(x - 6)^2 - 36 = 0\n\nb.)\nx^2 + 22x\nx^2 + 22x + 121 = 121\n(x + 11)^2 = 121\n(x + 11)^2 - 121 = 0\n\nc.)\nx^2 + 13x\nx^2 + 13x + (169/4) = (169/4)\n(x + (13/2))^2 = (169/4)\n(x + (13/2))^2 - (169/4)) = 0
34141
(3y+1)(-2)+y\n-6y-3+y\n-5y-3\n\n12s+(7-s)2\n12s+14-2s\n10s+14\n\n7x-3x(x+1)\n7x-3x^2-3x\n-3x^2+4x\n\n3t(t-5)+6t^2\n3t^2-15t+6t^2\n9t^2-15t\n\n4w^2-w(2w-3)\n4w^2-2w^2+3w\n2w^2+3w
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(x+3)(x^2+2x+4)\n\nStart by taking the first term in the first parenthesis and multiplying it by each term in the second parenthesis:\n\nx*x^2 + x*2x + x*4 = x^3 + 2x^2 + 4x\n\nNow take the second term and multiply it by each term in the second parenthesis:\n\n3*x^2 + 3*2x + 3*4 = 3x^2 + 6x +12\n\nNow add all of that together\n\nx^3 + 2x^2 + 4x + 3x^2 + 6x +12\n\nand combine terms\n\nx^3 + 5x^2 + 10x +12
106893
Solve by substitution.\nSolve for y^2 in the second equation by subtracting x^2 and 2x from both sides.\nx^2 + y^2 + 2x - x^2 - 2x = 19 - x^2 - 2x\ny^2 = 19 - x^2 - 2x  Now substitute this into the first equation in place of y^2\nx^2 + 2(19 - x^2 - 2x) = 33  Distribute the 2\nx^2 + 38 - 2x^2 - 4x = 33  Combine x^2 and -2x^2 and subtract 33 from both sides\n-x^2 - 4x + 5 = 0  Multiply both sides by negative one (I just do this because it is easier to factor when the x^2 is positive)\nx^2 + 4x - 5 = 0  Now factor\nWe are looking for factors of -5 that add up 4.  5 and -1 work because 5*(-1) = -5 and 5 - 1 = 4\nTherefore x^2 + 4x -5 = 0 becomes\n(x+5)(x-1) = 0  Set each part equal to 0\nx+5 = 0 or x-1 = 0\nTherefore,\nx = -5 or x = 1\nNow substitute each of these values into the y^2 equation from above.\ny^2 = 19 - x^2 - 2x\nFor x = -5\ny^2 = 19 - (-5)^2 - 2(-5)\ny^2 = 19 - 25 + 10\ny^2 = 4  Take the square root of each side to get\ny = -2 or 2\n\nFor x = 1\ny^2 = 19 - (1)^2 - 2(1)\ny^2 = 19 - 1 - 2\nY^2 = 16  Take the square root of each side to get\ny = -4 or 4\n\nThus, the solutions are\n(-5,-2), (-5,2), (1,-4), (1,4)\n\nIt is always good to check your answers, especially on a test.  To check your answers just plug each one into each equation to see if you get a true statement.
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ill give it a shot - im not sure , anyways i got 2 answers..\n\n(1st)\n\n(3-x)x+2 = 2 (x^2-2)\n\n3x-x^2+6-2x = 2x^2-4\n\nx-x^2+6 = 2x^2-4\n\nx-1/x+6 = 2/x-4\n\nx-1+6x = 2-4x\n\nx+4x+6x = 2+1\n\n11x = 3\n\nx = 3/11\n\n\n(2nd possible answer)\n\n\n(3-x)x+2 = 2 (x^2-2)\n\n3x-x^2+6-2x = 2x^2-4\n\nx-x^2+6 = 2x^2-4\n\nx-1/x+6 = 2/x-4\n\nx-1+6/x = 2-4/x\n\n-5 = -2/x\n\n-5x = -2\n\nx = -2+5\n\nx = 3
384567
1. (x-5)(x-3)\n2.(x+10)(x+2)\n3. (x+8)(x-4)\n4.(a-8)(a-7)\n5. ? your writing is confusing\n6. (a+6b)(a+5b)\n7.(-a+10)(a+5)\n8. I don't think this is factorable, 19 is a prime number\n9. (-x-7)(x-3)\n\nAlright, they key to factoring is pretty simple, once you get the hang of it it becomes second nature. The thing about factoring is you have to understand the multipliers of the 3rd number. \ni.e. #1: 15 has a couple factors, 1&15, and 3&5. Next, you must look at the factors it has, and see which add up or subract in any combination to form the second term. In this case, 5 and 3 combine to make 8. When the last term is positive and the middle term is negative, the two factors must both be subraction. Therfore the answer is (x-5)(x-3)= x2-3x-5x+15=x2-8x+15.\n\nHope I was able to help kid, I was you only a few years back secretly cussing about having to do this. Good luck in Algebra.
406320
x^2-6x+9=3+9\n(x-3)^2=+/-rt12\nx-3=+/-2rt3\nx=3+/-2rt3\n\n2.2(x^2+5x)=-11\n2(x^2+5x+25/4-25/4)=-11\n2(x+5/2)^2-25/2=-22/2\n2(x+5/2)^2=3/2\n(x+5/2)^2=3/4\nx+5/2=+/-(rt3)/2\nx=-5/2+/-(rt3)/2\n=1/2[-5+/-rt3]
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About Me\nMember since: October 03, 2006\nTotal points: 100 (Level 1)\nPoints earned this week:\nTotal answers:\nBest answers:\n\nI need math help again... PLEASE HELP! I m desperately in need.?\n\nAnswer PARTB and PART C in the space provided. Evaluation of your solution to each problem will be based on the following.\n\n*A correct mathematical method for solving the problem is shown\n\n*The final answer is accurate and a check of the answer is shown where asked for by the question.\n\nDetermine if the equation is a linear equation, and if it is, write in standard form.\n\na. 5x+1=3y\n\nb. y=2x   x=y/2\n\nc. 4(2y-x)=1  8y-4x=1\n\nd. 25x-2y+2=0\n\ne. 2x-5y/2=7x\n\n2. a. f(x)= x (exponent 2)+2x+1 f(0),f(2)\n\nb. f(x)=2x-5 f(3),f(-3)\n\nc. f(x)=(-3x)exponent 2 f(0),f(-1)\n\nd. f(x)=2x(exponent 3)+4 f(0),f(-1)\n\ne. f={(1,5),(3,7),(5,9),(7,11)} f(3),F(7)
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Wonderful.\nTHis is some sort of simultaneous equation.\nAccording to the principle of this equation, we try to rationalize one term of the equation so as to get one unknown term. We can either try finding the x term or the y term. For your question as to find the y term.\n\nFirstly, we rationalize the x terms by multipling them by the inverts. then it will result to \n                                    12x + 20y = 32\n                                    12x + 18y = 39\nThe we apply the subtraction rule. ie. we multiply all the terms beneath by the negative (-) sign. This will result to \n                                    12x + 20y = 32\n                                   -12x - 18 = -39\nWe later subtract so as to attain\n                                2y = -7\nWe apply the law of linear equation by dividing both sides by the coefficient which in this case is 2.\n                              2y/2 = -7/2\nThis will eventually cancel the 2 in the first term therby leavin y alone. \n The answer will be in either of the following cases through simplification.  -7/2, -3 1/2 or -3.5
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x^2-6xy+8y^2\nx^2-6xy+9y^2-9y^2+8y^2\n(x-3y)^2-y^2\n(x-3y+y)(x-3y-y)\n(x-2y)(x-4y)\n\naliter\nx^2-4xy-2xy+8y^2\nx(x-4y)-2y(x-4y)\n(x-4y)(x-2y)
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3(x^4+3x^2-11)\n\nseriously you need to pay attention in class instead of asking people. lol
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I can't believe I'm actually doing your homework for you.\n\nPretty funny how bored I am.\n\n2x{3x-2(3+x)} - x2\n= 2x(3x - 6 - 2x) - x2 =2x2 - 12x - x2 = x2 - 12x = x(x - 12)\n\n3/2(4x-6)-x(3-x)=1/3x(3x+6)\n6x - 9 - 3x + x2 = x2 + 2x\n(3x-9+x2) - (x2 + 2x) = (x2 + 2x) - (x2 + 2x)\nx - 9 = 0, x = 9\n\n4 (y-7)-2y(1-3y)=6y2\n4y - 28 - 2y + 6y2 = 6y2\n2y - 28 = 0\ny = 28/2 = 14\n\nr2-{4-r(3-r)}=2(2r-3)\nr2 - 4 +3r - r2 = 4r - 6\n0 = r - 2\nr = 2\n\n1=2x(x-2)-{5-2x(1-x)}\n1 = 2x2 - 4x - 5 + 2x - 2x2\n2x = - 6\nx = -3
441614
1. Evaluate 3x^2 + 5x - 8 when:\n\na) x = 2\n3(2)^2 + 5(2) - 8\n3(4) + 10 - 8\n12 + 10 - 8\n22 - 8\n14 \n\nb) x = -3\n3(-3)^2 + 5(-3) - 8\n3(9) -15 - 8\n27 - 15 - 8\n12 - 8\n4\n\nc) x = 0.62\n3(0.62)^2 + 5(0.62) - 8\n3(0.3844) + 3.1 - 8\n1.1532 + 3.1 - 8\n-3.7468\n\n2. Simplify: 4x - 3x^2 + 2x - x^2\ncollect like terms\n4x + 2x - 3x^2 - x^2 \n6x - 4x^2\n2x(3 - 2x)\n\n3. Simpilfy: (5 x 6^2 - 5 x 3^2)/( 3 + 2)\nUsing BODMAS or PEMDAS\nWe treat the problems in the brackets first\n{(5 x 6^2 - 5 x 3^2)}/(3 + 2)\n\nIn here we treat the multiplications first\n{(5 X 36) - (5 X 9)}/(3 + 2)\n(180 - 45)/5\n135/5\n= 27
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Use the slide method!\n2x^2 + 5x + 12\nMultiply the coefficient of x^2 with the the last term (the constant C), so it becomes:\nx^2 + 5x + 24\n(x + ) (x + )\nSince they're both positive, there are no factors of 24 you can plug into that would add up to 5.  Thus, you can't factor it.\n\n-4x^2 -9x - 10\nAgain, multiply the coefficient of x^2 with the constant.\nx^2 - 9x + 40 (it becomes positive because -10 * -4 = -40)\nThe plus sign after the 9x tells you that the signs are both going to be the same, and the subtraction sign after the x^2 tells you that they're both going to be subtraction signs.\n(x- )(x- )\nThen, think of what factors of 40 add up to 9.  None, so this can't be factored either.\n\n4a^2 - a - 5\nOnce again, multiply the coefficient of a^2 with the constant.\na^2 - a - 20\nYou can see from the - sign after the a that the signs will be different.\n(a+ ) (a- )\nThen think of factors of 20 that will equal -1.  \n(a+4) (a-5)\nFor it  to be -1, the bigger number has to be negative, so it will be a-5.\nBut then we have the problem of the -4 (coefficient of a^2) we got rid of earlier.  So, we have to divide the numbers by -4:\n(a + (4/-4))(a - (5/-4))\n(a + (4/4)) = (a - 1)\n(a - 5/-44)) doesn't divide cleanly, so the denominator goes in front of a. Thus:\n(a - 5/-4)) = (4a + 5)\n\nYes, no solution is definitely possible.  And you can use the discriminant to check:\nb^2 - 4 * a * c\na is the coefficient of x^2, b is the coefficient of x, and c is the constant.  If you plug those into the equation and you come up with a negative number, then there is no solution.\n\nYou should be able to do the last one.\nBasic rules of the slide method:\n-Multiply the coefficient of x^2 with the constant\n-Factor\n-Divide constants by coefficient: if it doesn't divide cleanly, simplify if possible and put whatever is left of denominator in front of the x.  For example:\n(x + 4/6) = (3x + 2) (because you can simplify both 4 and 6 by dividing by 2)