find the binomial to complete (x+1)(      )=x^2+7x+6 please show work thanks?
610706
Can't you think of another way to finish your homework, son?
708943
You have to complete the square.\n\ny = a(x^2 + b/a x) + c\ny = a(x^2 +b/a x + b^2/4a^2) + c - b^2/4a\ny = a(x + b/2a)^2 + (4ac - b^2)/4a\n\n[4ay - (4ac - b^2)]/4a = a(x + b/2a)^2\n[4ay - 4ac + b^2]/(4a^2) = (x + b/2a)^2\n\n+/- sqrt(4ay - 4ac + b^2)/2a = x + b/2a\n\nx = (-b +/- sqrt(4ay - 4ac + b^2))/2a
760641
x + 6 + 3x = 1 + 3x + 1\nCombine like terms.  x+3x = 4x and 1+1 = 2\n4x + 6 = 3x + 2\nSubtract 3x from each side\nx + 6 = 2\nSubtract 6 from each side\nx = -4
123892
3(x – 2)^2 – 3(x – 2) – 6\n3((x-2)^2 - (x-2) -2)\n3(x^2-4x+4-x+2-2)\n3(x^2-5x+4)\n\n3(x-4)(x-1)\n\nYay!!
825348
I prefer "long" multiplication but it doesn't line up as well...\n\n3x^2 + 7x - 6\n* _____x+12\n--------------------  Multiplying term by term\n_____36x^2 + 84X - 72\n3x^3 + 7x^2 - 6x\n--------------------------------  Add term by term\n3x^3 + 43x^2 + 78x - 72   <--  final answer
954164
grab a calculator and try 1.1 x = = = = =  until you get a number >2.\nthe first time you hit = is ^2, next is ^3 etc.  count.\n\nN=7
859057
2+(n-1)2+1/2(n-1)(n-2)1\n= 2 + 2n - 2 + (1/2)(n^2 - 2n - n + 2)\n= 2n + (1/2)(n^2 - 3n + 2)\n= 2n + (1/2)n^2 - (1/2)(3n) + (1/2)2\n= 2n + (1/2)n^2 - (3/2)n + 1\n= (1/2)n^2 + (1/2)n + 1
52440
[x+1]/x^2+10x+5=2x/x^2+x-2\n9x+7=[x+1]/x^2\n9x^3+7x^2=x+1\n9x^3+7x^2-x=1
385232
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.
831999
y = (x+3)(x-1)(x-2) = x^3 - 7x + 6
145326
i'm good at math, but what exactly is the quustion??? there isn't any... what is it, a equation?\nany way you can post it to sciences and mathematics > mathematics, im sure someone there can help you...\n\n>>>ok one minute i'm working on it...\n\n>>>there you go:\nlets do the first: (x-3)(x-1)(x+2)\nfirst we open the last two parentheses as in usual opening, and the (x-3) remains as it is. we get (x-3)[x^2+2x-x-2]\nequals (x-3)(x^2+x-2) then we open the two parentheses left. first we multiply by the first member (x) then the second (-3). then we get:\nx*x^2+x*x+x*(-2)-3*x^2-3*x-3*(-2)\nx^3 + x^2 - 2x -3x^2 -3x + 6\nand finally:\nx^3 -2x^2 -5x + 6\n\nthe second is the same way, i'm doing it, good luck :)\n\n>>>the second: (x-5)(x+1)(x-2)\nfirst we open the last two parentheses as in usual opening, and the (x-5) remains as it is. we get (x-5)[x^2-2x+x-2]\nequals (x-5)(x^2-x-2) then we open the two parentheses left. first we multiply by the first member (x) then the second (-5). then we get:\nx*x^2+x*(-x)+x*(-2)-5*x^2-5*(-x)-5*(-2)\nx^3 -x^2 -2x -5x^2 +5x +10\nand finally:\nx^3 -6x^2 +3x + 10\n\ni hope i helped...\n\n>>>oh i leave the first one to the last because i have to leave one of the three, and it really doesn't matter which one i leave. for example on the first i left the first parenthese (x-3)[(x-1)(x+2)] but i could leave the last [(x-3)(x-1)](x+2) or the middle (x-1)[(x-3)(x+2)] and i'd still get the same answer. i have to leave one of the three bcuz i can't open them all together, it's complicated and hard, and by opening two ane leave one it becomes much easier.
990847
isolate the variable\n\n-7x+1+8x=x+1\n-7x+8x=x\nx=x\nx=1\n\nthe answer is c) 1\nnow give me 10 points please!!!
333624
(ln x)^7=(7*(lnx)^6)*(1/x)=7*(lnx)^6/x
766363
YES...\n\n(7x+1)(x^2+1)=???\n\nyou could use these steps,\n1. multiply the first terms of the given factors, consider the signs\n(7x) * (x^2) = 7x^3\n\n2. multiply the inner terms of the given factors, consider the signs\n(1) * (x^2) = x^2\n\n3. multiply the outer terms of the given factors, consider the signs\n(7x) * (1) = 7x\n\n4. multiply the last  terms of the given factors, consider the signs\n(1) * (1) = 1\n\n5. then add these answers, simply write as one equation, considering the signs\n7x^3 + x^2 + 7x + 1\n\n6. if it happens that there are like terms, simply the equation by adding (or subtracting, whatever is applicable) these\n\nanswer: 7x^3 + x^2 + 7x + 1\n\n\nANOTHER METHOD:\n-you could simply distribute (multiply) the terms in the first factor on the terms on the second factor...\n\n\nfor (2x+6)(x-10) equals 2x-60\n1. first terms: 2x^2\n2. inner terms : 6x\n3. outer terms: -20x\n4. last terms: -60\n\nwriting in one equation: 2x^2+6x-20x-60\nsimplifying like terms: 2x^2-14x-60\nwe can further simplify this by dividing the equation by 2\n\nanswer: x^2-7x-30
848097
I'll haev a go at number 2.\n\na) So we prove that sum(k=1 to n)k + sum(k=1 to n-1) = n^2.\nSince the sum of the first n natural numbers is n(n+1)/2, this is equal to\nn(n+1)/2 + (n-1)n/2\n= (n^2 + n + n^2 -n)/2\n= 2n^2/2\n= n^2.  QED.\n\nb) We prove that (x^0 + x^1 + x^2 + ... + x^n)^2 = x^0 + 2x +... + (n+1)x^n + nx^(n+1) + ... + 2x^(2n-1) + x^(2n).\nThe LHS will generate n^2 terms x^k where k is an integer.  How many identical terms x^k will there be for a given k?\n\nCase 1) If k<=n, there are no restrictions on k so there are k+1 multiplications resulting in x^k which are x^0*x^k, x*x^(k-1), x^2*x^(k-2), ... ,x^k*1.  The coefficient is k+1.\nCase 2) If n<k<=2n, the products resulting in x^k are x^n*x(k-n), x^(n-1)*x^(k-n+1), ... x^(k-n+1)*x^(n-1), x^(k-n)*x^n, a total of (2n-k+1) terms.  The coefficient is 2n-k+1.\n\nSo the coefficient is 1 for x^0, increases by 1 up to (n+1)x^n, and then decreases by one up to x^2n.\n\nSetting x=10 demonstrates 1*1=1, 11*11=121, etc.\n\n\nc) n^2-n is divisible by 2 for all n.  Proof: if n is even then n^2 is even and n^2-n is also even.  If n is odd then n^2 is odd and n^2-n is odd.\n\nNow n^3-n = n(n^2-1) = n(n+1)(n-1).  This is the product of 3 consecutive numbers, one of which must be a multiple of 3.\n\nThe property doesn't hold for n^4-n, for example 2^4-2 = 14 which isn't divisible by 4.
762587
use remainder theorem\nhttp://en.wikipedia.org/wiki/Remainder_theorem\n"When p(x) is divided by (x-a), the remainder is p(a)"\nUR QUESTION:\np(x) = x^3+3x+k\n(x-a) = (x-1)\np(a) = 6\nhence, p(1)=6\nso, (1)^3 + 3(1) + k = 6\nhence, 1+3+k = 6\nso, k = 2\n\nhence, k = 2
599364
x^2 + x = x(x+1)\n\nIt can be simplified to x(x+1).