please help!can anyone show me how to solve this problem ... 2x^3+7x^2-77x-40 ... want to see the steps .. tnx
831395
Let f(x) = 2x^3 + 7x^2 - 77x - 40 = 0\n\nTry to find factors by remainder theorem\nSo test x = ±1, ±2, ±5, ±8 (factors of - 40) and try to find one factor.\nf(1) = 2 + 7 - 77 - 40 ≠0\nf(-1) = -2 + 7 + 77 - 40 ≠0\nf(2) = 16 + 28 - 154 - 40 ≠0\nf(-2) = -16 + 28 + 154 - 40 ≠0\nf(5) = 250 + 175 - 385 - 40 = 0 so x - 5 is a factor\n\nBy division\nf(x) = 2x^3 + 7x^2 - 77x - 40\n= (x - 5)(2x^2 + 17x + 8)\n= (x - 5)(2x + 1)(x + 8)\n= 0\nSo x = 5, -1/2, - 8
26076
=7(1-x)+2x(1-x)\n=7-7x+2x-2x^2\n=7-5x-2x^2
255329
Have patience :)  You got some answers the first time you asked this.  Here it is again:\n\nIt does not get eliminated, it gets split up into the 2 factors.\n\nOn this one, you can go ahead and divide by 2 to make things a little easier. You'll be left with\n\n3x^2 - 7x - 6 = 0\n\nNow to factor. You need 2 numbers whose product equals 3*-6, or -18 and whose sum equals -7. From this point on it's just sort of educated guessing. \n\n1*-18 = -18, but 1-18 = -17, so that won't work\n2*-9 = -18, and 1-9 = -7 -- THIS ONE WORKS!! yay!\n\nSo now you can rewrite your problem as\n\n3x^2 - 9x + 2x -6 = 0 and group the first 2 and last 2 terms\n\n(3x^2-9x) + (2x-6) = 0 -- you can factor a 3x out of the first group, and a 2 out of the second\n\n3x(x-3) + 2(x-3)=0 -- and now you can factor an (x-3) out\n\n(x-3)(3x+2)=0\n\n~~~~~~~~~~~~~~~~~~~\nOops, you're right. I was typing too fast.  Oh well, I still got the right final answer :)  Thanks
204269
isolate one of the radicals, then square both sides.\nhere you get 2x+3=5x-6.......3x=9.......x=3\n\nworst case is sqrt(something) + something = sqrt(something)\n\nhere you'd isolate a radical, square both sides, clean up the equation, isolate the remaining radical, square both sides, and finish.  AND CHECK YOUR ANSWERS.  when you square both sides, "extraneous" solutions sneak in that might have you, back in the original equation, taking the square root of something negative.\n\nexample:  sqrt(3x+3) + 1 = sqrt(5x+6)\n....... (3x+3)+2*sqrt(3x+3)+1=(5x+6) ..... left side is a binomial\n.........2*sqrt(3x+3) = 2x+2\n...........sqrt(3x+3) = x+1\n...............(3x+3) = x^2+2x+1 ....... right side is a binomial\n....................0 = x^2-x-2 ......... and it factors!\n....................0 = (x-2)(x+1)\n....................0 = (x-2) so x=2\nor .................0 = (x+1) so x=-1\n\nnow check: sqrt(3*2+3)+1 =? sqrt(5*2+6)\n.............sqrt(9) + 1 =? sqrt(16)\n...................3 + 1 = 4.............yes.\n\n..........sqrt(3*-1+3)+1 = sqrt(5*-1+6)\n...............sqrt(0)+1 = sqrt(1)\n.......................1 = 1 ............yes\n\nsigh.  when I wrote it the equal signs lined up.
55351
rt11guru is correct up to the point where he says:\n\nX = ((2x2)^15)x((3x3)^6 x 3)x((5x5)^2 x 5)x((7x7)x7)\n\nThe number of possibilities is then given by:\n\nN = (16 * 7 * 3 * 2) - 1 = 671\n\nYou need to subtract 1 to account for the case where the divisor has no prime factors (and so equals 1).  If 1 counts as a solution (perfect square and divisor) the number of solutions is 672.
93533
Well, the sum of two terms thing is stupid, but oh well. (You can immediatly make it (x-5)(x-9))\nx^2-9x-5x+45=x(x-9)-5(x-9) --> (x-5)(x-9)
931054
Group all elements with y together:\n\ny(x^2 + 6x +9) -x^2 + 2x + 3 = 0\n\nMove the rest to the other side:\n\ny(x+3)^2 = x^2 - 2x + 3 (1)\n\nx = -3 is obviously not the solution of this equation therefore we can divide both sides of (1) by (x+3)^2.\n\n(1) is equivalent to:\n\ny = (x^2 - 2x + 3) / (x+3)^2
974614
Simple. Get rid of the denominators on both sides by multiplying by the lowest common multiple (in this case, 8). So that would make the left side: 2(3x +7) and the right side: -x. Then just isolate and solve for x!
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!
810284
i got 2x^2-x-3 R -4\n\nfirst divide 2x+5 into 4x^3+8x^2-11x-19 and you get 2x^2 then you subtract and your left with -2x^2-11x-19 and then you divide that by 2x+5 and you get -x add that to 2x^2 and you get 2x^2 - x then you subtract from your original and get -6x-19then divide 2x+5 into that and you get -3 ((2x^2-x-3)) multiply that by 2x+5 and you subtract from -6x-19 and you get -4 you cant divide 2x+5 into that so -4 is your remainder
643397
(8y^3 + 27)/(2xy - 10y + 3x - 15)\n((2y + 3)(4y^2 - 6y + 9))/((2xy - 10y) + (3x - 15))\n((2y + 3)(4y^2 - 6y + 9))/(2y(x - 5) + 3(x - 5))\n((2y + 3)(4y^2 - 6y + 9))/((2y + 3)(x - 5))\nthe (2y + 3)s cancel out\n(4y^2 - 6y + 9)/(x - 5)\n\nANS : (4y^2 - 6y + 9)/(x - 5)
729449
It will not factorise in the rational number field so there are no rational solutions\n\nIrrational roots are -1.72107 and 0.269209 (approximately) using Quickmath
809607
I got 4x² + 2.  It was actually pretty straightforward once you lay it out like a long division problem.\n\nTake 4x² and multiply it by 7x+3i (since the first term will give us 26x³).\n\n4x² * (7x+3i) = 28x³ + 12ix², leaving us with a remainder of 14x + 6i.\n\nNow take (7x+3i) times 2 to get 14x + 6i.\n\nTherefore your answer is 4x² + 2.\n\n~ ♥ ~
1035131
i think you made an arithmetic error in the first step.\n\nstart with: |5x - 3| + 8 = 22\nthen subtract 8 from both sides, and you will get\n|5x-3| +8 - 8 = 22 - 8\n|5x-3|  = 14\nand not |5x - 3| = 24\n\nfrom there:\n|5x-3|  = 14\n5x-3 = 14 or 5x-3= -14\n5x-3+3=14+3  or 5x-3+3= -14+3\n5x = 17 or 5x = -11\nx = 17/5 or -11/5
541499
2x(x-5)-3(2x+1)\n2x^2-10x-6x-3\n2x^2-16x-3
811856
Okay multiply the top equation by 7 and the bottom equation by negative 3 to obtain 2 new equations with the x variable to eliminate:\n\n7(3x + 7y = -43)\n-3(7x + 6y = -59)\n\nthis becomes \n\n21x + 49y = -301\n-21x - 18y = 177\n\nNow cancel out the 21x's and add down.  Doing that, the equation becomes\n\n31y = -124   this can easily be solved for y\n\ny = -4\n\nNow plug -4 back into one of the original equations\n\n3x + 7(-4) = -43\n\n3x - 28 = -43\n\nadd 28 to both sides to yield\n\n3x = -15   and solve for x\n\nx = -5\n\nNow check x = -5 and y = -4 into both and see that you have solved this system correctly.
497089
I gues I have to read this as {3tan^2(2x)}-1=0\nand not as 3tan^2(2x-1)=0.\n{3tan^2(2x)}-1=0\ntan^2(2x) = 1/3\ntan(2x) = + or - 1/V3\n+case: 2x = 30 + 180n degrees. So x = 15 + 90n degrees.\n-case: 2x =150 + 180n degrees. So x = 75 + 90n degrees.\nThe ordered solutions are\n 15, 75, 105, 165, 195, 255, 285 and 345 degrees.