Given that p(x) = 4x^3 -9x^2 -10x + 3 and that P(3)=0 what is/ are the factor(s) of P(x)?
138480
ok well you know that, since P(3)=0, that (x-3) HAS to be a factor of it. so you divide the whole equation by x-3. \n\n''''________________________\nx-3 ) 4x^3 - 9x^2 - 10x + 3\n\nfirst step is to divide 4x^3 by x, andyou get 4x^2. then you take this and multiply it by x-3 and you get 4x^2-12x^2\nwrite the 4x^2 in the answer section, and the other thing under the polynomial and subtract.\n\n\n''''_________4x^2___________\nx-3 ) 4x^3 - 9x^2 - 10x + 3\n''''''4x^3 -12x^2\n''''''''''''''3x^2 - 10x\n\nnow divide 3x^2 - 10x by x-3 and do the same thing.\n\n''''_________4x^2 + 3x - 1_\nx-3 ) 4x^3 - 9x^2 - 10x + 3\n''''''4x^3 -12x^2\n''''''''''''''3x^2 - 10x\n''''''''''''''3x^2 - 9x\n'''''''''''''''''''''-x + 3\n'''''''''''''''''''''-x + 3\n''''''''''''''''''''''''''0\n\nso after you divide, you get 4x^2 + 3x - 1 and you can factor this easily. \n4x^2 + 4x - x - 1\n4x(x+1) - (x+1)\n(4x-1)(x+1)\n\nso the final factored form of 4x^3-9x^2-10x+3 is \n(x-3)(x+1)(4x-1). so then the factors of P(x) are x=3, x=-1,\nand x=1/4.\n\nhope that helps. laterz
548927
x= 3/5\n\nIf you factor by grouping you would get\n25x^2 - 15x -15x + 9 = 0\n\nthen you would factor the first two terms and the last two terms\n5x(5x-3)   -3(5x-3)    =  0\n\nuse the terms outside the parentheses are one factor and the one inside the parentheses as the second factor\n(5x-3)(5x-3) = 0\n\nset to zero\n5x-3 = 0\n\ndo algebra\nx=3/5\n\nor you could have just noticed that that equation is a perfect square
573287
factors\n1=1\n2=2x1\n3=3x1\n4=2x2=4x1\n5=5x1\n6=3x2=6x1\n7=7x1\n8=2x2x2=4x2\n9=3x3=9x1\n10=5x2=10x1\nMultiples\n\n1=1,2,3,4,5,6.......\n2=2,4,6,8,10,\n3=3,6,9,12,15\n4=4,8,12,16\n5=5,10,15,20\n6=6,12,18,24\n7=7,14,21,28\n8=8,16,24,32\n9=9,18,27,36\n10=10,20,30,40
664764
If the zeros are: -4, 0, and 2, our factored equation must look like this:\nf(x) = x(x+4)(x-2)\n\nFoil it out:\n\nf(x) = x(x2 + 4x - 2x - 8)\nf(x) = x(x2 + 2x -8)\n\nf(x) = x3 + 2x2 -8x\n\nAnswer number 3.\n\nHope that helps.
401623
if a is a zero x-a is a factor\n1)\nso factors are (x+2) ,x (x-5)(x-6)\n\nmultiply them and you get\nf(x) = x(x+2)(x-5)(x-6) \n\nmultiply them and you get leading coefficient 1\n\n2)\n factors are (x-(1-sqrt(5)) and (x-(1+sqrt(5))\n\nso (x-1-sqrt(5))(x-1+ sqrt(5)) (a+b a-b form )\n= (x-1)^2 - 5  \n= x^2-2x-4
315682
Do you mean real AND imaginary? Or just all imaginary?  Well, either way, since we \nhave a polynomial with four factors, we will factor by grouping to simplify.\n\nf(x)=(x^3)-(9x^2)+26x-24\n\nf(x)=[(x^3)-(9x^2)+26x]-24\n\nf(x)=x[(x^2)-9x-26)]-24\n\nf(x)=(x-24)[(x^2)-9x+26]\n\nWe have one real solution. Our linear factor is (x-24), so our Real solution is x=24.\n\nNow we need to use the quadratic formula:\n\n{-b(+or-)[The Square Root of:(b-4ac)]}ALL over 2a.\n\n[(x^2)-9x+26] a=1, b=-9, c=26\n\n-(-9) ± √(-9-(4*1*26))ALL divided by 2\n\n9 ± √(-9-104)ALL divided by 2\n\n9 ± √(-113)ALL divided by 2\n\nSo our two imaginary solutions are:\n\n [9+√(-113)]/2] and [9-√(-113)]/2]\n\nRecall that i=√(-1)\n\nSo, our first solution [9+√(-113)]/2]=9+i√(113)or 9+√(113)i\n\nand our second solution [9-√(-113)]/2]=9-i√(113)or 9-√(113)i\n\nTherefore, All Solutions, Real and Imaginary, for \n\nf(x)=(x^3)-(9x^2)+26x-24 are:\n\nx=24, x=9+√(113)i, and x=9-√(113)i\n\n:)
376210
C is correct but not complete... only the 3 is factored out... the real answer would be:\n\n3(x-1)(x+4)
210673
Given that f(x) is a polynomial of degree EIGHT such that\n\nf(m) = 1/m for m = 1,2,3,4,5,...,9\n\nFind f(10)  \n\n\n*Answer: 2/10\n\n*Solution:\n\nConsider another polynomial, g, defined as g(x) = x f(x) - 1\n\ng(1)=0, g(2)=0, …, g(9)=0\n\nSo g(x) = A(x-1)(x-2)…(x-9)\n\ng(0) = 0 f(0) - 1 = -1\n\nalso, g(0) = A(-1)(-2)…(-9), so A = 1/9!\n\ng(10) = A(9)(8)…(1) = (1/9!)(9!) = 1\n\ng(10) = 1 = 10 f(10) - 1, so f(10) = 2/10\n\n\n**Or try doing some research on "Rabin's method"
220771
(x+3)(2x+4) is incorrect.\n(x+3)(2x+4) = 2x^2 + 10x + 12\nSolution:\nMultiply the coefficient of the x^2 term with the constant.\n2(12) = 24\nYou are looking for factors of 24 that add the 11 (which is the coefficient of the x term).\n(3)(8) = 24\n3 + 8 = 11\nReplace 11x with 3x + 8x\n2x^2 + 11x + 12\n= 2x^2 + 3x + 8x + 12\nFactor by grouping\n(2x^2 + 3x) + (8x+12)\n= x(2x+3) + 4(2x+3)\n= (2x+3)(x+4)\nAnswer: (2x+3)(x+4)\nYou can easily check the answer by Foil.\n(2x+3)(x+4) = (2x)(x) + (2x)(4) + (3)(x) + (3)(4)\n= 2x^2 + 8x + 3x + 12\n= 2x^2 + 11x + 12
13951
Factor by grouping:\nx^2 (3x - 5) - 3 (3x - 5)\nAnswer: (3x - 5)(x^2 - 3)
90163
1) \nxy - x + 4y - 4\nx(y-1) + 4(y-1)\n(x+4)(y-1)\n\n2)note: x^3 means x cubed; x^2 means x squared\nx^3 -6x^2 + 2x - 12\nx^2(x-6) + 2(x-6)\n(x^2 + 2)(x-6)\n\n3)\n6tx + r - 3t - 2xr\n3t(2x - 1) - r(2x - 1)\n(3t - r)(2x - 1)\n\n4)Are you sure it wasn't -x^2 - 2x - xy - 2y?\n-x^2 - 2x - xy - 2y\n-x(x + 2) - y(x + 2)\n(-x - y)(x + 2)\n\nTo "factor by grouping," you must split the original polynomial that you are given into two different binomials by "grouping" them together.  This is done with perenthesis in the problems above.  Then, you factor out the "greatest common factor" from each of the binomials.  After factoring, the two quantities still within the perenthesis should be the same.  You could call these quantities the "common binomial."  Lastly, you factor out the common binomial to get the final answer.
536155
p(x) = 3x^4 + 3x^3 + 7x - 10\n\np(-2) = 3(-2)^4 + 3(-2)^3 + 7(-2) - 10\n= 48 - 24 - 14 - 10\n= 0 ← option B
165831
Factor it by "FOIL":\n\n(x^1/3 - 2)(x^1/3 - 3) = 0\n\nThen x^1/3 = 2 or x^1/3 = 3\n\nDo both sides of each equation to the third power :\n\nx = 2^3 or x = 3^3\n\nSo x = 8 or 27.
130751
are you trying to solve for x?\n\nFirst, find the factors for the equation.  (x+2)(x+1)\n\nCheck it to see if it's right.  x times x equals x squared. 1x plus 2x equals 3x. 2 times 1 equals 2.\n\nMake one of the parenthesis groups equal to zero.  For example, x+2 = 0   To solve this, x must equal -2.\n\nPut -2 back into the original equation and see if it works.\nNow make the parenthesis group equal to zero.  x+1=0  To solve this, x must equal -1.\n\nTherefore, the answer is using either -1 or -2 will make the equation zero.\n\nHope this helps.  ;)
467168
A recap on Remainder and Factor theorem:\nIf f(a) = 0, then (x-a) is a factor of f(x).\n\nWe know that the term in f(x) independent of x is a product of all the roots of f(x), so we can investigate the factors of 9 [+-1, +-3, +-9] in this case to find out the 1st factor of f(x).\n\nBy trial and error, f(3) = 81-90+9 = 0, so (x-3) is a factor of f(x)\nUsing long division f(x) is factorised as follows:\nf(x) = (x - 3)(x^3 + 3x^2 - x - 3)\n\nSo the problem reduces to finding factors for a g(x) = x^3 + 3x^2 - x - 3.\n\nThus similarly, by trial and error, f(1) = 1+3-1-3=0, so (x-1) is a factor of g(x) and therefore f(x) as well.\n\nSo our f(x) becomes f(x) = (x - 3)(x - 1)(x^2 + 4x + 3),\nwhere x^2 + 4x + 3is quadratic and so can be easily factorised to (x + 3)(x + 1).\n\nTherefore, f(x) = (x - 3)(x - 1)(x + 3)(x + 1)\n----------------------------------------------------------------------------------\n\nAlternatively, if you observe that f(x) consists of only even powers of x, you may try to use substitution method to factorise f(x).\n\nLet y = x^2, \n\ntherefore f(x) = x^4 -10x^2 +9 becomes f(y) = y^2 - 10y + 9, which is quadratic and can be easily factorised to give\n\nf(y) = (y - 9)(y - 1)\n\nSubstituting x back, we get\n\nf(x) = (x^2 - 9)(x^2 - 1) = (x + 3)(x - 3)(x + 1)(x - 1), \n\nusing the property a^2 - b^2 = (a + b)(a - b).
228335
I think u are talkin abt this method....\n\n\nif u hav a cubic like  x^3 + 2x^2 - 2x - 1 = 0\n\nto find out whether (x-1) is a factor....\n\nfirst write the coeff of the eq in this way n write the factor ( here it iz 1 since x-1=0 or x=1) on the left hand side Like this\n\n{one}  1     2     -2     -1\n--------  0     1      3      1\n_______________________________\n{sum}  1     3      1      0\n\n\nHere what i hav done is write the 0 under the coeff of x^3\nnow add that coeff n the zero\nu get 1...\nnow multiply this one wid the number written on the left hand side(here it is also one)\nwite what u get under the coeff of x^2....\nas above add the new no and the coeff of x^2\nin a similar way multiply one and the sum u got n write it under the coeff of x...\nadd the no with coeff of x n then write it below....\nfinally u multiply the sum wid one and write it under the constant term.... n then add...\nif u get 0 as the sum then whatever no u hav taken in da LHS is on of the root.... here we get 0 in the last case... so (x-1) is a fctor.....\n\nnow this eq can b written as \n\n    (x-1) * (1*x^2 + 3 * x + 1) = 0\nhere 1 , 3 and 1 of the quadratic are the sum u got in the above method......
435542
P(1/3)=3(1/3)^4-(1/3)^3-21(1/3)^2-11(1/3)+6\nP(1/3)=3(1/81)-(1/27)-21(1/9)-11(1/3)+6\nP(1/3)=(1/27)-(1/27)-(7/3)-(11/3)+6\nP(1/3)=0\n\nP(-2)=3(-2)^4-(-2)^3-21(-2)^2-11(-2)+6\nP(-2)=3(16)-(-8)-21(4)-11(-2)+6\nP(-2)=48+8-84+22+6\nP(-2)=0\n\nNow factor (x+2) and (x-1/3) out of the original equation (I used synthetic division). That leaves you with:\n3x^2-6x-9=0\n3(x^2-6x-9)=0\n3(x-3)(x+1)=0\n\nSo, the other two zeros are -1 and 3.