How do you factor f(x)=x^4+x^3-7x^2-13x-6? and x^5+4x^4+6x^3-8x^2-7x-2?
920579
use synthetic  divsion method \ni think this link would help\nhttp://www.purplemath.com/modules/synthdiv.htm
276641
f(x)=3x^2-2x+8 for x=6\n\nf(6)=3*(6^2)-2*6+8\nf(6)=3*(36)-12+8\nf(6)=108-12+8\nf(6)=104
371573
If i is a root, you may express it into x = i, a root of x^3 + i = 0.\nsince x = i, x - i = 0\nTherefore, x^3 + i has a root x - i. to get the other 2 factors, divide.\n(x^3 + i)/(x - i) = 0\nDivide it using long division, and you'll get\nx^2 + ix - 1 = 0\nActually, \nx^3 + i factors as (x - i)(x^2 + ix - 1) \ntherefore, \nx^2 + ix - 1 = 0\nuse the quadratic formula:\na = 1\nb = i\nc = -1\nx = [-b +or- sqrt(b^2 - 4ac)]/2a\nx = [-i +or- sqrt(i^2 - 4(1)(-1))]/2(1)\nx = [-i +or- sqrt(-1 + 4)/2\nx = (-i +or- sqrt -3)/2\nx = (-i +or- i sqrt3)/2\nTherefore, the other roots are\nx = (-i + i sqrt3)/2\nx = (-i - i sqrt3)/2\n\n^_^
653663
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:)
294274
well, (1/3)x^3 equals x\n\nso you can simplify it to (8/3)x + 2 = f(x)\n\nthat is a good start.
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
935264
Actually, the answer is *not* 4+x.\n(f/g)(x) = (16-x^2)/(4-x)\nNow, factorise the top:\n(4+x)(4-x)/(4-x).\nSo *if x is not 4* then (f/g)(x) = 4+x. However, the function is not defined at x = 4.
526107
You are only allowed to factor if the equation is equal to zero; your's is equal to 4.  You need to solve "x^3 - 6x^2 +9x - 4 = 0"  using synthetic division.
1095294
5) find the factors\nsquareroot of x * x * x * x * x * x /  7*7*5*5*2*2\nthe factors can come out of the square root by pairs only\nyou have 3 pairs of x's = x^3\nyou have 1 pair of 7's so it comes out of the square root as a 7\nx^3/7*5*2\nx^3/70 is it's simplified term\n\n__________________\nif u want all the correct answers then that would tarenirator's answers
879979
f(x)=(x^2-2x)(x^2+2.44x+1.4884)\n =(x^4+2.44x^3+1.4884x^2-2x^3-4.88x^2- 2.9768x)\n =(x^4 + 0.44x^3 - 3.3916x^2 - 2.9768x)\n\nF(x)=(x^5)/5+(0.44x^4)/4-(3.3916x^3)/3-             (2.9768x^2)/2+c\n\nam i really asked a Q?
821200
(x²+4x+4) / (2x-3) / (x²-4) / (2x²-7x+6)\n\nYou have 4 terms and three divisions, so it's a little unclear if you mean (a/b) / (c/d) or what.  I will try to help to the extent I can.\n\nLet's assume you mean:\n\n[ (x²+4x+4) / (2x-3) ] / [ (x²-4) / (2x²-7x+6) ]\n\nFactor your x² equations to see if you can cancel terms:\n\nx²+4x+4 = (x+2)(x+2)\nx²-4 = (x+2)(x-2)\n2x²-7x+6 = (2x-3)(x-2)\n\nWoohoo - things are looking good!  Now we can rewrite our original problem using these factors:\n\n[ (x²+4x+4) / (2x-3) ] / [ (x²-4) / (2x²-7x+6) ] = \n\n[ (x+2)(x+2) / (2x-3) ] / [ (x+2)(x-2) / (2x-3)(x-2) ] =\n \n[ (x+2)(x+2) / (2x-3) ] * [ (2x-3)(x-2) / (x+2)(x-2) ] =\n\nnotice we can cancel out three terms: (2x-3), (x-2) and (x+2)\n\n[ (x+2)(x+2) / (2x-3) ] * [ (2x-3)(x-2) / (x+2)(x-2) ] =\n\n(x+2)\n\nHope that helped!\n\n~ ♥ ~
694817
f-8-3f = 3-7f+4\nf-3f+7f = 8+3+4\n8f-3f = 15\n5f = 15\nf=15/5\nf=3
650997
f(3(x - 4)) \n= f(3x - 12) \n= (3x - 12 + 1)^2 - 4 (3x  - 12)\n= (3x - 11)^2 - 12x - 48
387334
ok. well. the um first one is 15c^4 - 26c^2 - 21. if you let x = c^2, then you can rewrite this equation as \n15x^2 - 26x - 21. and to factor this, you find 2 numbers that multiply to 15*-21   and add to -26. so it is -35 and 9.\nwhich means you can rewrite 15x^2 - 35x + 9x - 21\n= 5x (3x-7) + 3(3x-7)\n= (5x+3)(3x-7)\n\nand since x = c^2, you can rewrite this as \n(5c^2 + 3)(3c^2 - 7) and if you multiply that out, youll see you get your first equation.\n\n2. so same thing here. put x= h^2.\n12x^2 - 29x - 8\nthe numbers in this case, 12*(-8) = -96. so 3*(-32) = -96. so the 2 numbers are -32 and 3.\n\n12x^2 - 32x + 3x - 8\n4x (3x-8) + 1(3x-8)\n(4x-1)(3x-8)\n(4h^2 -1) (3h^2-8)\nso there.\n\n3. in this case, the highest is squared, which you know how to do.\n2x^2 + 7x + 6\n2*6 = 12. and um 4*3=12, 4+3=7. so the numbers are 4 and 3\n\n2x^2 + 4x + 3x + 6\n2x (x+2) + 3(x+2)\n(2x+3)(x+2)\n\nso there. hpe that helps. laterz
993104
take the equation x^2 - 7x + 12 and do the following  (x ) (x) now...see that there is a - and a + in the equation....you need to do this ( x -  ) ( x - ) 'cause you put the minus in the first factor in the second factor you take the - and the + and multiply them....So you get (x- ) (x -) the other 2 terms for each equation comes from the 7x and the 12.  What you do is to find 2 numbers that multiplied gives you 12 and added give you 7....The numbers are 3 and 4 ...Then put them into your factors (x-3)(x-4).\n\nFor the other 2 equations you do the same process\n\n(x^2 - 8x + 7) = (x-  ) (x-  ). In this one, you have to find 2 numbers that multiplied give you 7 and added give you 8. They are 1 and 7. Then (x^2 - 8x + 7) = (x- 1 ) (x- 7 ).\n\nw^2 - 8w + 15 = (w - ) (w- ). As you saw in one and two, three is the same....so the factors are 3 and 5. Then, w^2 - 8w + 15 = (w -3 ) (w-5 )\n\nFor the solve...\n1. x^2 - 6x = 0\n2. x^2 - 2x - 3 = 0\n\nTHe 1 is like this...you have to complete your quadratic equation then you do the following x^2 - 6x = 0 --> x^2 - 6x + 9 =0+ 9...The 9 in the right side was added in order to keep the two sides equal...IN the left side you got a perfect quadratic equation, hence it will be (x-3)^2= 9....Now you pass the 9 of right to the left ...then it will be [(x-3)^2]-9=0\n\n\nThe other equation x^2 - 2x - 3 = 0....This one is quite simple, it is the same as the part 1...The difference is that when you find the factors, they will be 2 numbers that multiplied give you 3 but substracted gives you 2.....The #s are 3 and 1....Here is the thing....when you break the equation into factors, you will have the following (x- ) (x+ )...Since you have a -2 that means that it is the -3 and the +1...Hence your factors will be the following (x-3)(x+1)
836052
You can start by factorizing 3x^4 from everything\nm(x) = 3x^4 (2x^3 + 5x^2 - 3x - 3)\n\nI think there's something wrong with your problem, none of the possible roots given by the Ruffini's rule are working. Perhaps, if instead of 6x^7 + 15x^6 - 9x^5 - 9x^4 it was 15x^7 + 15x^6 - 9x^5 - 9x^4, then it could be factorized very nicely in\nm(x) = 3x^4 (x+1)(5x^2-3) \nor if instead you wrote 6x^7 + 6x^6 - 9x^5 - 9x^4 then it'd factorized to\nm(x) = 3x^4 (x+1)(2x^2-3)
461122
Since f(x) is a polynomial of degree two in x,\nLet f(x) = ax^2 + bx + c, where a,b,c are constants.\n\nLet's find f(x) first.\n\nf(0) = 1 implies, c = 1.\n\nf(1) = 3 implies, a + b + c = 3\nor  a + b = 2 , or b = 2 - a  ...(1)\n\nf(-1) = 5 implies, a - b + c = 5\nor a - b = 4 ...(2)\n\nSubstituiting (1) into (2), we get,\n\na - [ 2 - a  ] = 4\n2a = 6 or a = 3.\nThus, b = 2 - a = 2 - 3 = -1.\n\nHence, f(x) = 3x^2 - x + 1\n\nNote that by long division or by the identity below,\n\n3x^2 - x + 1 = 3(x + 1)^2 - 7x - 2\n             \nWe get,\n\n[ 3x^2 - x + 1 ] /[1 + x]^2 = 3 - [7x + 2]/ [1 + x]^2   \n\nAnd that together with below,\n\n[7x + 2]/ [1 + x]^2  = 7 /(x + 1) - 5 /(x+1)^2\n\nGives,\n\n[ 3x^2 - x + 1 ] /[1 + x]^2 = 3 - 7 /(x + 1) + 5 /(x+1)^2\n  \nNow, we have an "integratable form".\n\n∫ [f(x) / (1 + x)^2] dx\n\n= ∫ [3x^2 - x + 1] / (1 + x)^2] dx\n\n= ∫ [ 3 - 7 /(x + 1) + 5 /(x+1)^2 ] dx\n \n= 3x - 7 log(x + 1) - (5/[x + 1]) + C\n\nwhere C is any arbitary constant.\n\nNote: Sometimes log(x + 1) can also be written as ln (x + 1)\n\nRemark. It is generally accepted that log and ln are both in base "e", the Euler number. If its in a different base, then usually its said or stated. For example logarithm of x in base 2 can be written as log_2(x).\nJust symbols :)\n\nRegards