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?
832515
(1) any of the following may be the root of the expression,+1,-1,+2,-2,+3,-3,+6 or -6.\nNow putting the value of x=-1,see ifthe value of the expression becomes zero or not?If it becomes zero,then x=1 is one factor of the expression,if it does not become zero try with =1,then -2.+2 and so on.\nLuckily for us,with x=-1,the value of the expression becomes zero\nTherfore,x+1 is one factor\nx^4++x^3-7x^2-13x-6\n=x^3(x+1)-7x(x+1)-6(x+1)\n=(x+1)(x^3-7x-6)\n=(x+1){x^2(x+1)-x(x+1)-6(x+1)}\n=(x+1)(x+1)(x^2-x-6)\n=(x+1)^2(x-3)(x+2)\n(2)Will you please check up the second one once more and if there is any correction send it to me through Yahoo answers.
735529
f(x)=-2x^4+6x+7\nf(-x)=-2(-x)^4+6(-x)+7\n=-2x^4-6x+7\nnether odd nor even
776562
f(g(x)) = 3(2x-7)+5\n= 6x-21+5\n=6x-16\n\ng(f(x)) = 2(3x+5)-7\n=6x+10-7\n=6x+3
982146
Just add and substract terms in both sides.\n\nSince \ny-x^2+8x\nis equal to 7, \nyou can add x^2 and substract 8x to both sides.\n(y-x^2+8x) +x^2-8x = y\n7 +x^2-8x = x^2-8x+7\n\nso y=x^2-8x+7\n\nby the way, if you factorize in terms of x then you'd get\ny=(x-1)(x-7)
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
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:)
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!
1070097
(x+3) (5x-3) (7x-1)\n\nYou can do this manually by figuring out the factors of 35 and 9.\n\nFor 35:  {1, 5, 7, 35, -1, -5, -7, -35)\nFor 9: {1, 3, 9, -1, -3, -9)\n\nI started assuming that (x + n) would be a factor.  I tried (x + 1), (x - 1), then (x + 3).\n\nIf you do long division, you get:\n(x + 3) ( 35x^2 - 26x^2 + 3)\n\nThen you just repeat the process for ( 35x^2 - 26x^2 + 3)... you need factors of 35 and factors of 3 that will add up to -26.  \n-3 x 7 = -21 and -1 x 5 = -5, so (5x - 3)(7x - 1) works.\n\nThat gives the final answer of: \n(x+3) (5x-3) (7x-1)
522506
All you do is plug in that value for x...\n\nSo, (a)f(0)\n\nf(0) = (0)^2 - 7(0) + 10\nf(0) = 10\n\n(b)f(5)\n\nf(5) = (5)^2 - 7(5) + 10\nf(5) = 25 - 35 + 10\nf(5) = 0\n\n(c)f(-2)\n\nf(-2) = (-2)^2 - 7(-2) + 10\nf(-2) = 4 + 14 + 10\nf(-2) = 28\n\nHope that helps :)
868344
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).
1119411
1. Add 3 to both sides to set equal to zero, then factor to get something that adds to three and multiplies to 28---7 and -4\n(x-4)(x+7) = 0\nx=4, 7\n\n2. Add 4 to both sides to set equal to zero, then factor to get something that adds to 2 and multiplies to -15---5 and -3\n(x+5)(x-3) = 0\nx = -5, 3
532925
9c(4+5c)\nFactors of 12 that add to 7\n(x+3)(x+4)\nFactors of -44 that add to -7\n(x-11)(x+4)\nFirst factor out 5: 5(x^2 + 8x + 16)\nLook for factors of 16 that add to 8\n5(x+4)(x+4)
547775
f(x)= x^3+x^2-22x-40\nf(x)= (x+4)(ax^2 + bx + c)\n\nComparing coefficients,\nx^3: 1=a\nconstant: -40=4c => c=-10\nx^2: 1=b+4a => 1=b+4 => b=-3\n\nf(x) = (x+4)(x^2-3x-10) = (x+4)(x+2)(x-5)\n\nThe other factors are (x+2) and (x-5).
741149
1.f(x)=x^2+4x-7\nx=[-4+/-rt(16+28)]/2\n=[-4+/-2rt(11)]/2\n=-2+/-rt11\n\n2.3x^2-11x-20=0\n=3x^2-15x+4x-20=0\n=(3x+4)(x-5)=0\nx=-4/3 or 5
570705
Factoring out X gives:\nX(-X^2 + X + 6)=0\nUsing the quadratic formula to factor the terms inside the brackets gives:\n-X^2 + X + 6 = (X+2)(-X+3)\nSo we have:\nX(X+2)(-X+3)=0 and the zeros are X=0, X=-2, X=3
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
1002101
factor 9x^2y^2-9y^2\n9y^2(x^2-1)\n9y^2(x+1)(x-1)