factor the polynomial
199391
a^3 - a^2*b + a*b^2 - b^3 =\n= (a^3 + a*b^2) - (a^2*b + b^3) =\n= a(a^2 + b^2) - b(a^2 + b^2) =\n= (a - b)*(a^2 + b^2)
569437
t^2 - 12t + c\nc = 36\n\nThats if you want to have a perfect square polynomial.\n\nIf you would like to tell me a little more, like maybe what one of the factors are, or zeros are, and i can tell you what c is.
136946
This is the best I could come up with:\n\n(a)  (x^2-4)(x^4+4x^2+16)\n      (x-2)(x+2)(x^4+4x^2+16)\n\n(b)  (x^3-8)(x^3+8)\n     (x-2)(x^2+2x+4)(x+2)(x^2-2x+4)\n\n(c)  If you take out the factors that are the same, what's left should be equal hence:\n     x^4+4x^2+16=\n(x^2+2x+4)(x^2-2x+4)\n\n(d)  x^(a^2)+2ax^a+4a^2=\n(x^a+ax+2a)(x^a-ax+2a)
434430
You typed exponents just fine!\n\nx^3 - x^2 -x + 7 divided by x+1 is a "long division" type problem rather than factoring because x+1 doesn't go evenly into the polynomial\n\nThe answer is x^2 -2x + 1 + 6/x+1
289766
I think you set the equation to zero then you have to factor it.  Go to the link below if you wanna know how to do it. OR look your your math book.
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
463805
I suppose you need the second factor of this polynomial. In order to get the answer, we have to factorise the polynomial.\n6a^2 - 5a + 1 = 6a^2 -3a - 2a + 1\n                      = 3a(2a - 1) -1(2a -1)\n                      = (3a -1)(2a - 1)\nThe answer is 2a - 1
244661
If those are the zeros, you just do it like this for the first one:\n\ny=(x-5)(x-3i)(x+3i)\n\n\nIf your teacher wants it in general form, you have to multiply out all the terms. \n\ny=(x-5)(x^2+9)\n\ny=x^3+4x^2-45\n\n\nThe second set doesn't make sense because you have an i without a partner. In polynomial equations, the complex factors always come as conjugate pairs, like the -3i and +3i.\n\nAnyway, I hope this helps. Basically, you set up a bunch of factors that are x-each root.\n\nI just changed it because I realized that the 1. meant problem number 1.
389812
2x - 2 = x^2 - 4x + 3\nSet one side equal to 0\n0 = x^2 - 6x + 5\nFactor\n0 = (x - 5)(x - 1)\nSet each factor equal to 0\n0 = x - 5              0 = x - 1\nSolve\n5 = x                  1 = x
205504
By the remainder theorem, substituting a value b into a polynomial gives the remainder if you divided the polynomial by (x - b). So if you can find a value that goes into the polynomial that gives a value of 0, then you know that (x - b) is a factor, since there is no remainder. \n\nNow (x - b) can only be a factor of the polynomial if b is a factor of the constant at the end, in this case 5. So the possibilities to test are 5, -5, 1 and -1.\n\nTesting +/- 1 first as they are easiest\n\nPutting x = 1 gives 8 + 52 + 66 + 31 + 5 = 162.\nPutting x = -1 gives 8 - 52 + 66 - 31 + 5 = =4\nSo (x - 1) and (x + 1) aren't factors.\nPutting x = 5 clearly gives a positive value\nPutting x = -5 gives 0, so (x + 5) is factor.\n\nThe next step is long polynomial division.\n\n(x + 5)   )    8x^4 + 52x^3 + 66x^2 + 31x + 5\n\n8x^4 divided by x gives 8x^3 \n8x^3 * (x + 5) = 8x^4 + 40x^3\nSubtract this from the polynomial to find the remainder:\n12x^3 + 66x^2 + 31x + 5\n\nThe quotient so far is 8x^3\n\nRepeat this process to find the final quotient of:\n\n8x^3 + 12x^2 + 6x + 1\n\nIf you can recognise that this is of the form of a cubic:\n\na^3 + 3.a^2.b + 3.a.b^2 + b^3\nwhere a = 2x and b = 1, you have your solution.\n\nIf not, we again have to try to factorise by testing factors. Now it should be obvious that (x - b) can no longer be factor, since we know (x + 1) and (x - 1) can't be factors. We need to try something of the form (ax - b), by observing that x^3 has a coefficient of 8, so (2x + 1) is a reasonable factor to try, since 2 is a factor of 8 at the start of the polynomial and 1 is a factor of 1 at the end of the polynomial. You can test if it's factor, because if (2x + 1) is factor, then x = -1/2 must be solution to the polynomial being equal to 0, which it is.\n\nPerforming polynomial division again gives 4x^2 + 4x + 1, with the factors now at (x + 5)(2x + 1). From here you can use trinomial factorising techniques to get the final solution of (x + 5)(2x + 1)^3
199400
So, for a) & b) the two things that you should notice are:\n1 - The exponents inside the second polynomial alternate, meaning that as the powers of the x variable decrease, the powers of the y variable increase, both by increments of one.\n2 - The answer will only contain the variables x and y raised to the same power, which is one more than the leading exponent on the x variable.  \n\nSince (x + y)(x^2 - xy + y^2) = x^3 + y^3\n and  (x + y)(x^4 - x^3y + x^2y^2 - xy^3 + y^4) = x^5 + y^5,\nthen for any given polynomial the pattern is written out...\n(x + y)(y^2n - x^(2n-1)y + x^(2n-2)y^2 + ... + x^2y^(2n-2) - xy^(2n-1) + y^2n = x^(2n+1) + y^(2n+1)
345973
1.\n3x + 6/((x + 1)(x + 2)) Factor 3 out of the top.\n[3(x + 2)]/[(x + 1)(x + 2)] x+2's cancel\n3/(x + 1) <-- This is simplified.\n\nI don't have the energy to do the other ones right now.  :/
395786
Notice that both terms have a 'factor' of (p+q) \nThe first term is (p+q)*(p+q) and the second is (5)*(p+q)\nSo 'factor out' the (p+q) from each :\n= (p+q) * [ (p+q) + (5) ]\n= (p+q) * (p+q+5)\nis the factorization.
21064
The instruction is to write a polynomial function with the zeros 0, 1, and 1+i. Please keep in mind that when a zero consists radicals or imaginary numbers, the conjugate of that zero is also a zero. For exmaple, here 1+i is a zero. Then 1-i must also be a zero.\n\nThis problem is like working backwards. Instead of factoring the polynomial, you are asked to do it backwards, starting with the zeros. Since 0, 1, and 1+i and 1-i are zeros, you would expect the last step of the factoring to look something like this:\n\nx(x-1)[x-(1+i)][x-(1-i)]\n\nNow simplify this expression a bit. It becomes:\n\nx^2-x[(x-1-i)(x-1+i)]\n\nSimplify this expression even further and you get:\n\nx^4-3x^3+4x^2-2x
217152
Factor each one.\n(x^2-3x-18) / (x^2-4x-32) divided by (x^2-2x-24) / (x^2-x-20)\n= (x^2-3x-18) / (x^2-4x-32) times (x^2-x-20)/(x^2-2x-24)\n= [(x-6)(x+3)]/[(x-8)(x+4)] times [(x-5)(x+4)/(x-6)(x+4)]\n= [(x-6)(x+3)(x-5)(x+4)] / [(x-8)(x+4)(x-6)(x+4)]\nYou can now cancel anything on top with anything on bottom\nx-6 and x+4 cancels\n= [(x+3)(x-5)]/[(x-8)(x+4)]
264544
You simply divide each term in the polynomial by that integer.. For your example here it goes..\n\n(3x^2+8x-2) divided by 2 \n\n> {3(x^2)/2} + {8x/2} - {2/2}\n\n>> Answer: 1.5 x^2 + 4x -1    or   (3/2)x^2 + 4x -1
442605
.more kids homework... factoring?\n1. rewrite the middle term as the sum of two terms and then factor completely 15x^2 -21x +6\n\n3(5x^2-7x+2)\n3(5x^2-5x-2x+2)\n3(5x-2)(x-1)\n\n2. rewrite the middle term as the sum of two terms and then facto by grouping. x^2 - 14x +45 \n\nx^2-9x-5x+45\n(x-9)(x-5)\n\n3. rerewrite the middle term as the sum of two terms and then facto by grouping x^2 + 19x +90\n\nx^2-10x+9x+90\n(x+10)(x+9)\n\n4. rewrite the middle term as the sum of two terms and then facto by grouping x^2 -2x -63\n\nx^2-9x+7x-63\n(x-9)(x+7)\n\n5.solve x^2 = -6x\n\nx=-6