How do you factor this completely? a to the fourth power- b to the fourth power?
817461
a^4 - b^4\n\n= (a^2 + b^2)(a^2 - b^2)\n\n= (a^2 + b^2)(a + b)(a - b)
445887
you have to take out an x because there are x's that are there that are common to both terms. x(1-x^4) then -x(x^4-1)\nso factoring is easier. Then -x(x^2+1)(x^2-1). Then it becomes -x(x^2+1)(x+1)(x-1). That's the answer.
283601
1) First we move all the factors with x to one side, and the ones without x to the other.  Result will be:\n\n4x/3 - 2x + x = 3/2 + 1/5\n\n2)  Then, we factor out an x from the left side and get:\n\nx(4/3 - 2 + 1) = 3/2 + 1/5\n\n3)  Calculate the actual numbers on both sides\n\nx(4/3 - 6/3 + 3/3) = 15/10 + 2/10\nx(1/3) = 17/10\n\n4)  Divide both sides by 1/3 and get a final result of...\n\nx = (17/10)/(1/3)\nx = (10/17)*(3/1)\nx = 30/17\n\n5)  Comform that I PWN n00bs.  And... I'm pretty sure I pwn n00bs.  Therefor, we are done!  ^-^
640891
I'm guessing you meant 17x^3 - 10x^2 - 6x +4 and 2x^2- 6x +2\nIf that's right, you just add like powers of x and their coefficients to get 17x^3 - 8x^2 -12x +6.
1051369
Given a+b+c=0\na+b=-c\nCubing both sides, we get\n(a+b)^3=(-c)^3\na^3+3a^2 b+3ab^2+b^3=-c^3\na^3+b^3+c^3=-3a^2 b-3ab^2=-3ab(a+b)=-3ab(-c)=3abc\nTherefore a^3+b^3+c^3=3abc
610613
In 6ab + 4a + 3b + 2, we group the first 2 terms and the last 2 terms:\n\n6ab + 4a + 3b + 2 =\n(6ab + 4a) + (3b + 2)\nthen factor out the common factor 2a in (6ab + 4a)\n(Note: to factor out is the same as dividing each term with the common factor)\n\nwe get:\n2a(3b + 2) + (3b + 2)\nNotice that in each term of the above we have a common factor of (3b + 2)\nBy dividing each term by (3b + 2) or factoring out (3b + 2) we have:\n(3b + 2)(2a + 1)
200291
The factor common to the first 2 terms in #1 is 3x^4 because both terms are divisible by that (take the GCF of the numbers in front, and the lower of the two exponents).\n\nFor the second pair, the common factor is 4\n\nSo you'd get 3x^4(3x + 2) - 4(3x - 2) if you have to actually factor them\n\nProblem 2, the common factor of the first 2 terms is 4x^3 and for the second pair, 1.\n\nSo you'd get 4x^3(5x + 4) - 1(5x + 4)  which becomes\n\n(4x^3 - 1)(5x + 4).
767628
I'm assuming you want to factor this\n\n5z^2 + 4z -12 = (5z -6)(z+2)
448387
4b(5 - a) - 3c(5 - a)\n= (5 - a)(4b - 3c)\n\nThe answer is D. Since both have the factor (5 - a), you factor it out, and what's left is (4b - 3c).
141990
(assuming the second 2 is an exponent) 2*2*2*7*5*g*h*h \nyou really should learn to do this if you are ever going to get through school though
919174
first multiply the first two factors and multiply the product by the third factor\n\nTo multiply something complicated like (4x + 6)(5x - 3)(15 - x), just do FOIL on two of the binomials and then distribute the answer onto the remaining binomial
91753
factorize\nx^2 + x = x(x+1)\n\nfactorize\nx^2 - 1 = .... (you should know)\n\n2 = 2 (!!!)\nTake the lcm of the three denominators (each different factor (=parenthesys) once and with the highest power)\n\nIt is 2x(x+1)(...)   (fill the dots)\n\n.....\n\nnow if you can't go on, you need to open your book again and study, or pay more attention at school at your math classes, and ask immediately for further explanation to your teacher when you feel you've not understood.\n\nGood work and good study
1136055
Here's how we can solve the problem logically, without doing all the multiplication.\n\nFirst of all, the coefficients aren't relevant to the problem; we don't care what the terms are specifically, just how many of them there are.\n\nLet's create two new temporary variables: let a = x + 5y, and b = 3z + w.  That reduces the problem, for the moment, to (a + b)^6.\n\nNow, any time you take a binomial to a power -- such as (a+b)^n -- there will be n+1 terms in the final expansion: (a+b)^2 has 3 terms, (a+b)^3 has 4 terms, etc.\n\nSo in (a+b)^6 there will be seven terms, which will have various coefficients but generally be of the forms a^6, a^5b, a^4b^2, a^3b^3, a^2b^2, ab^5, and b^6.  Note that, in each term, the sum of the powers is 6.\n\nNow, if we reinstate the original variables, substituting back in for a and b, we can consider each of those seven terms.\n\na^6 becomes (x+5y)^6, which will have 7 terms.\na^5b becomes (x+5y)^5 (3z+w), which will have 6 * 2 = 12 terms.\na^4b^2 becomes (x+5y)^4 (3z+w)^2, which will have 5 *3 = 15 terms.\na^3b^3 becomes (x+5y)^3 (3z+w)^3, which will have 4 * 4 = 16 terms.\na^2b^4 becomes (x+5y)^2 (3z+w)^4, which will have 3 *5 = 15 terms.\nab^5 becomes (x+5y) (3z+w)^5, which will have 2 *6 = 12 terms.\nAnd, finally, b^6 will have 7 terms.\n\n7+12+15+16+15+12+7 = 84.\n\nHope that made sense.
844053
I'll rewrite the question for you, because we usually denote ^ as "to the power of" for simplicity.\n\nx^2 + y^2 - 4x + 2y = 4\n\nWe have to do a process called completing the square, for both the x and y variables.  So first, let's group all the x terms together and y terms together.\n\nx^2 - 4x + y^2 + 2y = 4\n\nTo complete the square, we have to add 4 to the x terms and 1 to the y terms.  Since we're adding 4 and 1 to the left hand side, we have to do the same to the right hand side.\n\nx^2 - 4x + 4 + y^2 + 2y + 1 = 4 + 4 + 1\n\nNow, we factor those now square polynomials.\n(x-2)^2 + (y+1)^2 = 9\n\nThe center of the circle is given by the values in the brackets, but the negative of them.  In the above case, the center is at (2,-1), since you take the negative of -2 (the value next to the x) and 1 (the value next to the y).\n\nTo get the radius, you take the square root of the right hand side, so the square root of 9 is 3.
727905
(a + b)^6 - (a - b)^6\n\nNow, the previous answerers chose to factor the difference of 2 squares first, but I prefer to expand first.\n= (a^6 + 6a^5 b + 15a^4 b² + 20a³b³ + 15a²b^4 + 6ab^5 + b^6) - (a^6 - 6a^5 b + 15a^4 b² - 20a³b³ + 15a²b^4 - 6ab^5 + b^6)\n\nNow, we distribute the negative sign:\n= a^6 + 6a^5 b + 15a^4 b² + 20a³b³ + 15a²b^4 + 6ab^5 + b^6 - a^6 + 6a^5 b - 15a^4 b² + 20a³b³ - 15a²b^4 + 6ab^5 - b^6\n\nWe combine like terms\n= 12a^5 b + 40a³b³ + 12ab^5\n\nWe factor out 4ab\n= 4ab(3a^4 + 10a²b² + 3b^4)\n\nWe factor out the biquadratic\n= 4ab(3a² + b²)(a² + 3b²)\n\nTherefore, the completely factored form is\n(a + b)^6 - (a - b)^6 = 4ab(3a² + b²)(a² + 3b²)\n\n^_^
333726
Perform operations in this order:\ni)   brackets\nii)  powers\niii) divide\niv)  multiply\nv)   add\nvi)  subtract\n\nSo work it out like this:\n\n2 + 2 (4 - 6 * 7) + 4\n= 2 + 2 (4 - 42) + 4\n          By doing multiplication 6 * 7 in brackets first.\n= 2 + 2 (-38) + 4\n          By doing subtraction 4 - 42 in brackets.\n= 2 + -76 + 4\n          By doing multiplication 2 * -38\n= 2 - 76 + 4\n          Since + and - sign in front of 76 combine into a -\n= -74 + 4\n          By doing subtraction 2 - 76\n= -70\n          By doing addition -74 + 4
843761
well assuming you just want to break it down...\n1: you put the -2 power on both the 4 and a^3.  then it'd be (4^-2)(a^3^-2).\n2: remember that when it's an exponent to an exponent you multiply the numbers together so it becomes (4^-2)(a^-6)\n3: a negative exponent means it is one over itself so it would be 1/(4^2)(a^6)\n4: just simply from there to make it 1/ (16a^6)\n\nhopefully that was what you were asking!!