how do you factor a^4-7a^2+10  ?????
545883
(a^2  - 5) * (a^2  - 2)
531417
(a+b)(b^6-ab^5+a^2b^4-a^3b^3+a^4b^2-a^5b+a^6)\nGood Night, and Good Luck!
708943
You have to complete the square.\n\ny = a(x^2 + b/a x) + c\ny = a(x^2 +b/a x + b^2/4a^2) + c - b^2/4a\ny = a(x + b/2a)^2 + (4ac - b^2)/4a\n\n[4ay - (4ac - b^2)]/4a = a(x + b/2a)^2\n[4ay - 4ac + b^2]/(4a^2) = (x + b/2a)^2\n\n+/- sqrt(4ay - 4ac + b^2)/2a = x + b/2a\n\nx = (-b +/- sqrt(4ay - 4ac + b^2))/2a
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.
208740
EDIT: ** Baah. When I typed this answer I was first. Then at the end, I'm like the 3rd. Oh well..**\n\nThere are two ways. Which one is easier depends greatly on your thinking skills.\n\nYou can take the expression x2+7x+12 and assume it should break up in two factors:\n(x+ ? )(x+ ?)\nThen, search for two numbers that add up to the 7, but when multiplied equal 12.\nIt helps if you break down 12 into it's components, 12= 2*2*3. So, obviously, the answers could be:\n\n2 * (2*3) = 2 * 6 (but 2+6 is not =7)\n3 * (2*2) = 3 * 4, voila! 3+4 also = 7, so:\n\n(x+3)(x+4).\n\nThe second way could be just solving the cuadratic formula for second degree equations:\n\nb=(-b +/- SQRT(b2 - 4ac))/2a\n\nWhere ax2 + bx + c, so you'll have: a=1, b=7, c=12.\n\nReplace, then solve twice: Once for the + value of the SQRT, another for the - value of the SQRT. You should get something like:\n\nFirst run:\nx= -3\n\nSecond run:\nx= -4\n\nx=-3 is the same thing as (x+3)=0\nand \nx=-4 is also (x+4)=0\n\nNotice the two factors are precisely (x+3) and (x+4)?\n\nHope this helped..
858168
1. 14x^2 +21x\n\na good way to start would be to factor out any common terms:\n\n7(2x^2 + 3x)\n\n7x(2x + 3)
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)
302542
4t - 7 = 10t - 25\n4t - 7 - 10t = - 25\n4t - 10t = 7 - 25\n(4 - 10)t = -18\n-6t = -18\nt = 3
214070
If using modulo is permitted,\n\n3 + 7 - (8 % 4)\n= 3 + 7 - 0\n= 10
824178
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
977376
at the first 4 you use a^3-b^3=(a-b)(a^2+ab+b^2)\n and a^3+b^3=(a+b)(a^2-ab+b^2)\n\nfor the next 2 :\nthe first 2 factors and the second 2 factors have something in comon\n\nthe next ones use the above formulas plus one more \na^2-b^2=(a-b)(a+b)
460347
When you have groups of terms like your problem you factor out the things that each group of terms share. \n\nFor your first part: (a^2c^2 + b^2c^2 ) the terms share a c^2 so you factor that out. \n\nyou have c^2 (a^2 + b^2)\n\nThe second terms (-4a^2d^2-4b^2d^2) share 4d^2 so you factor that out. \n\nSo you're left with c^2(a^2+b^2) - 4d^2 (a^2 + b^2). \n\nWhen you've done this part you can actually go farther because if you look at it both sections share (a^2 + b^2). \n\nso you can simplify the whole problem with \n\nc^2 - 4d^2 (a^2 + b^2).
1174116
8x^2 -5 -3x^2 + 7\n8x^2 - 3x^2 -5 + 7\n5x^2 + 2
927974
Multiply the coefficients and then add and subtract the exponents for the like bases. \n(7x^3y^-5)(4xy^3)= 28x^(3+1)y^(-5+3)= 28x^4y^-2
954300
ok\nso its factoring.this is hella easy.10x+40 is 10(x+4)\nso the answer is 10(x+4)there you go
275675
15a^2bc^2 - 9a^2b^4c\n\nLook to see what is in both terms, that can be factored out.  If you don't know, try rewriting it as:\n\n15*a*a*b*c*c - 9*a*a*b*b*b*b*c\n\nIn fact, the 9 and 15 can be broken down even further\n\n3*5*a*a*b*c*c - 3*3*a*a*b*b*b*b*c\n\nNow, there is a 3 in both terms, 2 a's in both terms, 1 b, and 1 c.  Pull those out to the front\n\n3*a*a*b*c*(...) now see what is leftover\n\n3*a*a*b*c*(5*c - 3*b*b*b) and rewrite in exponent form\n\n3a^2bc(5c - 3b^3)\n\nThe 3a^2bc is your greatest common factor.
17048
= x^2 - 7x + 4x -28\n= x^2 - 3x - 28