10xsquaredyz minus 22x cubed yz. factor the polynomial?
402918
I think you wrote 10x*2yz - 22x*3yz.  To factor this oyu must take out like terms by dividing.  You can see that both terms share a 2, y, z, and an x*2.  After the  division you get  2x*2yz( 5 - 11x).  I hope this helps.
896553
4x^2+8x-x-2\n(4x-1)(x+2)
1033909
14x^4y^4/18x^2y^5\n\nSince x^1 and y^1 must remain in any reduction, at lowest terms this would be:\n\n7x^3y/9xy^2\n\nRead: seven x cubed y over nine x y squared
773695
Look in your text book under "factoring polynomials"... or you could try this web site that explains this fairly thoroughly...\n\nhttp://faculty.ed.umuc.edu/~swalsh/Math%20Articles/Factoring.html
175133
2x-1(xsquared-x+10)squared
178683
yeah, I suppose you want me to factorise x^n - y^n. the general formula is\n\nx^n - y^n = (x-y)(x^n-1.y + x^n-2.y^2 + x^n3.y^3+.........+x^2.y^n-2 + x^n-1.y^1)\n\nTry putting n=2 or 3 or4 in this expression and you'll see what I mean.\n\nYou'll arrive at nothing but\n\nx^2 - y^2 = (x-y)(x+y)\nand so on..\n\nAnd please, the next time, a question without exponents. i'm SICK of '^'. \n\nAnd you know what, mr. 'timmobilelaptop' , the next time try using your brain. It's better than any T-189 calculator.
917477
where else?\n\nhttp://www.factoring-polynomials.com/
907035
You have \n100 +4X^2 -16y -40x\n\nFirst let's factor out a 4, since 4 is a factor of each term, that gives\n\n4*(x^2 -10X +25) -16Y\n\nOf course I rearranged the terms so the highest powers of X are\non the left.  I didn't factor 4 out of -16Y because I want the \nquadratic (X^2-10X+25), which I know I can factor.\n\nsince (X^2-10X+25) = (X-5)*(X-5) = (X-5)^2, we have\n\n4*(X-5)^2 -16Y\n\nThe secret is being able to recognize that \n(x^2 -10X +25) = (X-5)^2
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:)
801272
x³y - 3x²y - 10xy --- Take x and y out of each term...\nxy(x² -3x - 10) --- Factor the x² - 3x - 10\nxy(x - 5)(x + 2)\n\n*Note: To show exponents, such as x², use the ^ key (such as x^2). This expression looks like you basiclly wrote:\n3xy - 6xy - 10xy
774014
2^3+2*2^2-3(2)-10=8+8- 6 -10 =0\nTherefore x-2 is a factor of the given polynomial
617264
Since each term has at least x in it, divide out the x by distributive property:\n\nx(x^2 - 4xy + 4y^2)\n\nThen factor the rest by FOIL:\n\nx(x-2y)(x-2y)
913334
x^2 + 2x + 7 Remainder 10\n\ni can add the steps if you don't think i did it by hand
699742
-(3x+y)(3x-y)\n-(9x^2-y^2)\ny^2-9x^2
865479
First of all is the radical in the numerator over just the 10 or over the entire 10y^3?\n\nThe only this you can factor out and cancel is a radical 2 on the top and bottom.  Other than that, you can't really simplify this problem.  Are you sure you wrote it down correctly?
781910
not possible to factor easily ... try the quadratic formula.
697928
First you group the 'like terms'; in your example, the first two and the last two. Then you see the common factor and pull it out\n\ny^2(x^2-1) - 15 (x^2-1)\nNow you see the common term in both and pull that out\n(x^2-1) (y^2-15)\n\nthus you have your two factors of the polynomial