If (a+b)^20 is multiplied out, the coefficient of ab^19 is?
538839
It's 20C19 which is 20.  So d).
600427
It is already factored, you can write it more concisely as (x-y)^2.\n\nThe EXPANDED form is x^2-2xy+y^2.
875476
(a + b - c)(a + b + c)\n((a+b)+c)((a+b)-c)\n(a+b)^2-c^2\na^2+2ab+b^2-c^2
573482
Literal coefficient: -ab\nNumerical coefficient: -1
1161280
2√5  \n\nif you have √20 its √4*5\n√4 is 2\nso you have 2√5
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!
868234
P = 2W + 2L\nReplace W with 10 and L with 19\nP = 2(10) + 2(19)\nP = 20 + 38\nP = 58 in
793106
Easy since  x square has a coefficient bigger than 1, then we multiply this number by -13. The result is -26. No what two numbers multiplied gives -26,  and added gives -25? The two numbres are -26 and 1. Now we factor by grouping\n\n2xx - 26x +x -13\n2x(x-13)  1(x-13)\n(2x+1) (x-13)
672272
I have taken the liberty of calling it the (a+b)^2=a^2+b^2+2ab ; (a+b+c)^2=a^2+b^2+c^2+2ab+2ac+... ; (a+b+c+d)^2=a^2+b^2+c^2+d^2+2a...\n+2cd etc... theorem. A bit long winded but exact.
962288
When you have a quadratic equation Nx² + Mx + P, you can sometimes factor it into (ax+r)(bx+s).  Here's a way to determine a, b, r and s:\n\nWrite down N and M and P, then look at what two numbers a & b when multiplied will equal N, and what two numbers r and s when multiplied will equal P and at the same time when all multiplied together (as + rb) will equal M.  It sounds scarier than it is!  Here's an example with your first equation:\n\nb² + 4b - 12\n\nSo we have our coefficients (N,M,P):  1  4  -12\n\nFor 1 = ab our only choices are a=1, b=1\n\nFor -12 = rs, we have several choices 1,12 ... 2,6 ... 3,4  with one factor being positive and one negative so we can get -12.\n\nWhich two factors have a difference of 4?  Well it's 2 and 6.  And since 4 is positive we want +6 and -2.   So putting it all together we get:\n\nb² + 4b - 12 = (b-2)(b+4)\n\nLet's try one more (and then I have to go to a meeting so I can't help you anymore right now, sorry!) ...\n\n2b² - 13b +15\nOur coefficients are:  2  -13  +15\nFactors of 2:  1,2\nFactors of 15:  1,15 ... 3,5\nHow can we combine factors of 2 and factors of 15 to get -13?\nNotice that -15+2 = -13, so our factors are:\n(2b-15)(b+1) = 2b² - 13b +15\n\nHope that helped a little.  Good luck!\n\n~ ♥ ~
836453
a^24b^24\njust multiply the exponents\n\nhope thie helps!!!
580919
You substitute 10 for a, and 20 for b, so you have 10^2 + 20^2 = c^2, then you do the math, so you have 500= c^2\n\nThen you square 500 so c=22.36
139883
(a+b)(a-b) \n= a^2 - ab + ba - b^2\n= a^2 - b^2 -ab +ab\n= a^2 - b^2
813834
It didn't disappear.  Expand (x + b/2a)^2 and you will find it.
814891
a^2 - 2ab + b^2 = (a - b) ^2\n\nthen...\n\n(a - b)^2 / (a + b) = (a -b) (a - b) / (a + b)
550482
-2a(b+2)-6b=\nor\n-2(ab+2a+3b)
503153
(x+7)(x+3)\n(x-11)(x+2)\nforget this one.  its AC factoring though\nthis ones AC too