How do you find the term with b^2 in the expression (a+b^2)^10?
606538
What a lot of longwindedness for a simple question.\n\nThe expansion of (x + y)^10 is\n\nx^10 + 10x^9y + 45x^8y^2 + 120x^7y^3 ... etc.\n\nSo the term you are looking for is the second in the expansion,\n\n10a^9b^2\n\nWell, whosdogareyou and the always reliable blahb have the answer. The others should cease answering math questions and try questions on dating and relationships, or cooking.
780298
first u distribute the - to the stuff in the parenthesis so you have\n7x^2-3+x^2\nthen combine like terms\n8x^2-3
551906
Add each equation\nA + B = 3x^2 + 2x - 2\nA - B = -x^2 + 4x - 8\n2A = 2x^2 + 6x - 10\nDivide each side by 2\nA = x^2 + 3x - 5\nNow plug into one of the original equations to get B\nB = 3x^2 + 2x - 2 - (x^2 + 3x - 5)\nB = 2x^2 - x + 3
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
419498
First, group the terrms with x in parenthesis and the terms with y in parenthesis:\n(x^2 + 6x)  + (y^2 - 8y) = -9\nLet's think for a moment, we have two squared terms, x^2 and y^2. Think of the two groups in parenthesis as if you were completing the square to solve a quadratic equation.\nFor both, take 1/2 the middle term and square it, adding it to both sides.:\n(x^2 + 6x +(3)^2) + (y^2 - 8y + (4)^2) = -9 + (3)^2 + (4)^2\nFactor and combine like terms:\n(x + 3)^2 + (y - 4)^2 = 16\nNow we have the standard form of a circle\nThe center is the OPPOSITE of the numbers in parenthesis, according to PLNR (positive left, negative right)\nThus the center is (-3, 4) and radius is sqrt(16)= plus/minus 4\nRemember that the radius must extend in negative directions: down and left. That's why it is plus/minus.
36273
(A^3 + B^3)=(A+B)(A^2 - AB + B^2)
16175
=9x^2+6x+1\n\nsince (a+b)^2=a^2+2ab+b^2\n\nhttp://www.algebrahelp.com/lessons/
798648
(x+3)(x^2+2x+4)\n\nStart by taking the first term in the first parenthesis and multiplying it by each term in the second parenthesis:\n\nx*x^2 + x*2x + x*4 = x^3 + 2x^2 + 4x\n\nNow take the second term and multiply it by each term in the second parenthesis:\n\n3*x^2 + 3*2x + 3*4 = 3x^2 + 6x +12\n\nNow add all of that together\n\nx^3 + 2x^2 + 4x + 3x^2 + 6x +12\n\nand combine terms\n\nx^3 + 5x^2 + 10x +12
1027133
x^2-9x-10=0\n\nx^2 +(-1 +10)x -10=0\n\nx^2 -x +10x -10=0\nx(x-1) +10(x-1)=0\n\n(x+10)(x-1)=0\nthus x=10 , -1
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
913334
x^2 + 2x + 7 Remainder 10\n\ni can add the steps if you don't think i did it by hand
641478
a^2 + b^2 = c^2\n\nPythagorean theorem
538044
hmmm...very tough one!\n\nill sovle by the formula!\n       _________\n-b +- V (b^2)-4ab\n------------------\n       2a\n       _________\n-10 +-V 100+512\n------------------\n       2\n\n       _________\n-10 +-V 612\n------------------\n       2\n\n-10 +- 24.7\n------------\n     2\n\n14.7/2\n\n34.7/2\n\n(7.35, 17.35)
1147051
Use pythagorean theorem\n(2x)^2 + x^2 = 10^2\n4x^2 + x^2 = 100\n5x^2 = 100\nx^2 = 20\nx = sqrt(20)\nx = 2sqrt(5)\nArea = (1/2)(2x)(x)\n= (1/2)(2*2sqrt(5))(2sqrt(5))\n= (1/2)(4*2*5)\n = 20
295680
You can't answer it unless it equals something (i.e. v^2 + 20v + 100 = 1) so I'm assuming you want to simplify it.\n\nFirst check that you can't factor something out (which you can't)\n\nThen the first thing you need to do is get rid of v^2 so you put it in the following form\n\n(v +/- _)(v +/- _)\n\nSince all the numbers are positive, you can figure that it's going to be v+_\n\n(v + _)(v + _)\n\nWhat you want now is to find two numbers whose sum is 20 and whose product is 100\n\nYou can take the long way out on this and write out all the possibilities but rather than going through that long tedious process, you can also take a look at the numbers and see if something comes out.\n\nIn this case, the number 10 works.\n\n10 x 10 = 100\n10 + 10 = 100\n\nSo you plug it into the equation.\n\n(v + 10)(v + 10)\n\nThe you can simplify that to say\n\n(v + 10)^2\n\nAnd there's your answer.\n\nYou can check it by foiling:\nv^2 + 10v + 10v + 100\n= v^2 + 20v + 100
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.
3037
This is not presented as an equation, so the goal appears to be to combine the rational expression --> x+10 \nThat's all you can do with the question as it is written.