Rewrite the middle term as the sum of two terms and then factor completely. x^2 + 19x + 90?
87719
x^2 + 9x + 10x + 90\n\n(x + 9)(x + 10)
913289
You end up with a remainder, so it's not pretty, but you should be able to do long division and get:\n\n3x^2 - 4x + (16/3) - 23/(3*(3x+2))
783870
7x^2+16x over (x-2)(x+2)(x+3)
1083459
Use this formula:\n\nSn=n/2[t1+tn]\n\nwhere\nSn --> ur required sum\nn --> no. of terms in the given progression \nt1 --> first term (in ur case: 3)\ntn -->last term (in ur case: 803)
605867
(x^2)/2 - 3(x^(3/2))/(3/2) + c = (x^2)/2 - 2x^(3/2) + c
631080
You can noy factor it out... it makes it easier to figure out when you put it into lowest terms... so it would be \n\n   x^2+7x-25..... so you cant factor it.
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)
651993
Here's the best I can do to help with the first one.\n   __x_+9___\nx-2)X^2+7x-2\n    x^2-2x\n    ------\n        9x-2\n        9x-18\n        -----\n           16\nFollow the example for the rest
570705
Factoring out X gives:\nX(-X^2 + X + 6)=0\nUsing the quadratic formula to factor the terms inside the brackets gives:\n-X^2 + X + 6 = (X+2)(-X+3)\nSo we have:\nX(X+2)(-X+3)=0 and the zeros are X=0, X=-2, X=3
904822
Hello Dear Blue \n\n2x^3+8x^2 - 5x + 5 ÷ x-2\n\n▪ Step1\n( 2x^2 ) ( x-2 ) = 2x^3 - 4x^2 ; { let Q1 = 2x^2 }\n2x^3+8x^2 - 5x + 5 - ( 2x^3 - 4x^2 ) = \n2x^3+8x^2 - 5x + 5 - 2x^3 + 4x^2 = +12x^2 - 5x + 5 {remainder} \n\n▪ Step 2\n+12x^2 - 5x + 5 ÷ x - 2 \n(x - 2) (12x) = 12x^2 - 24x ; { let Q2 = +12x }\n+12x^2 - 5x + 5 - ( 12x^2 - 24x  ) =\n+12x^2 - 5x + 5 - 12x^2 + 24x = +19x +5 {remainder}\n\n▪ Step 3 \n19x +5 ÷ x -2\n(x -2) (+19) = +19x - 38; { let Q3 = +19}\n19x +5  - (+19x - 38) =\n19x +5  -19x + 38 = +43 {remainder}\n\n▪ Step 4\nSum Q1, Q2, Q3\nQ1 + Q2 + Q3\n2x^2 +12x +19\n\nSo;\n(2x^3+8x^2-5x+5) ÷ (x - 2)    = >\n\nQ(x) OR quotient = 2x^2 +12x +19 \nR(x) OR Remainder is + 43 \n\nGood Luck dear.
517819
start pulling out common factors\n\n2 (x^3 + 27)\n\n2(x^3 + 3^3)
938878
x^2 - 2(3-x^2)\nDistribute the -2\nx^2 + (-2)(3) + (-2)(-x^2)\nSimplify\n= x^2 - 6 + 2x^2\nCombine like terms\n= 3x^2 - 6
636732
8x^2y^3+5y^2 when "FACTORED COMPLETELY"=\n\ny^2 (8yx^2+5). \n\nThis should be right. My dad is a complete genius and he did this in like 2 seconds. I double checked it on my calculator too by plugging in numbers, so I can assure you he's right.\n\nHope this helps! Have a nice day :)
1006943
Rewrite as a power series, since there is no way to do this using traditional methods:\n\nx^2 - (1/2)x^4 + (3/8)x^6 - (37/144)x^8 + (1031/5760)x^10 -.....\n\nThen integrate:\n\n(1/3)x^3 - (1/10)x^5 + (3/56)x^7 - (37/1296)x^9 + (1031/63360)x^11 -.......
255329
Have patience :)  You got some answers the first time you asked this.  Here it is again:\n\nIt does not get eliminated, it gets split up into the 2 factors.\n\nOn this one, you can go ahead and divide by 2 to make things a little easier. You'll be left with\n\n3x^2 - 7x - 6 = 0\n\nNow to factor. You need 2 numbers whose product equals 3*-6, or -18 and whose sum equals -7. From this point on it's just sort of educated guessing. \n\n1*-18 = -18, but 1-18 = -17, so that won't work\n2*-9 = -18, and 1-9 = -7 -- THIS ONE WORKS!! yay!\n\nSo now you can rewrite your problem as\n\n3x^2 - 9x + 2x -6 = 0 and group the first 2 and last 2 terms\n\n(3x^2-9x) + (2x-6) = 0 -- you can factor a 3x out of the first group, and a 2 out of the second\n\n3x(x-3) + 2(x-3)=0 -- and now you can factor an (x-3) out\n\n(x-3)(3x+2)=0\n\n~~~~~~~~~~~~~~~~~~~\nOops, you're right. I was typing too fast.  Oh well, I still got the right final answer :)  Thanks
208112
use the order of operations, so first the multiplcation. so\n\n2 + 2 - 3 + 96/2\n\nThen divide by two\n\n2 + 2 - 3 + 48\n\nAdd the twos\n\n4 - 3 + 48\n\nSubtract\n\n1 + 48\n\nAdd and get\n\n49
77583
add 5 to both sides:    x^2 = 5\ntake the sqrt of both sides:  x= + or - sqrt of 5