Ok can one of you tell me how to do this ( Factor Each Expression 10X+40)?
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
6182
You can use what is called the Master Product Method to factor this and other quadratic trinomials.\n\nFor something of the form ax^2 + bx + c, the master product is a*c.  In this example, a is 2, b is -5 and c is 2, so the master product is 2*2 = 4.  Now you must find two numbers which multiply to give you the master product of 4, but add to give the middle coefficient of -5.\n\nThis requires a bit of trial and error, but you should be able to quickly discover that the numbers -1 and -4 will work since -1*-4 = 4 and -1+-4 = -5.  Now you rewrite the the entire trinomial replacing the middle term -5x with the equivalent -1x - 4x:\n\n    2x^2 - 5x + 2\n= 2x^2 - 1x - 4x + 2\n\nWe factor by grouping the first pair and last pair together and pulling out what's common to each pair:\n\n= (2x^2 - 1x) + (-4x + 2)\n= x(2x - 1) -2(2x - 1)\n\nNotice that (2x-1) appears in the factorization of both pairs.  We finish by factoring out the (2x-1) to get\n\n= (x - 2)(2x -1)
684474
You have to factor each part first, then cross off (cancel) identical factors from the numerator and denominator of either expression.  If the factors are exact opposites, like x-8 and 8-x, cancel and put a -1 in the numerator.\n\n\nIn your example only the first expression is factorable:\n(x-8)(x+3)/(2x)(3+x) times 5x^2/(8-x)\n\nThe (x-8) cancels the (8-x) to become -1, then (x+3) cancels the (3+x), and the x from the (2x) cancels one of the x's from (5x^2)\n \nAnswer: -5x over 2
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.
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.
781910
not possible to factor easily ... try the quadratic formula.
751075
Yes, you've got half the equation right. It should look like this:\nx = n+20
597439
=[(4+3x)/x] + (2/3)[(5+3x)/x]\n=[(4+3x)/x] + [(10+6x)/3x]\n=[(12+9x+10x+6x)/3x]\n=[(15x+22)/3x]
669578
I have corrected the expression. Well here I redo the factorization\n\nThe given expresion \n= 12x^3+60x^2y+75xy^2\n= 3x(4x^2+20xy+25y^2)\n= 3x[(2x)^2+2(2x)(5y)+(5y)^2]\n= 3x(2x+5y)^2
945505
{(x^2+5x+4)/(x+3)^2}/(x+1/(x^2 -9))\nFactor everything that you can factor.\nx^2 + 5x + 4 = (x+1)(x+4)\nx^2 - 9 = (x+3)(x-3)\nYou have\n{(x+1)(x+4)/(x+3)^2} * {(x+3)(x-3)/(x+1)}\nYou can cancel the (x+3)'s and the (x+1)'s leaving you with\n(x+4)(x-3)/(x+3)
92952
(8x+3)(4x+3)\n32x2+24x+24x+9\n32x2+48x+9
791576
I'm going to assume the placement of parenthesis.\n(x^2 – 3x + 2)/(8x - 8) ÷ (x^2 – 4)/(5x +10)\n\nStep 1: Flip the second fraction, so that you have multiplication instead of division.\n(x^2 – 3x + 2)/(8x - 8) * (5x +10)/(x^2 – 4)\n\nStep 2: Factor.\nx^2 – 3x + 2 = (x - 2)(x - 1)\n8x - 8 = 8(x-1)\n5x +10 = 5(x + 2)\nx^2 – 4 = for a^2 - b^2, the factors are (a + b)(a - b). So: (x + 2)(x -2)\n\n(x - 2)(x - 1) / 8(x-1) * 5(x + 2) / (x + 2)(x -2)\n\nStep 3: Cancel out, multiply\n(x - 2) / 8 * 5 / (x -2)\n5 / 8 (solution!)
310788
You need to factor everything, then cancel common factors.\n4x+8 factors as 4(x+2)\n6x-3x^2 factors as 3(2-x)\n3x^2-4x-4 factors as (3x+2)(x-2)\n9x^2-4 factors as (3x+2)(3x-2)\n\nCancel the (3x+2)'s, and the (x-2) can cancel with the (2-x) leaving a factor of -1.  The factors that are left are:\n[-4(x+2)]/[3(3x-2)].
586333
First, FUNNYPLAYA12'S answer is a partial of what I got, and my formula looked like this:\n\n-8X^3 + 10X^2 - 2X = 0\n-----------------------------------\nFactor out one X and you get:\n\n-8X^2 + 10X - 2 = 0\n\nFactor this and FOIL-check\n\n(-4X + 1)\n(2X - 2)\n___________\n-8X^2 (8X + 2X) -2\n\n-8X^2 +10X - 2\n\nX = {1/4, 1)\n\n\nTo verify:\n\n-8(1) + 10(1) -2(1) = 0\n\n-8 + 10 -2 = 0; and finally:\n\n10 - 10 = 0, which is very true!\n----------------------------------------\n\n-8(1/4) + 10(1/4) -2(1/4) = 0\n\n-2 + 2.5 - 0.5 = 0\n\n-2.5 + 2.5 = 0, which is also very true!\n\n\nOTBPOC
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 :)
770441
take 2(x-2) as common factor\n\n= 2(x-2)((x+3) + (x-3))\n= 2(x-2)(2x)\n=4x^2-8x
764068
2x - 2 = x^2 - 4x + 3\nSet one side equal to 0\n0 = x^2 - 6x + 5\nFactor\n0 = (x - 5)(x - 1)\nSet each factor equal to 0\n0 = x - 5              0 = x - 1\nSolve\n5 = x                  1 = x