what is the factored form of this trinomial? y^2-11y+30?
439650
(y-6)(y-5)\n\n-6 and -5 add up to -11\n-6 and -5 multiply to 30
1068964
simplify expression: -2(y+3)+6(y+1)\n\nsimply distribute then add.
960995
I'm going to assume that you "forgot" to include parentheses.\n3/(y^2 -3y +2) + 7/(y^2 -1)\nFactor each denominator.\ny^2 - 3y + 2 = (y-2)(y-1)\ny^2 - 1 = (y-1)(y+1)\nYour common denominator (CD) will be (y-2)(y-1)(y+1)\nThe first term is missing the (y+1) part, so you need to multiply the numerator and denominator by (y+1)\n[3/(y^2 -3y +2)] (y+1)/(y+1) = 3(y+1)/[(y-2)(y-1)(y+1)]\n= (3y +3)/[(y-2)(y-1)(y+1)]\nThe second term is missing the (y-2) part.\n7/(y^2 -1) = 7(y-2)/[(y-2)(y-1)(y+1)]\n= (7y-14)/[(y-2)(y-1)(y+1)]\nWe now have a common denominator, so we can add the numerators.\n(3y +3)/[(y-2)(y-1)(y+1)] + (7y-14)/[(y-2)(y-1)(y+1)]\n= (3y + 3 + 7y - 14)/[(y-2)(y-1)(y+1)]\n= (10y -11)/[(y-2)(y-1)(y+1)]
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)
904579
ok sorry, i'll work it out this time\n\n3x-6y=-30\n-8x+20y=104\n\n(8)(3x-6y)=-30(8)\n(3)(-8x+20y=104(3)\n\n24x-48y=-240\n-24x+60y=312\n\n12y=72\n\ny=6\n\n3x-6(6)=-30\n\n3x-36=-30\n\n3x=6\n\nx=2\n\ndoes that help?
643397
(8y^3 + 27)/(2xy - 10y + 3x - 15)\n((2y + 3)(4y^2 - 6y + 9))/((2xy - 10y) + (3x - 15))\n((2y + 3)(4y^2 - 6y + 9))/(2y(x - 5) + 3(x - 5))\n((2y + 3)(4y^2 - 6y + 9))/((2y + 3)(x - 5))\nthe (2y + 3)s cancel out\n(4y^2 - 6y + 9)/(x - 5)\n\nANS : (4y^2 - 6y + 9)/(x - 5)
517819
start pulling out common factors\n\n2 (x^3 + 27)\n\n2(x^3 + 3^3)
1002101
factor 9x^2y^2-9y^2\n9y^2(x^2-1)\n9y^2(x+1)(x-1)
683580
y^2 + 7y + 3y + 21\ny^2 + (7 + 3)y + 21\ny^2 + 10y + 21\n(y + 7)*(y + 3)
503348
No, the previous answer is wrong.\n\nGroup it like this:\n\n2p + pq -6q -3q^2\n\nthen factor like this:\np(2+q) -3q(2+q)\n\nfactor out the 2+q to get:\n(2+q)(p-3q)\n\nFOIL it out and you will see it is correct.
948364
Well, this is a standard square trinomial, which you can tell because it is written in the form a^2x^2 + 2abx + b^2, the square root of which is ax + b.  In this case, that means that your new equation is:\n(4y - 1)^2 = 0\nTake the square root of both sides to get:\n4y - 1 = 0\n4y = 1\ny = 1/4\n\nOne rational root
944317
Im guessing your solving for y... so make the problem 58=29Y.  Divide both sides by 29 and you get y=2
444843
It's a perfect square trinomial.  Take the square root of the first term and the square root of the last.  If you multiply them and then double it, and it equals the middle term, then it's a perfect square trinomial.\n\nIn this case, it factors into (x/3 + 6)^2.  Using FOIL, you'll see that it expands into your original expression.
827502
1.6a^2b^2(1+3ab)\n\n2.(y-4)(y-1)\n\n3.(2b-1)(b+7)\n\n4.3(x^2-9y^2)\n=3(x+3y)(x-3y)
706243
-2y+3=7y-42\n-2y+3-3=7y-42-3\n-2y=7y-45\n-2y-7y=7y-7y-45\n-9y=-45\n-9y/-9=-45/-9\ny=5
783104
(y-11) -(y+6)=6y\ny - 11 - y - 6 = 6y\ny - y - 6 - 11 = 6y\n0 - 17 = 6y\n6y = -17\ny = -17/6\nThe answer is D) -17 over 6
166772
2x - 3y = 5        \n2x^2 + 3y^2 = 35 \n\n2x = 5 + 3y\nx = (5 + 3y)/2\n\n\n2 [(5 + 3y)/2]^2 = 5 + 3y\n2 [(25+30y+9y^2)/4] = 5 + 3y\n25+30y+9y^2 = 10+6y\n9y^2-24y+15=0\n(3y-3)(3y-5)=0\n\ny=1; y=5/3\n\nif y=1\n2x - 3(1) = 5\nx = 4\n\nif y = 5/3\n2x - 3(5/3) = 5\nx = 5