what is the factored form of this trinomial? y^2-11y+30?
460061
y^2-11y+30\nThe discriminant (b^2 - 4ac in ax^2 + bx + c) is a perfect square, 1. So, it could be factored.\n(y - 5) (y - 6)\nThe first two y's make up 1y^2.\n(-6y) + (-5y) = -11y\n-5 x -6 = 30.\n\nSo, the factored form is (y-5)(y-6) = 0
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.
235893
(15y^3 + 12y^2)+ (5y+4)\nGCF 3y^2(5y+4)+1(5y+4)\n\n(5y+4)(3y^2+1)
287717
-4y^+8y+3y^2-6\n-4y(y^2-2)+3(y^2-2)\n(-4y+3)(y^2-2)\n\ndoes that make sense?
439650
(y-6)(y-5)\n\n-6 and -5 add up to -11\n-6 and -5 multiply to 30
807151
you can factor by using this:\n(y + 4)( y + 4)=0\n\nnow you have\ny + 4 = 0\n\nsubtract 4 from each side:\ny + 4 = 0 \n   - 4    -4\nyou'll end up with:\ny = -4\n\nPS please award best answer
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
517819
start pulling out common factors\n\n2 (x^3 + 27)\n\n2(x^3 + 3^3)
586962
(6y^2-3y)(y+7)-->(2y)(y-1)(y+7)\nif that helps
175207
Next time use ^2 instead of typing out "to the 2". Anyway, write it in standard form, so that the variables are in descending order.\n\n14w^3 + 49w^2 - 8w - 28\n\nYou can't factor this. If it was 7w instead of 8w, then you could factor out a 7 by dividing all the coefficients by 7 and pulling it outside the parentheses. Alternatively, you could factor the first two terms separately from the second two:\n\n7w^2(2w + 7) - 4(2w + 7)\n\nNow it's in a great form for factoring. Group them.\n\n(2w+7)(7w^2-4)\n\n---\n\nDo the same thing for the second example.\n\n44t^3 - 4t^2 + 66t - 6\n\nFor this one, you can factor out a 2.\n\n2(22t^3 - 2t^2 + 33t - 3)\n\nOr, again, you can factor the terms separately:\n\n4t^2(11t - 1) + 6(11t - 1)\n\nGroup.\n\n(11t-1)(4t^2+6)\n\n\n\nHope this helped.
627033
6y + 3 = -4y^2\n4y^2 + 6y + 3 = 0\n\na=4, b=6, c=3\n\ny = -b+-SQRT(b2-4ac) / 2a\ny = -6+-SQRT[36-(4)(4)(3)] / [(2)(4)]\ny = -6+-SQRT[-12] / 8\n\nBut SQRT[-12] is not possible, unless we are looking for imaginary values, then:\n\ny = -6/8 +- SQRT[-12]/8\ny = -3/4 +- [2SQRT3]i/ 8\ny = -3/4 + [(SQRT3)/4]i or -3/4 - [(SQRT3)/4]i
815517
OK First you take out the common factor which is y. After you have done that.you have (y^2-12y+36). After this, it becomes simple factoring. The answer is\ny(y-6)(y-6)\nor also known as y(y-6)^2
387063
X=[-(30y-433)]/1232\nY=[-(1232x-433)]/30
865479
First of all is the radical in the numerator over just the 10 or over the entire 10y^3?\n\nThe only this you can factor out and cancel is a radical 2 on the top and bottom.  Other than that, you can't really simplify this problem.  Are you sure you wrote it down correctly?
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
683580
y^2 + 7y + 3y + 21\ny^2 + (7 + 3)y + 21\ny^2 + 10y + 21\n(y + 7)*(y + 3)
1148686
(2y-8)(y-4)\n\n2y^2-8y-8y+32\n\n2y^2-8y^2+32