Rewrite the middle term as the sum of two terms and then factor by grouping. x^2 – 14x + 45?
93533
Well, the sum of two terms thing is stupid, but oh well. (You can immediatly make it (x-5)(x-9))\nx^2-9x-5x+45=x(x-9)-5(x-9) --> (x-5)(x-9)
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.
690127
First pull out a common factor in each term.\n\n4y(2n^2 +9n +10)\n\nThen break down the binomial.\n\n4y (2n + 5) (n + 2)\n\nVoila!
503153
(x+7)(x+3)\n(x-11)(x+2)\nforget this one.  its AC factoring though\nthis ones AC too
968026
I got   3(x² - 5x + 4) or 3( x - 4 )( x - 1 )\n\nLet me show you my work...\n\n3( x - 2 )² - 3(x - 2) - 6\n\nSimplify\n\n3( x -2 )( x - 2 ) - 3( x- 2 ) - 6\n\nDistribute\n\n3( x² - 4x + 4) - 3x + 6 - 6\n\nDistribute and Simplify\n\n3x² - 12x + 12 - 3x\n\nSimplify\n\n3x² - 15x + 12\n\nYou can now simplify by factoring out 3\n\n3(x² - 5x + 4)\nor\n3( x - 4 )( x - 1 )
402918
I think you wrote 10x*2yz - 22x*3yz.  To factor this oyu must take out like terms by dividing.  You can see that both terms share a 2, y, z, and an x*2.  After the  division you get  2x*2yz( 5 - 11x).  I hope this helps.
725839
I think it is this:\n\nFirst multiply through the groups..\nso you get:\n6n^7-4n^6-16+4n^6+2n^7+15\n\nThen gather like terms\n\n=8n^7-1\n\n(The -4n^6 and the +4n^6 cancel each other out)\n\nI don't think you can go further than that.\n\nGood luck.
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)
495916
3x^2+9x-84 \nGCF = 3\n3(x^2 + 3x - 28) now we factor the binomial\n\nwe could do this by multiplying the first coefficient (1) by the last coefficient/term (-28). we get -28\nnow we need to select two numbers that multipli to -28 and add to the middle coefficient, 3.\n\nthe two numbers are (7) and (-4)\nwe rewrite the problem with these two terms\n(x^2 + 7x - 4x - 28) i did not write the GCF 3 just yet\nfactor by grouping\n\nx(x + 7) - 4(x + 7)\n\nfactor completely\n\n(x + 7)(x - 4)\n\nfinal answer, don't forget about the 3\n\n3(x + 7)(x - 4)\n\n----------------------------\n4x^2 - 4x - 3   \n1. check for GCF - none\n2. first and last (4)(-3) = -12\n\ntwo numbers that multiply to -12 and add to -4 = 2 & -6\n\n4x^2 + 2x - 6x - 3\n\nfactor by grouping\n\n2x(2x + 1) - 3(2x + 1)\n\n=(2x + 1)(2x - 3)\n\n----------------------------\n3x^2-17x+22 \n\n1. check for GCF - none\n2. first and last   (3)(22) = 66\n3. two numbers are (-6) & (-11)\n4. rewrite\n\n3x^2 - 6x - 11x + 22\n\n5. factor by grouping\n3x(x - 2) - 11(x - 2)\n\n(x - 2)(3x - 11)\n\n----------------------------\nx^2-9 \n\ndifference of squares\n\na^2 - b^2 = (a + b)(a - b)\n\n1. GCF - none\n2. factor\nx^2-9 = (x)^2 - (3)^2 = (x + 3)(x - 3)\n\n---------------------\n\n-3x^2+48?\n\nalso difference of squares\n1. GCF = -3\n\n= -3(x^2 - 16)  \n\n2. factor: x^2 - 16 \nx^2 - 16 = (x)^2 - (4)^2 = (x + 4)(x - 4)\n\n3. final answer = -3(x + 4)(x - 4)
123389
distribute the x to all terms inside the equation\n\n2x^2-26x
685531
[(2x^3+2x^2+3x+3)/(x^2-9)]*\n[(x^2+2x-15)/(4x^2+6)]\n\nFactor each expression individually.\n\n(2x^3+2x^2+3x+3) = (x+1)(2x^2+3)\n\n(x^2-9) = (x+3)(x-3)\n\n(x^2+2x-15) = (x+5)(x-3)\n\n(4x^2+6) = 2(2x^2+3)\n\nSubstituting these in, our expression becomes\n{[(x+1)(2x^2+3)]/[(x+3)(x-3)]}\n*{[(x+5)(x-3)]/[2(2x^2+3)]}\n\nCombining these into one larger expression, we get\n[(x+1)(2x^2+3)(x+5)(x-3)]/\n[2(x+3)(x-3)(2x^2+3)]\n\nCancel the quantities common to the numerator and denominator and we get our final answer\n[(x+1)(x+5)]/[2(x+3)]\n\nGood luck.
947283
You can do this using "synthetic division"\n....... ______________________\nx - 5 ) x^4 - 0x^3 - 3x^2 + 0x + 6\n\nI'm going to simplify this leaving out the x-terms and only showing the coefficients, but just imagine they are there:\n\n.............. 1 . 5 .22 110\n1 -5 ) 1 . 0 .-3 . 0 . 6\n......... 1. -5\n......... ------\n.............. 5 .-3\n.............. 5 -25\n.............. -------\n................... 22. 0\n....... ........... 22 -110\n....... ............----------\n....... ........ ........ 110  6\n....... ........ ........ 110 -550\n....... ........ ........ ------------\n....... ........ ........ ....... 556\n\nSo if I did it correctly the answer should be:\n1 5 22 110 remainder 556\n\nThis equates to:\n(x^3 + 5x^2 + 22x + 110) + 556/(x-5)
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.
32760
You can use the AC Method to answer this question:\n\nFirst, rewrite the expression so it's in the form Ax^2 + Bx + C where A,B, and C are constants (numbers)\n\n16x^2 + 8x - 15\n\nThere's no common factors we could take out first, so...\n\nMultiply A x C, which in this case is 16 x -15 or -240.  Then look for factors of this number that add up to B.  You're looking for two numbers that fill in the blanks below:\n__ x __ = -240\n__ + __ = 8\n\nPrime facorization will help give you some ideas, or make an ordered list of numbers that multiply to -240.  \n240 and -1\n120 and -2\n80 and -3\n60 and -4\n48 and -5\n40 and -6\n30 and -8\n24 and -10\n20 and -12\n\nSee how the last pair of numbers satisfies the conditions?\n\nRewrite your polynomial, substituting the 2 values in for B to have 4 terms:\n\n16x^2 + 20x - 12x -15\n\nThen factor by grouping\n(16x^2 - 12x) + (20x - 15)\n4x (4x - 3) + 5 (4x - 3)\n(4x + 5)(4x - 3)\n\nThis method works for most polynomials of the form Ax^2 + By + C!  \n\nYou can check your answer of (4x + 5)(4x - 3) by using the FOIL method and see that it does work.\n\nGood luck!
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
665603
you put (x+2) in for x in the equation first\n\n((x+2)^3)+(2(x+2)^2)-4\n\nthen simplify \n\nx^3 + 8x^2 + 20x + 12\n\nand multiply the entire thing by 3\n\n3(x^3 + 8x^2 + 20x + 12)\n\n3x^3 + 24x^2 + 60x + 36
888714
If you are multiplying both terms it's 49x^14 if you are dividing it's x^6