Factoring - Can someone show me? - Extra credit work?
962288
When you have a quadratic equation Nx² + Mx + P, you can sometimes factor it into (ax+r)(bx+s).  Here's a way to determine a, b, r and s:\n\nWrite down N and M and P, then look at what two numbers a & b when multiplied will equal N, and what two numbers r and s when multiplied will equal P and at the same time when all multiplied together (as + rb) will equal M.  It sounds scarier than it is!  Here's an example with your first equation:\n\nb² + 4b - 12\n\nSo we have our coefficients (N,M,P):  1  4  -12\n\nFor 1 = ab our only choices are a=1, b=1\n\nFor -12 = rs, we have several choices 1,12 ... 2,6 ... 3,4  with one factor being positive and one negative so we can get -12.\n\nWhich two factors have a difference of 4?  Well it's 2 and 6.  And since 4 is positive we want +6 and -2.   So putting it all together we get:\n\nb² + 4b - 12 = (b-2)(b+4)\n\nLet's try one more (and then I have to go to a meeting so I can't help you anymore right now, sorry!) ...\n\n2b² - 13b +15\nOur coefficients are:  2  -13  +15\nFactors of 2:  1,2\nFactors of 15:  1,15 ... 3,5\nHow can we combine factors of 2 and factors of 15 to get -13?\nNotice that -15+2 = -13, so our factors are:\n(2b-15)(b+1) = 2b² - 13b +15\n\nHope that helped a little.  Good luck!\n\n~ ♥ ~
386814
They are correct.  Observe the order of operations or PEMDAS\n\nParenthesis\nExponents\nMultiplication-Division\nAddition-Subtraction\n\nso first multiply   -89 * 658 =   -58562\nThen add 2   =                    -58560\nThen add 6   =                    -58554\nThen subtract 89 (or add -89) =   -58643
508408
It's called factoring by grouping. Take a factor out of two of the four terms (and the same for the other two) so they share a common factor.\n(2x^3 - x^2) - (18x - 9)\nx^2(2x - 1) - 9(2x - 1)\n(x^2 - 9)*(2x - 1)\n(x + 3)*(x - 3)*(2x - 1)\n(ungkog WAS correct)
945604
1.adding-2f-7\n-6f-2f+7-7<2f-2f-9-7\n-8f<-16\ndividing by -8\nx>2\n\n2.5/8g-4>2/3g+6\nadding 4\n5/8g>2/3g+10\nmultiplying by 24\n15g>16g+240\nadding -16g\n-g>240\ndividing by -1\ng<-240\n\n3.sum incomplete\n\n4.-2 1/9d+3/5<-3 2/9d-4 2/5\n-19/9d+3/5<-29/9d-22/5\nmultiplying by45\n-95d+27<-145d-198\nadding 145d\n50d+27<-198\nadding -27\n50d<-225\ndividng by 50\nd<-5 1/2
833702
Do you mean (2x^2 + 3x + 1/ 2x^2 + x - 3) for your first fraction?  I'm going to assume you do.\n\nFirst, factor each numerator and denominator:\n\n2x^2 + 3x + 1 = (2x+1)(x+1)\n\n2x^2 + x - 3 = (2x-3)(x+1)\n\n4x - 4 = 2(2x-2)\n\nNow let's look out our multiplied fraction, rewritten using our factors we just found:\n\n[ (2x+1)(x+1)2(2x-2) ] / [ (2x-3)(x+1)(2x+1) ]\n\nNotice we can cancel out the (2x+1)'s and the (x+1)'s, to get:\n\n2(2x-2)/2x-3\n\nANSWER:   2(2x-2)/2x-3
703982
1.\n3x + 6/((x + 1)(x + 2)) Factor 3 out of the top.\n[3(x + 2)]/[(x + 1)(x + 2)] x+2's cancel\n3/(x + 1) <-- This is simplified.\n\nI don't have the energy to do the other ones right now.  :/
10610
X^2-9X+18\n\nShort and simple just factor it FOIL
610613
In 6ab + 4a + 3b + 2, we group the first 2 terms and the last 2 terms:\n\n6ab + 4a + 3b + 2 =\n(6ab + 4a) + (3b + 2)\nthen factor out the common factor 2a in (6ab + 4a)\n(Note: to factor out is the same as dividing each term with the common factor)\n\nwe get:\n2a(3b + 2) + (3b + 2)\nNotice that in each term of the above we have a common factor of (3b + 2)\nBy dividing each term by (3b + 2) or factoring out (3b + 2) we have:\n(3b + 2)(2a + 1)
81890
it's not factored completely if you leave it as a^2-25 so the proper answer is 5(a-5)(a+5)
1123580
factor the numerator, factor the denomerator\n\n= (3+b)(3-b) / (b+2)(b-3) \n\nrewrite (3-b) as -(b-3) and cancel (b-3) from top and bottom:\n\n= -(3+b)(b-3) / (b+2)(b-3) = -(3+b)/(b+2)\n\nFor the second problem, use FOIL method (first, outside, inside, last).\n\n(5x+4)(3x+1) = (5x)(3x) + (5x)(1) + (4)(3x) + (4)(1) \n\n= 15x^2 + 5x + 12x + 4\n= 15x^2 + 17x + 4\n\nTo factor this from the beginning is different.\nassume two factors,  (ax + c)(bx + d)\nthen ab = 15, cd = 4, and ad + cb = 17\nthrough trial and error, (using 5 and 3 for a,b is a natural place to start), only one combination works.  a=5, b=3, c=4, d=1
84658
i got it!\n\nit's 3(2c+7)(c-4d)(c+d)!!!!\n\n6c^3 + 21c^2 - 18dc^2 - 63cd - 24cd^2 - 84d^2\nfirst we take out factor 3:\n3(2c^3 + 7c^2 - 6dc^2 - 21cd - 8cd^2 - 28d^2)\nthen we take out (2c+7) from each pain inside the parenthesses:\n3[(2c+7)c^2 - (2c+7)3cd - (2c+7)4d^2)]\nwe take (2c+7) out:\n3(2c+7)[c^2 - 3cd - 4d^2]\nwe factor [c^2 - 3cd - 4d^2] trinomially:\nc^2 - 3cd - 4d^2 = (c-4d)(c+d)\nthen we get:\n3(2c+7)(c-4d)(c+d)\n\nam i right??\n\n:) yeah i didn't believe myself...\nbut if u really write down all your factors, it's much more easier to take them out. and it WAS fun :))
548405
please dont hate math\nshow interest\nthen you do well\n\nFOIL\nFirst-Out-Inner-Last\n\nThis method is used for multiplying two binomials.\nsay (a+b)(x+y)\n\n( a + b ) (x + y)\n\nFirst------first terms in the two binomials. they are to be multiplied. so ax\nthen Outer a and y. so ay\nInner-----bx\nLast-----last terms in the binomials. so 'by'\nthen add all the terms.\nax+ay+bx+by\n\nyou may consult me
102673
If you can't get an answer here, try asking your Algebra teacher for extra help after school tomorrow.  It's sure to add to your grade that you asked for the help.
76424
You can reduce you work by knowing that if your polynomial looks like  Ax^n + ... + B that if you have a factor that looks like (ax + b) that a has to be a divisor of A and b has to be divisor of B. That means that the only possibilitys in your example are\n(x + 1), (x - 1), (-x + 1) and (-x -1)\nSince (-x - 1) = -1(x+1) and (-x + 1) = -(x - 1)\nyou only have to look at two cases (x+1) and (X-1)\n\nTo find out if either of the two is a factor, you can do long division on the polynomial just like you would with regular numbers. If you don't know how to do that look at the link below for an example.\nI can't show you here because it won't format correctly.
600095
You can factor out 15bc^2\n15bc^2(5b + 4c^4)
149556
4x^3+36x^2+32x----using common factor which in this case would be 4x so, 4x(x^2+9x+8) and then u use guessing method \nand the complete factorization would be 4x(x+8)(x+1).
526107
You are only allowed to factor if the equation is equal to zero; your's is equal to 4.  You need to solve "x^3 - 6x^2 +9x - 4 = 0"  using synthetic division.