How do you factor this Binomial?
816170
You have done some mistake while copying the problem.Please check it up.It may be 36l^2-36lq+9q^2 or something else.However I am doing ipresuming it to be 36l^2-36lq+9q^2\n\ngiven exp.=9(4l^2-4lq+q^2)\n=9{2l)^2-2.2l.q+(q)^2}\n=9(2l-q)^2\nThe answer may also be written as {3(2l-q)}^2 or (6l-3q)^2
796585
1.factor\n(2x+1)(x-4)=0\n2.set it to 0\n2x+1=0\nor\nx-4=0\n3.solve\n{x=-1/2 or x=4} <=answer
767628
I'm assuming you want to factor this\n\n5z^2 + 4z -12 = (5z -6)(z+2)
1119411
1. Add 3 to both sides to set equal to zero, then factor to get something that adds to three and multiplies to 28---7 and -4\n(x-4)(x+7) = 0\nx=4, 7\n\n2. Add 4 to both sides to set equal to zero, then factor to get something that adds to 2 and multiplies to -15---5 and -3\n(x+5)(x-3) = 0\nx = -5, 3
667985
You can factor this by grouping:\n\nx^3 + 2x^2 - 9x = 18 \nx^3 + 2x^2 - 9x -18 = 0\nx^2(x + 2) + (-9)(x + 2) = 0\n(x^2 -9)(x + 2) = 0\n(x - 3)(x + 3)(x + 2) = 0\n\nx = { 3, -3, -2}\n\nFirst subtract 18 from both sides, then facter each binomial, and finally use the difference of squares.
768805
Factor out the 3:\n3(x^2+3x+2)\nUnfoil:\n3(x+1)(x+2)\nTake the roots:\nx=-1 or -2
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.
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.
893470
x^2-4x-5 = (x+1)(x-5) if that helps any.  I just factored the trinomial into two binomials.  A graphing calculator would help you solve this in no time.  I have my own homework to do, so I'm not gonna do this for you, but I hope that that will get you started.  Good luck.
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
159156
factoring, then:\n\n(2x-3)(x+1)=0\n\nyou can assume that\n\n2x-3 = 0, and\n\nx+1 = 0\n\nSolve for first x,\n\n2x=3 , x=3/2\n\nsecond x,\nx= -1\n\nThe sum would be 3/2 - 1, or 3/2 - 2/2 =1/2
1054627
assuming this is (3a+1)/(a^2-1) - 1/(a+1)\n\nfactorise the (a^2-1) to give (a+1)(a-1)\n\nmultiply the second fraction by (a-1)\n\n-->  (3a+1)/[(a+1)(a-1)] - 1(a-1)/[(a+1)(a-1)]\n\nfactorising this gives [(3a+1)-(a-1)]/[(a+1)(a-1)]\n\nwhich equals (2a+2)/[(a+1)(a-1)]\n\n= [2(a+1)]/[(a+1)(a-1)] \n\nthe (a+1)'s cancel out...\n\ntherefore = 2/(a-1)
614345
(x+10)(x-10)\n(x-4)(x²+4x+16)\n(x+w)(x²-wx+w²)\n\nUnfortunately you just have to memorize the formulas.  There is no other way to figure out the factoring of the cubic expressions.
757625
1) \n\n1+2+3+4+5+....n +(n-1)+(n-2)+...+3+2+1...regrouping\n\n=2(1+2+3+4+5+....(n-1)) +n...regrouping the term in parenthesis\n\n=2( (1+n-1)+(2+n-2)+(3+n-3)...)+n\n=2(n+n+n+...(n-1) terms)+n\n=2(n)(n-1)/2 +n\n=n^2-n+n\n=n^2\n\n\n2)\n 1=10^0 and 11=(10^1+10^0) and 111=(10^2+10^1+10^0)...\n\n(10^n+10^(n-1)+....10^2+10^1+10^0) times itself\n=(10^0+10^2+10^3+....10^n)^2\n=10^2n+2*10^(2n-1)+.....\n\n\n3)\n\nn^2-n=n(n-1)...one of them must be even and is therefore divisible by two.\n\nn^3-n=n(n^2-1)=n(n+1)(n-1)=(n-1)(n)(n+1) or three consecutive integers.\n\nIf the product is three consecutive integers, then one of them must be divisible by three and therefore the product is divisible by three.
898442
You just factor\n(2x-1)(x-2)=0\nx=2, 1/2
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.
1136055
Here's how we can solve the problem logically, without doing all the multiplication.\n\nFirst of all, the coefficients aren't relevant to the problem; we don't care what the terms are specifically, just how many of them there are.\n\nLet's create two new temporary variables: let a = x + 5y, and b = 3z + w.  That reduces the problem, for the moment, to (a + b)^6.\n\nNow, any time you take a binomial to a power -- such as (a+b)^n -- there will be n+1 terms in the final expansion: (a+b)^2 has 3 terms, (a+b)^3 has 4 terms, etc.\n\nSo in (a+b)^6 there will be seven terms, which will have various coefficients but generally be of the forms a^6, a^5b, a^4b^2, a^3b^3, a^2b^2, ab^5, and b^6.  Note that, in each term, the sum of the powers is 6.\n\nNow, if we reinstate the original variables, substituting back in for a and b, we can consider each of those seven terms.\n\na^6 becomes (x+5y)^6, which will have 7 terms.\na^5b becomes (x+5y)^5 (3z+w), which will have 6 * 2 = 12 terms.\na^4b^2 becomes (x+5y)^4 (3z+w)^2, which will have 5 *3 = 15 terms.\na^3b^3 becomes (x+5y)^3 (3z+w)^3, which will have 4 * 4 = 16 terms.\na^2b^4 becomes (x+5y)^2 (3z+w)^4, which will have 3 *5 = 15 terms.\nab^5 becomes (x+5y) (3z+w)^5, which will have 2 *6 = 12 terms.\nAnd, finally, b^6 will have 7 terms.\n\n7+12+15+16+15+12+7 = 84.\n\nHope that made sense.