How do you factor b^2-6b+10?
611167
You can't factor that equation.
825816
You have a quadratic equation whose factors are (z-2) and (z-6); that is, (z-2) * (z-6) = z^2-8*z+12.\nHow does one get this answer? \nWell you have the z^2 term, which tells you there are two factors that we'll call z+a and z+b; now we have to find a and b. What do we know about a and b? We know that the product of z+a and z+b is z^2 + a*z + b*z + a*b, and this means that a*b=12 and a+b=-8. It's easy to see that a=-2 and b=-6 satisfies these equations, but there is a method that works for all cases (even when the answer is not easy to see) called "completing the square" (see ref.). Read the first example at that web page, and I'll list below the equations to substitute for each equation in the example:\nz^2-8*z+12=0 \nz^2-8*z=-12    (eq. 2)\nz^2+2*b*z+b^2=(z+b)^2 (binomial formula)\nz^2-8*z+16=(z-4)^2 (binomial, completed square where b=-4)\nz^2-8*z+16=-12+16=4 (adding 16 to both sides of eq. 2)\n(z-4)^2=4 (from above 2 eqs)\nz-4=+2 and z-4=-2\nz=6 and z=2\nz^2-8z+12=(z-6)*(z-2)\nI hope this helps.
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
862819
5(2+b) - (a+3b)\n= 10+ 5b - a - 3b\n= 10 +2b -a
1053403
Write it as:\n2x^2-4x+3x-6\n=2x[x-2]+3[x-2]\n=[2x+3][x-2]\nSimple middle-term factorisation!!!
395515
6x+2y=10\nsubract 6x fro both sides\n2y=-2x+10\ndivide both sides by 2\ny=-x+5
913334
x^2 + 2x + 7 Remainder 10\n\ni can add the steps if you don't think i did it by hand
838654
I would factor each quadratic and then see if I could somehow cancel some stuff out.\n\nx² - x - 6 = (x-3)(x+2)\ny² + 3y - 10 = (y+5)(y-2)\ny² - 7y + 10 = (y+5)(y+2)\n3x² + x - 10 = (3x-5)(x+2)\nx² - 2x - 3 = (x-3)(x+1)\n9x² - 25 = (3x+5)(3x-5)\n\nRewrite your problem with these factors and you will see that you can cancel out a lot of these terms!
767438
b1=(-b+ (b^2-4ac)^1/2 )/2a=(-8+(64-48)^1/2)/6=(-8+4)/6\n    =-4/6=-2/3\nb2=(-b- (b^2-4ac)^1/2 )/2a=(-8-4)/6=-12/6=-2
193438
the 6^2 will divide out, and then you have to multiply the 2 with all the exponents on the inside....oh and r^0 is just one....so i think the answer would be t^-10 \nPS i just changed this answer, realized i made some stupid mistkae :-)
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.
902478
wow its been a long time.. i use a calculator to do that now.. but here goes one thing i remember...\n\nlets say we have a polynomial like this\n\nx^2 -3x-10  and we need to factor it..\n\nwe take a look at what factors of -10 we could use that would add up to -3  in this case there are two way to get -10 (-2*5) or (2*-5)  if we add the two sets together one gives us -3 and the other gives us 3 so we can throw out the -2*5 terms.\n\nSo now we lay out the factors.... we need two x's to get x^2 so we need two x's in the first terms (remember FOIL?)\n\nlay the terms out so they give us the correct values\n\n(x+2)(x-5)  now we can use FOIL (firs, outer, inner, last) to check ourselves...\nX*X = x^2\nX*-5 = -5X\n2*x =2x\n2*-5=-10\n\nso we get [x^2] +[-5x] +[2x] +[-10]\nor x^2 -5x+2x-10\ncombine like terms now\n-5x+2x =-3x\n\nwhich gives us\nx^2-3x-10\n\nhope that helps
10610
X^2-9X+18\n\nShort and simple just factor it FOIL
167287
= (lnx + lnx)-lny = 6+6-2 = 10
863683
I suppose you need the second factor of this polynomial. In order to get the answer, we have to factorise the polynomial.\n6a^2 - 5a + 1 = 6a^2 -3a - 2a + 1\n                      = 3a(2a - 1) -1(2a -1)\n                      = (3a -1)(2a - 1)\nThe answer is 2a - 1
791576
I'm going to assume the placement of parenthesis.\n(x^2 – 3x + 2)/(8x - 8) ÷ (x^2 – 4)/(5x +10)\n\nStep 1: Flip the second fraction, so that you have multiplication instead of division.\n(x^2 – 3x + 2)/(8x - 8) * (5x +10)/(x^2 – 4)\n\nStep 2: Factor.\nx^2 – 3x + 2 = (x - 2)(x - 1)\n8x - 8 = 8(x-1)\n5x +10 = 5(x + 2)\nx^2 – 4 = for a^2 - b^2, the factors are (a + b)(a - b). So: (x + 2)(x -2)\n\n(x - 2)(x - 1) / 8(x-1) * 5(x + 2) / (x + 2)(x -2)\n\nStep 3: Cancel out, multiply\n(x - 2) / 8 * 5 / (x -2)\n5 / 8 (solution!)
551906
Add each equation\nA + B = 3x^2 + 2x - 2\nA - B = -x^2 + 4x - 8\n2A = 2x^2 + 6x - 10\nDivide each side by 2\nA = x^2 + 3x - 5\nNow plug into one of the original equations to get B\nB = 3x^2 + 2x - 2 - (x^2 + 3x - 5)\nB = 2x^2 - x + 3