Help with factoring an equation with variables and exponents...?
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.
735282
Seems like you are missing some parentheses and an exponent. Please check to make sure yiu have copied the problem correctly and resubmit.
1085664
Good job!  You're halfway there!\n\nAs you are writing it out on your homework, underline the (4x+1) in each of the two terms.  That helps you see what you are factoring.\n\nFactoring out the (4x+1) from each term, you should get your final answer.\n\n=(5x + 4y)(4x+1)
800980
1. Move the constant to the right side of the equation:\n4x^2 + 2x = 1\n\n2. Divide out the x^2 coefficient:\nx^2 + 1/2x = 1/4\n\n3. Divide the coefficient of x (in this case, 1/16) by 2 and square it, then add it to both sides:\nx^2 + 1/2x + 1/16 = 5/16\n\n4. Factor the left side of the equation to (x + a)^2, where a is the number you just squared:\n(x + 1/4)^2 = 5/16\n\n5. Take the square root of both sides, remembering that this will produce both a positive and a negative root:\nx + 1/4 = +/-sqrt(5/16)\n\n6. Move the constant on the left to the right and simplify:\nx = -1/4 +/-sqrt(5/16) = -1/4 +/-(sqrt 5)/4 = (-1 +/- sqrt 5)/4
10610
X^2-9X+18\n\nShort and simple just factor it FOIL
636732
8x^2y^3+5y^2 when "FACTORED COMPLETELY"=\n\ny^2 (8yx^2+5). \n\nThis should be right. My dad is a complete genius and he did this in like 2 seconds. I double checked it on my calculator too by plugging in numbers, so I can assure you he's right.\n\nHope this helps! Have a nice day :)
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!
736710
Please show your exponents using the ^ operator: It will make these problems much clearer. Also, I would have put this in 'Mathematics' rather than Chemistry.\n\nFirst equation:\n(6s^5)(2s^2) \n\nGroup your integer coefficients:\n6*2 = 12\n\nGroup your s terms:\ns^5 * s^2 = s^7 (multiplying like bases, just add the exponents)\n\nSo you have 12s^7 for the first part.\n\nSecond equation:\n(3s^4)(4s^3)\n\nMultiply integers:\n3 * 4 = 12\n\nMultiply s terms (add exponents):\ns^4 * s^3 = s^7\n\nThis is also 12s^7.\n\nSubtracting:\n12s^7 - 12s^7 = 0\n\nD) None of the above
1135408
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.
835073
okay well\n\n1) combine two to figure out the first variable\nx-2y+z=4\n-x+2y+3z=-4\nso 4z=0; z=0\nthen figure out the next variable with a different set of equations\nlike and multiply so the ys will cancel [take Zs out cause zero]\nx-2y+z=4\n2(2x+y)=6\n\ntherefore\nx-2y=4\n4x+2y=6\n5x=10\nx=2\n\nthen substitute for y\n\n-x+2y=-4\n-2+2y=-4\n2y=-2\ny=-1\n\nso y=-1, x=2, z=0\n\n\ncontinue the process with the rest.
1177677
shouldn't it be x^2??? if it IS x-squared, then sure, you could use the factorization method. \n\nit goes like this:  (6x + 1) (x - 1) = 0\n\ntherefore, x = -1/6 , 1
576570
algebra...usually (well i do) set up a equation for everything, and solve for the variable. If u need help u can always get help in www.algebra.com .....
589722
Step 1: Solve one equation for one variable in terms of the other.\nStep 2: Plug that into the other equation and solve for the second variable.\nStep 3: Plug the second variable into the first equation and solve for the first variable.
208169
Factoring, I guess, but it really depends on the type of polynomial question. Can you give a sample?
600427
It is already factored, you can write it more concisely as (x-y)^2.\n\nThe EXPANDED form is x^2-2xy+y^2.
669578
I have corrected the expression. Well here I redo the factorization\n\nThe given expresion \n= 12x^3+60x^2y+75xy^2\n= 3x(4x^2+20xy+25y^2)\n= 3x[(2x)^2+2(2x)(5y)+(5y)^2]\n= 3x(2x+5y)^2
824178
You typed exponents just fine!\n\nx^3 - x^2 -x + 7 divided by x+1 is a "long division" type problem rather than factoring because x+1 doesn't go evenly into the polynomial\n\nThe answer is x^2 -2x + 1 + 6/x+1