how do you solve 16^2=49 using factoring?
117122
If you mean 16x^2=49 then here is the answer. Subtract 49 from both sides. 16x^2-49=0. This is the difference between squares so it factors to (4x+7)(4x-7)=0. Set each factor equal to 0. \n            4x+7=0\n            4x-7=0\n Then solve. The answers are x=-7/4 and x=7/4
815411
Write the ratios as fractions:\n1 ... x\n-- = --\nx .. 64\n\nCross multiply:\nx² = 64\n\nTake the square root of both sides:\nx = ±8\n\nTechnically the answer is ±8, but given your answers, the correct answer is:\n\nB) x = 8
794955
It's simple arithmetic.  Let's break it up into different parts:\n\n7^2=49\nsq(16)=4, since 4^2=16\n\n49*4=196\n196/11 is your answer since you cannot simplify this any more.  Eleven is prime, and 196 is not divisible by 11.\n\n\nHope this helps.
528350
(5x + 4)(6x + 4)\n30x^2 + 20x + 24x + 16\n30x^2 + 44x + 16\nANS : 16
837436
Your Algebra teacher is trying to kick your butt, right?  I have the answer:\n\nReorder the terms:\n3+64x+16x^2=0\n\nStart with the 16 - divide all numbers by 16\n\n3/16+64x/16+16x^2/16=0/16\n\nResults are:\n\n.1875+4x+ x^2=0\n\nNot done yet...Next, take out the .1875.\n\n.1875+(-.1875)+4x+x^2=0+.1875\n\nResults from this calculation:\n\n4x+x^2=-.1875\n\nThen, add a 4 into the equation\n\n4+4x+x^2=-.1875+4\n\nAdd like terms:\n\n4+4x+x^2=3.8125\n\nThen, you need to factor the problem out.\n\n(x+2)(x+2)=3.8125\n\nTake the square root of 3.8125\n\n=1.952562419\n\nAfter that, take one of the factors you created above and use the square root to arrive at the answer...one positive, one negative\n\n2 + x = 1.952562419\n\nx = 1.952562419 - 2\n\nx = -.0474437581\n\nand\n\n2 + x = -1.952562419\n\nx = -1.952562419 - 2\n\nx = -3.952562419\n\nx = (-.047437581, -3.952562419)\n\nYou might have to plot that on a graph.  You have just calculated a point of off the x axes...the horizontal line.\n\nHope that helps. Maybe someone can give you fewer steps.
812111
x^2 = 96\n\nx= +/- sqrt(96) = +/- 4 sqrt(6)
980031
Use your calculator to note the last two digits of the powers of 16 until you see them start to repeat:\n\n16^1 = 16\n16^2 = x56 \n16^3 = x96\n16^4 = x36\n16^5 = x76\n16^6 = x16\n\nAha - note the last two digits for 16^6 are the same as for 16^1.  So, the last two digits of 16^7 will be the same as for 16^2, etc.\n\nTo solve your problem now, the last two digit of 16^198 will be the same as for 16^3 - see why?  :)
890098
I enjoy algebra, and this is easily my favorite type of problem.  So I hope you, too, one day enjoy them for their inner beauty.\n\nFor now, let's tackle this "factoring problem."  We may wish to use the method of "synthetic division" which is arguably easier than "long division of polynomials."\n\nThink back to grade school where you learned that 6 = 2 * 3.  This simple statement enabled you to 'deconstruct' the number 6 as either "two times three" or "three times two".  Even if you forgot one of the 'factors' you could conjure it up again quickly by asking yourself, "What is six divided by two?" for example.\n\nIn problem (a), you are essentially asking, "What are the coefficients of P(x) if the only factor we know is (x-1)?"\n\nLet's set up synthetic division (see link below), by placing all the coefficients of P(x), and using 1 (the zero implied by (x-1)) as the divisor:\n\n1 ||  a  b  c  d\n___________\n.....a\n\n1 ||  a  b  c  d\n_____a_____\n.....a (a+b)\n\n1 ||  a  b        c       d\n_____a___(a+b)____\n.....a (a+b) (a+b+c)\n\n1 ||  a  b        c                d\n_____a___(a+b)___(a+b+c)_\n.....a (a+b) (a+b+c) (a+b+c+d)\n\nNote that this last term must equal 0 since it is assumed that (x-1) is a factor ... thus, the "remainder" term must equal zero!\n\nThere's the result you were after:  a+b+c+d = 0\n\nHope this helps you tackle the others!
563831
4! + 4! + 4/4 =49\n\n4!= factorial(4) =24
986254
(z^4)^2 - 14z^4 +16 =0\nQuadratic factoring:\nz^4 = (14+sqrt(196-64))/2 = 7+sqrt(33) =\n12.7446 so z = 1.89 or -1.89\nOR\nz^4 = 7-sqrt(33)= 1.2554\nso z= 1.059 or -1.059
910180
3ab(2ab^2)^4\n3ab(16a^4b^8)\n48a^5b^9\ntrust me!!!! i just got done doing this in class a week ago.
734091
This particular quadratic equation will require the quadratic formula:\n\n-b +/- sqrt(b^2-4ac) all divided by 2a.\n\nHere, a=1, b=18, and c=49\n\n-18 +/- sqrt[(18^2 - 4(1)(49)] all divided by 2.\n\n18^2 - 4 x 49 = 128.....\n\nLet's do the top first:\n\n-18 +/- sqrt(128)\n\nWhat is sqrt(128)? You need to factor 128 = 64 x 2\nTherefore, sqrt(128) = sqrt(64 x 2) = 8 x sqrt(2).\n(we took 64 out of the root function, so comes out as the sqrt, which is 8.)\n\nSet it all up:\n\n-18 +/- 8sqrt(2) divided by 2:\n\n-9 +/- 4sqrt(2)\n\nAnswer: -9 - 4sqrt(2), -9 + 4sqrt(2).\n\nNot easy due to the answer not a set of integers.
954300
ok\nso its factoring.this is hella easy.10x+40 is 10(x+4)\nso the answer is 10(x+4)there you go
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
902011
Ha! You had an answer to that 10 min before you even asked :). Check your previous question.\nBy factoring x out we have \nx(12x^2 + 3x -42)=0\n\nx1=0 \nsee reference and use equations provided\nx2=14/8=7/4\nx3=-16/8=-2
897152
let x = smaller number\n2x + 8 = larger number\n2(2x + 8) + 3x = 65\n4x + 16 + 3x = 65\n7x + 16 - 16 = 65 - 16\n7x = 49\n7x/7 = 49/7\nx =7
786976
16x^2 + 12x\n\nIn order to solve, you would need an equal sign.  I think maybe you mean finding the zeros?\n\nFirst we would need to factor it.  We could pull an x and a 4 out of each.\n\n4x(4x + 3)\n\nWe get two factors.\n4x\n4x + 3\n\nThen we set each factor equal to 0 to find the zeros of the equation.\n\n4x = 0\nx = 0\n\n4x + 3 = 0\n4x = -3\nx = -3/4\n\nOur zeros are 0 and -3/4.\n\nI hope I helped. :)