What is the factored form of 8x^2+6x+8?
148359
No true factored form.\n\na=8 b=6 c=8\nx= (-b +- squareroot[b^2-4ac])/2a\nx= (-6 +- squareroot[36-256])/16\nx= (-6 +- squareroot[-220])/16\nx= (-6 +- 2i{squareroot[55]})/16\nx= (-3 +- i{squareroot[55]})/8
32760
You can use the AC Method to answer this question:\n\nFirst, rewrite the expression so it's in the form Ax^2 + Bx + C where A,B, and C are constants (numbers)\n\n16x^2 + 8x - 15\n\nThere's no common factors we could take out first, so...\n\nMultiply A x C, which in this case is 16 x -15 or -240.  Then look for factors of this number that add up to B.  You're looking for two numbers that fill in the blanks below:\n__ x __ = -240\n__ + __ = 8\n\nPrime facorization will help give you some ideas, or make an ordered list of numbers that multiply to -240.  \n240 and -1\n120 and -2\n80 and -3\n60 and -4\n48 and -5\n40 and -6\n30 and -8\n24 and -10\n20 and -12\n\nSee how the last pair of numbers satisfies the conditions?\n\nRewrite your polynomial, substituting the 2 values in for B to have 4 terms:\n\n16x^2 + 20x - 12x -15\n\nThen factor by grouping\n(16x^2 - 12x) + (20x - 15)\n4x (4x - 3) + 5 (4x - 3)\n(4x + 5)(4x - 3)\n\nThis method works for most polynomials of the form Ax^2 + By + C!  \n\nYou can check your answer of (4x + 5)(4x - 3) by using the FOIL method and see that it does work.\n\nGood luck!
863644
take out 4x first\n4x(x2-6x+8)\n4x(x-4)(x-2)
976886
It is a difference of two squares, in the form a^2-b^2\n\n\nTake the square root of both terms, which will be 3k and 5b.  Then write them in the form\n\n(a-b)(a+b)\n\nYour factored form will be \n\n(3k+5b)(3k-5b)
293335
8+9+6+(1+2+3+4+5+7)^0 = 24\n\nX^(0) = 1
504285
It has been a long time, here are suggestions:\n1. 5x^3 + 11x^2 + 3x + 1  \n2. 4x^3 - 13x^2 - 8x + 7 \n3. 2x^4 - x^3\n4. 21x^2 + 26 x + 8
751945
For this question, you need to use the factorization of a^3 - b^3.\n\na^3 - b^3 = (a - b)(a^2 + ab + b^2)\n\nFor your question, let a = (8x^3 + 5x^2)^(1/3) and b= 2x. Then the expression you are given is the a - b term. You want to multiply by (a^2 + ab + b^2).\n\n(8x^3 + 5x^2)^(1/3) - 2x = [((8x +5x^2)^(1/3) - 2x)((8x^3 + 5x^2)^(2/3) + (2x)(8x^3 + 5x^2)^(1/3) + (2x)^2)]/((8x^3 + 5x^2)^(2/3) + (2x)(8x^3 + 5x^2)^(1/3) + (2x)^2)\n\nNote that I multiplied and divided by the same term, to keep the value of the expression unchanged. It's the same as writing an equivalent fraction. Simplifying the numerator, we get\n\n(8x^3 + 5x^2) - (2x)^3 = 8x^3 + 5x^2 - 8x^2 = 5x^2\n\nIn the denominator, I want to pull out an x^2 out of each of the terms. It becomes\n\n((8x^3 + 5x^2)^(2/3) + (2x)(8x^3 + 5x^2)^(1/3) + (2x)^2) = (x^2)(8 + 5/x)^(2/3) + (2x^2)(8 + 5/x)^1/3 + 4x^2\n\nThe x^2 in the numerator cancels with the x^2 in the denominator. The final expression is this:\n\n5/((8 + 5/x)^(2/3) + 2(8 + 5/x)^(1/3) + 4)\n\nThe limit as this expression approaches infinity is simple. The only x's occur in the 5/x parts. 5/x approaches zero as x approaches infinity. The answer is\n5/(8^(2/3) + 2(8^(1/3)) + 4)= 5/(4 + 4 +4) = 5/12.
219588
3x^3 + 2x^2 + x - 6\n= 3x^3 + 5x^2 - 3x^2 + 6x - 5x - 6\n= x(3x^2 + 5x + 6) - 1(3x^2 + 5x + 6)\n= (3x^2 + 5x + 6)(x - 1) \n\nThe factors are : (3x^2 + 5x + 6) and (x - 1) \n\nAlternatively, using polynomial division : \n(3x^3 + 2x^2 + x - 6) divided by (x-1) yeilds \n(3x^2 + 5x + 6)
327770
FOIL the left side\n(x+8)(x+1) = x^2 + x + 8x +8\n= x^2 + 9x + 8\nThus, \nx^2 + 9x + 8 = -12  Add 12 to each side\nx^2 + 9x + 20 = 0  Look for factors of 20 that add up to 5\n(x+4)(x+5) = 0  Set each one equal to zero\nx+4 = 0 or x+5 = 0\nAnswer:  x = -4 or x = -5
6182
You can use what is called the Master Product Method to factor this and other quadratic trinomials.\n\nFor something of the form ax^2 + bx + c, the master product is a*c.  In this example, a is 2, b is -5 and c is 2, so the master product is 2*2 = 4.  Now you must find two numbers which multiply to give you the master product of 4, but add to give the middle coefficient of -5.\n\nThis requires a bit of trial and error, but you should be able to quickly discover that the numbers -1 and -4 will work since -1*-4 = 4 and -1+-4 = -5.  Now you rewrite the the entire trinomial replacing the middle term -5x with the equivalent -1x - 4x:\n\n    2x^2 - 5x + 2\n= 2x^2 - 1x - 4x + 2\n\nWe factor by grouping the first pair and last pair together and pulling out what's common to each pair:\n\n= (2x^2 - 1x) + (-4x + 2)\n= x(2x - 1) -2(2x - 1)\n\nNotice that (2x-1) appears in the factorization of both pairs.  We finish by factoring out the (2x-1) to get\n\n= (x - 2)(2x -1)
387267
Factor a^2 + 8a + 16\na^2 + 8a + 16 = \n(a+4)(a+4) =\n(a+4)^2\nNow \n(a^2)+8a+16-(b^2)\n= (a+4)^2 - b^2\nThis is a difference of squares.\n(a+4+b)(a+4-b)
570705
Factoring out X gives:\nX(-X^2 + X + 6)=0\nUsing the quadratic formula to factor the terms inside the brackets gives:\n-X^2 + X + 6 = (X+2)(-X+3)\nSo we have:\nX(X+2)(-X+3)=0 and the zeros are X=0, X=-2, X=3
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!
1174486
First look at the highest powers of a and b which in this case is "squared" and the product 6ab is present, so factoring will likely involve (a+b)^2\nIn fact 3a^2+6ab+3b^2 = 3(a^2+2ab+b^2) = 3(a+b)^2\nSo your problem reduces to:\n\n3(a+b)^2 +7(a+b) - 6\n\nIf we choose to factor -6 as -2 x 3 then the it is easy to factor, and the solution is:\n\n(3(a+b) -2)((a+b)+3)
779993
if a is a zero x-a is a factor\n1)\nso factors are (x+2) ,x (x-5)(x-6)\n\nmultiply them and you get\nf(x) = x(x+2)(x-5)(x-6) \n\nmultiply them and you get leading coefficient 1\n\n2)\n factors are (x-(1-sqrt(5)) and (x-(1+sqrt(5))\n\nso (x-1-sqrt(5))(x-1+ sqrt(5)) (a+b a-b form )\n= (x-1)^2 - 5  \n= x^2-2x-4
252087
(12x^3-18x^2)-8x+12=\n\n6x^2(2x-3)-4(2x-3)=\n\n(2x-3)(6x^2-4)=\n\n(2x-3)*2*(3x^2-2)=\n\n2(2x-3)(3x^2-2)
10610
X^2-9X+18\n\nShort and simple just factor it FOIL