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Fundamental Factors For battlefield play4free redeem codes 2014 - Insights

fundamental be good
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find all zeros of P(x) = x^5 + 9x^3 + 8x^2 + 72

Factor Theorem If r is a zero of the polynomial P(x), then x  r is a factor of P(x). Conversely, if x  r is a factor of P(x), then r is a zero of P(x). The relationship between zeros, roots, factors, and x intercepts is fundamental to the study of polynomials. With the addition of the factor theorem, we now know that the following statements are equivalent for any polynomial P(x): 1. r is a root of the equation P(x)  0. 2. r is a zero of P(x). 3. x  r is a factor of P(x). If, in addition, the coefficients of P(x) are real numbers and r is a real number, then we can add a fourth statement to this list: 4. r is an x intercept of the graph of P(x). SO,By using this theory u can do your problem easily.
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how to solve for x with fractions

Simplifying x3 + 3x2 + -4x = 0 Reorder the terms: -4x + 3x2 + x3 = 0 Solving -4x + 3x2 + x3 = 0 Solving for variable 'x'. Factor out the Greatest Common Factor (GCF), 'x'. x(-4 + 3x + x2) = 0 Factor a trinomial. x((-4 + -1x)(1 + -1x)) = 0 Subproblem 1 Set the factor 'x' equal to zero and attempt to solve: Simplifying x = 0 Solving x = 0 Move all terms containing x to the left, all other terms to the right. Simplifying x = 0 Subproblem 2 Set the factor '(-4 + -1x)' equal to zero and attempt to solve: Simplifying -4 + -1x = 0 Solving -4 + -1x = 0 Move all terms containing x to the left, all other terms to the right. Add '4' to each side of the equation. -4 + 4 + -1x = 0 + 4 Combine like terms: -4 + 4 = 0 0 + -1x = 0 + 4 -1x = 0 + 4 Combine like terms: 0 + 4 = 4 -1x = 4 Divide each side by '-1'. x = -4 Simplifying x = -4 Subproblem 3 Set the factor '(1 + -1x)' equal to zero and attempt to solve: Simplifying 1 + -1x = 0 Solving 1 + -1x = 0 Move all terms containing x to the left, all other terms to the right. Add '-1' to each side of the equation. 1 + -1 + -1x = 0 + -1 Combine like terms: 1 + -1 = 0 0 + -1x = 0 + -1 -1x = 0 + -1 Combine like terms: 0 + -1 = -1 -1x = -1 Divide each side by '-1'. x = 1 Simplifying x = 1 Solution x = {0, -4, 1}
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(8x)[(2x)^-2=8

Simplifying 8x = 2x2 + 8 Reorder the terms: 8x = 8 + 2x2 Solving 8x = 8 + 2x2 Solving for variable 'x'. Reorder the terms: -8 + 8x + -2x2 = 8 + 2x2 + -8 + -2x2 Reorder the terms: -8 + 8x + -2x2 = 8 + -8 + 2x2 + -2x2 Combine like terms: 8 + -8 = 0 -8 + 8x + -2x2 = 0 + 2x2 + -2x2 -8 + 8x + -2x2 = 2x2 + -2x2 Combine like terms: 2x2 + -2x2 = 0 -8 + 8x + -2x2 = 0 Factor out the Greatest Common Factor (GCF), '2'. 2(-4 + 4x + -1x2) = 0 Factor a trinomial. 2((-2 + x)(2 + -1x)) = 0 Ignore the factor 2. Subproblem 1 Set the factor '(-2 + x)' equal to zero and attempt to solve: Simplifying -2 + x = 0 Solving -2 + x = 0 Move all terms containing x to the left, all other terms to the right. Add '2' to each side of the equation. -2 + 2 + x = 0 + 2 Combine like terms: -2 + 2 = 0 0 + x = 0 + 2 x = 0 + 2 Combine like terms: 0 + 2 = 2 x = 2 Simplifying x = 2 Subproblem 2 Set the factor '(2 + -1x)' equal to zero and attempt to solve: Simplifying 2 + -1x = 0 Solving 2 + -1x = 0 Move all terms containing x to the left, all other terms to the right. Add '-2' to each side of the equation. 2 + -2 + -1x = 0 + -2 Combine like terms: 2 + -2 = 0 0 + -1x = 0 + -2 -1x = 0 + -2 Combine like terms: 0 + -2 = -2 -1x = -2 Divide each side by '-1'. x = 2 Simplifying x = 2 Solution x = {2, 2}
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solve by factorisation the following equation (1).5x^2+2x-15=0 (2).21x^2-25x=4 (3).10x^2+3x-4=0

  (1) 5x^2+2x-15=0 I suspect that this should be 5x^2+22x-15=0, because the equation as written does not factorise. Write down the factors of the squared term coefficient and the constant term. We write these as ordered pairs (a,b) squared term (c,d): (1,5) and constant: (1,15), (15,1), (3,5), (5,3). The sign of the constant (minus) tells us that we are going to subtract the cross-products of  factors. If it had been plus we would be adding the cross-products. Now we create a little table: Quadratic factor table a b c d ad bc  |ad-bc| 1 5 15 1 1 75 74 1 5 1 15 15 5 10 1 5 3 5 5 15 10 1 5 5 3 3 25 22 The column |ad-bc| just means the difference between ad and bc, regardless of it being positive or negative, just take the smaller product away from the larger. If the coefficient of the x term is in the last column, then that row contains the factors you need. If it isn't in the last column, then the quadratic doesn't factorise, or you've missed some factors. If 22x is the middle term, then the factors are shown in the last row and (a,b,c,d)=(1,5,5,3). Now we look at the sign of the middle term. Whatever the sign is we put it in front of the number c or d for the larger cross-product. The cross-products are bc and ad. In this case, bc is bigger than ad so the + sign goes in front of c. The sign in front of the constant tells us whether the sign in front of d is different or the same. If the sign is plus it's the same, otherwise it's the opposite sign. So in this case, it's minus, so the minus sign goes in front of d and we have (ax+c)(bx-d) (note the order of the letters!) or (x+5)(5x-3), putting in the values for a, b, c and d. If (x+5)(5x-3)=0, then x+5=0 or 5x-3=0. So in the first case x=-5 and in the second case 5x=3 so x=3/5. (2) 21x^2-25x-4=0 (moving 4 over to the left to put the equation into standard form). Quadratic factor table a b c d ad bc  |ad-bc| 1 21 1 4 4 21 17 1 21 4 1 1 84 83 1 21 2 2 2 42 40 3 7 1 4 12 7 5 3 7 4 1 3 28 25 3 7 2 2 6 14 8 The row in bold print applies. The sign in front of 4 is minus so the signs in the brackets will be different. 28 is the larger product, so since we have -25x, the minus sign goes in front of c (4) and plus in front of d (1): (3x-4)(7x+1)=0. So the solution is x=4/3 or -1/7. (3) 10x^2+3x-4=0. Quadratic factor table a b c d ad bc  |ad-bc| 1 10 1 4 4 10 6 1 10 4 1 1 40 39 1 10 2 2 2 20 18 2 5 1 4 8 5 3 2 5 4 1 2 20 18 2 5 2 2 4 10 6 (2x-1)(5x+4)=0, so x=1/2 or -4/5.  
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sqrt(3a+10) = sqrt(2a-1) + 2

sqrt(3a+10)=sqrt(2a-1)+2 To remove the radical on the left-hand side of the equation, square both sides of the equation. (~(3a+10))^(2)=(~(2a-1)+2)^(2) Simplify the left-hand side of the equation. 3a+10=(~(2a-1)+2)^(2) Squaring an expression is the same as multiplying the expression by itself 2 times. 3a+10=(~(2a-1)+2)(~(2a-1)+2) Multiply each term in the first group by each term in the second group using the FOIL method. FOIL stands for First Outer Inner Last, and is a method of multiplying two binomials.  First, multiply the first two terms in each binomial group.  Next, multiply the outer terms in each group, followed by the inner terms.  Finally, multiply the last two terms in each group. 3a+10=(~(2a-1)*~(2a-1)+~(2a-1)*2+2*~(2a-1)+2*2) Simplify the FOIL expression by multiplying and combining all like terms. 3a+10=(~(2a-1)^(2)+4~(2a-1)+4) Remove the parentheses around the expression ~(2a-1)^(2)+4~(2a-1)+4. 3a+10=~(2a-1)^(2)+4~(2a-1)+4 Raising a square root to the square power results in the expression inside the root. 3a+10=(2a-1)+4~(2a-1)+4 Add 4 to -1 to get 3. 3a+10=2a+3+4~(2a-1) Since a is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation. 2a+3+4~(2a-1)=3a+10 Move all terms not containing ~(2a-1) to the right-hand side of the equation. 4~(2a-1)=-2a-3+3a+10 Simplify the right-hand side of the equation. 4~(2a-1)=a+7 Divide each term in the equation by 4. (4~(2a-1))/(4)=(a)/(4)+(7)/(4) Simplify the left-hand side of the equation by canceling the common terms. ~(2a-1)=(a)/(4)+(7)/(4) To remove the radical on the left-hand side of the equation, square both sides of the equation. (~(2a-1))^(2)=((a)/(4)+(7)/(4))^(2) Simplify the left-hand side of the equation. 2a-1=((a)/(4)+(7)/(4))^(2) Combine the numerators of all expressions that have common denominators. 2a-1=((a+7)/(4))^(2) Expand the exponent of 2 to the inside factor (a+7). 2a-1=((a+7)^(2))/((4)^(2)) Expand the exponent 2 to 4. 2a-1=((a+7)^(2))/(4^(2)) Simplify the exponents of 4^(2). 2a-1=((a+7)^(2))/(16) Multiply each term in the equation by 16. 2a*16-1*16=((a+7)^(2))/(16)*16 Simplify the left-hand side of the equation by multiplying out all the terms. 32a-16=((a+7)^(2))/(16)*16 Simplify the right-hand side of the equation by simplifying each term. 32a-16=(a+7)^(2) Since (a+7)^(2) contains the variable to solve for, move it to the left-hand side of the equation by subtracting (a+7)^(2) from both sides. 32a-16-(a+7)^(2)=0 Squaring an expression is the same as multiplying the expression by itself 2 times. 32a-16-((a+7)(a+7))=0 Multiply -1 by each term inside the parentheses. 32a-16-a^(2)-14a-49=0 Since 32a and -14a are like terms, add -14a to 32a to get 18a. 18a-16-a^(2)-49=0 Subtract 49 from -16 to get -65. 18a-65-a^(2)=0 Move all terms not containing a to the right-hand side of the equation. -a^(2)+18a-65=0 Multiply each term in the equation by -1. a^(2)-18a+65=0 For a polynomial of the form x^(2)+bx+c, find two factors of c (65) that add up to b (-18).  In this problem -5*-13=65 and -5-13=-18, so insert -5 as the right hand term of one factor and -13 as the right-hand term of the other factor. (a-5)(a-13)=0 Set each of the factors of the left-hand side of the equation equal to 0. a-5=0_a-13=0 Since -5 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 5 to both sides. a=5_a-13=0 Set each of the factors of the left-hand side of the equation equal to 0. a=5_a-13=0 Since -13 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 13 to both sides. a=5_a=13 The complete solution is the set of the individual solutions. a=5,13
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adx/(b-c)yz=bdy/(c-a)zx=cdz/(a-b)xy

(1) I am assuming this is not a calculus question where dx, dy and dz have a different meaning. Variable d can be removed as a common factor: ax/((b-c)yz)=by/((c-a)zx)=cz/((a-b)xy) Take the first pair and remove common factor z: ax/((b-c)y)=by/((c-a)x); ax^2(c-a)=by^2(b-c); ax^2=by^2(b-c)/(c-a); by^2=ax^2(c-a)/(b-c) The second pair have common factor x: by^2(a-b)=cz^2(c-a); by^2=cz^2(c-a)/(a-b); cz^2=by^2(a-b)/(c-a) First and last have common factor y: ax^2(a-b)=cz^2(b-c); ax^2=cz^2(b-c)/(a-b); cz^2=ax^2(a-b)/(b-c) We have a pair of equations for each of the three quantities ax^2, by^2, cz^2. So, ax^2=by^2(b-c)/(c-a)=cz^2(b-c)/(a-b), by^2/(c-a)=cz^2/(a-b) or (A) by^2/(cz^2)=(c-a)/(a-b) by^2=ax^2(c-a)/(b-c)=cz^2(c-a)/(a-b), ax^2/(b-c)=cz^2/(a-b) or (B) ax^2/(cz^2)=(b-c)/(a-b) cz^2=by^2(a-b)/(c-a)=ax^2(a-b)/(b-c), by^2/(c-a)=ax^2/(b-c) or (C) by^2/(ax^2)=(c-a)/(b-c) The relative sizes of a, b and c can be: (1) a0 because b0 implies ac>0 and a>0 because c>0 (1C)<0 implies ab<0 and b<0 so ba: INCONSISTENT (2A)<0 implies bc<0 and b>0, c<0 because b>c (2B)<0 implies ac<0 and a>0 because c<0 (2C)>0 implies ab>0 and b>0 but c<0 and a>0 making ca: INCONSISTENT (3A)>0 implies bc>0 (3B)<0 implies ac<0 and a<0, c>0 and b>0 because a0 and so b>a, but b0 because b0 (4C)>0 implies ab>0, true, because a and b are both negative, but c0: INCONSISTENT (5A)>0 implies bc>0 (5B)<0 implies ac<0 and c<0, b<0, a>0 because a>c (5C)<0 implies ab<0, but b>a which cannot be if b<0: INCONSISTENT (6A)<0 implies bc<0 and c<0, b>0 because b>c (6B)>0 implies ac>0 and a<0, but b0 which cannot be: INCONSISTENT So there are no solutions because there is inconsistency throughout. (2) I am assuming this is a calculus problem. Assume a, b and c are constants and x, y and z are variables. adx/((b-c)y)=bdy/((c-a)x); b(b-c)ydy=a(c-a)xdx Integrating: b(b-c)y^2/2=a(c-a)x^2/2 (constant to be added later) a(c-a)x^2-b(b-c)y^2=k a constant. A similar result follows for the other two pairs of variables (x,z and y,z) by separation of variables and using the other two pairs of equations: a(a-b)x^2-c(b-c)z^2=p and b(a-b)y^2-c(c-a)z^2=q  
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graph theory

  We make a list of the letters and their frequencies in the message MISSION SUCCESSFUL, including the space. S=5; I=U=C=2; M=O=N=space=E=F=L=1. The number after the equals sign gives the "weight", the frequency of the letter(s). The next step is to make a tree with the highest frequency letters near the top of the tree and the lowest at the bottom. These are the leaves. The idea is to find an optimal binary code that is shortest for the highest frequency letter and longest for the lowest. The branches of the tree bifurcate to two "child" branches we call left and right. Left is denoted on the code by a zero and right by a 1. We start at the base of the tree with all the letters with a frequency of 1. We group them into pairs. Since there are seven we take three pairs: FL, spE, ON and join each pair to a "node", which is simply a point where each letter branches off to the left or right. We give the node the combined weight of the letters that branch off from it. So for ON, spE and FL the combined weight is 2. So we have three nodes. We take a pair of these (spE node and FL node) and create another node that joins them. What we can do now is create another node that joins the spare ON node and the spare letter M. We end up with a type of family tree where at each node including the topmost, we have exactly two children at every node. In this tree there is only one "parent" per pair of "children", a sort of monogenesis. Each child can only parent two children. At the end of the generation chain are the letters, the leaves of the tree. To find out which unique code each letter is associated with we use the digit 1 when we branch to the right and 0 if we branch to the left. Here are the codes I came up with. S=0, C=110, I=1110, U=1111, M=1000, F=10010, L=10011, space=10100, E=10101, O=10110, N=10111. The message is these codes following one another in sequence, so it goes: 1000|1110|0|0|1110|10110|10111|10100|0|1111|110|110|10101|0|0|10010|1111|10011 The vertical bar has been inserted for legibility; it isn't part of the code. This is not the only way the message can be encoded, and the decoding depends on the receiver possessing the tree from which the codes are derived.
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how do you solve 3t^4+2t^3-300t-50=0

The expression does not factorise and the roots seem to be t=-0.16669 and 4.48653 approx. Check that you haven't missed out a term (e.g., t^2). If the equation was meant to factorise completely, then the product of the factors=-50. The factors must therefore consist of 1, 2, 5, 5. The factors of the 3t^4 must consist of 3t, t, t, t. The minus sign in front of the constant 50 implies that there must be an odd number of pluses and minuses amongst the factors: one minus and 3 pluses, or three minuses and one plus.  Let the coefficient of the apparently missing t^2 be k. The equation then reads: 3t^4+2t^3+kt^2-300t-50=0. If (t-1) is a factor then 3+2+k-300-50=0 and k=345. If (t+1) is a factor then 3-2+k+300-50=0 and k=52-303=-251. If (t-2) is a factor then 48+16+4k-600-50 and k=(1/4)(650-64)=293/2. If (t+2) is a factor then 48-16+4k+600-50 and k=(1/4)(66-648)=-291/2. It seems improbable that (t-2) or (t+2) could be a factor because k would be a mixed number (or improper fraction), so k=345 or -251. If (t-5) is a factor then 1875+250+25k-1500-50=0 and k=(1550-2125)/25=-23. If (t+5) is a factor then 1875-250+25k+1500-50=0 and k=(300-3375)/25=-123. We know that (3t+f) must be included as a factor, but if f=-5 or +5 then, since we need two 5's within the set of factors, so if one contains 3t, the other can only contain t. That means that (t-5) or (t+5) must be a factor. There's no point trying other factors like (t-25) or (t+10), because they would require (t-2) or (t+2), or (t-5) or (t+5), and (t-1) and (t+1) as other factors, which have already been covered. The exercise shows that no one value of k allows us to find more than one root. Had any two values of k above been the same we would have had at least two roots and a quadratic, which could have been solved, if it had had real roots. The conclusion is that finding one root only leaves us with a cubic equation that has no rational, real roots. This means that the question has been wrongly stated, and the inclusion of kt^2 doesn't help to solve it satisfactorily.
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Factoring by grouping?

Is there a typo here? Maybe it should be 0 = x^3 + 6x^2 -3x - 18 .  This can be solved by grouping like so:   0 = (x^3 + 6x^2) + (-3x - 18 )  This is the grouping step.  I groupred the first two terms, and the last two terms.   0 = x^2(x+6) - 3(x+6)   In this step, I factored out the greatest common factor from each set of parenthesis.   0 =  ( x+6 )(x^2 - 3)  In this step, I factored out the common (x+6) binomial factor from the right side of the equation. (a challenging notion to follow, but it goes something like this:  In  x^2(x+6) - 3(x+6) , think of x^2 as "a", 3 as "b", and (x+6) as "c",  Then x^2(x+6) - 3(x+6) is of the form  ac - bc.  I factored out the "c" to get c(a-b).  Substituting back, using  x^2 as "a", 3 as "b", and (x+6) as "c", get c(a-b) = (x+6)(x^2-3)   Continuing on to solve the equation, if 0 =  ( x+6 )(x^2 - 3) , then 0 = x+6   or    0 = x^2 -3 Solving these equations, get x = -6, x =sqrt(3), x = - sqrt(3)  (remember when taking the square root of each side you must include both the positive and negative results)
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