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is 2=1 by applying factoring?

prove 2=1 by applying knowledge of factoring  

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Special Factoring: Differences of Squares | Purplemath


Special Factoring: Differences of ... the x 2 – 1 factor — is itself a difference of squares, so I need to apply the formula again to get the fully-factored form.
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Solved: Solve 2x2 – 8x ­+ 1 = 0 by factoring and applying th ...


Solutions for Chapter 1 Problem 5CT. Problem 5CT: Solve 2x2 – 8x ­+ 1 = 0 by factoring and applying the zer... 5022 step-by-step solutions; Solved by professors ...
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Applying for factoring is easy and fast - Factoring Companies


Applying for factoring is easy and fast. Find out more about the process here. Skip to content. ... Cashflow (2) Distribution (1) Factoring (21) Factoring Articles (50)
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3 x 1 2 Section 07 Factoring Polynomials 078 h Apply the ...


3 x 1 2 Section 07 Factoring Polynomials 078 h ... 3 x 1 2 section 07 factoring polynomials 078 h apply ... 0.8.1 SECTION 0.8: FACTORING RATIONAL ...
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Factoring in Algebra - Math is Fun


Factoring in Algebra Factors. Numbers have factors: And expressions (like x 2 +4x+3) also have factors: Factoring. Factoring (called "Factorising" in the UK) is the ...
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1. Factoring the expression and applying the Zero Product ...


1. Factoring the expression and applying the Zero Product Property 2 ... Ex. 1 Solve by factoring and applying the Zero Product Property. xx2 2480 ...
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West Texas A&M University ; College Algebra -- Tutorial 7 ...


Factor a sum or difference of cubes. Apply the factoring strategy to factor a polynomial completely. ... Step 2: Factor out a GCF from each separate binomial.
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two or more polynomials. 2.1 Factoring by Grouping


2.1 Factoring out the Greatest Common Factor and ... so after applying the distributive property (or FOIL), the middle terms ... − + −2 1 ( ) ( ) ( )( ) ...
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Apply Today for Capital! Invoice Factoring Applications for ...


Invoice Factoring Applications. If your company generates new accounts receivables of at least $15,000 per month from other businesses, ...
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Suggested Questions And Answer :


how to factorise fully

I can offer tips: Look out for constants and coefficients that are multiples of the same number, e.g., if all the coefficients are even, 2 is a factor. If 3 goes into all the coefficients, 3 is a factor. Place the common numerical factor outside brackets, containing the expression, having divided by the common factor. In 2x^2+10x-6 take out 2 to give 2(x^2+5x-3). In 24x^2-60x+36 12 is a common factor so this becomes: 12(2x^2-5x+3). This also applies to equations: 10x+6y=16 becomes 2(5x+3y)=2(8). In the case of an equation, the common factor can be completely removed: 5x+3y=8. Now look for single variable factors. For example, x^2y^3+x^3y^2. Look at x first. We have x^2 and x^3 and they have the common factor x^2, so we get x^2(y^3+xy^3) because x^2 times x is x^3. Now look at y. We have y^2 and y^3 so now the expression becomes x^2y^2(y+x). If the expression had been 2x^2y^3+6x^3y^2, we also have a coefficient with a common factor so we would get: 2x^2y^2(y+3x). If you break down the factors one at a time instead of all at once you won't get confused. After single factors like numbers and single variables we come to binomial factors (two components). These will usually consist of a variable and a constant or another variable, such as x-1, 2x+3, x-2y, etc. The two components are separated by plus or minus. In (2) above we had a binomial component y+x and y+3x. These are factors. It's not as easy to spot them but there are various tricks you can use to help you find them. More often than not you would be asked to factorise a quadratic expression, where the solution would be the product of two binomial factors. Let's start with two such factors and see what happens when we multiply them. Take (x-1) and (x+3). Multiplication gives x(x+3)-(x+3)=x^2+3x-x-3=x^2+2x-3. We can see that the middle term (the x term) is the result of 3-1, where 3 is the number on the second factor and 1 the number in the first factor. The constant term 3 is the result of multiplying the numbers on the factors. Let's pick a quadratic this time and work backwards to its factors; in other words factorise the quadratic. x^2+2x-48. The constant term 48 is the product of the numbers in the factors, and 2 is the difference between those numbers. Now, let's look at a different quadratic: x^2-11x+10. Again the constant 10 is the product of the numbers in the factors, but 11 is the sum of the numbers this time, not the difference. How do we know whether to use the sum or difference? We look at the sign of the constant. We have +10, so the plus tells us to add the numbers (in this case, 10+1). When the sign is minus we use the difference. So in the example of -48 we know that the product is 48 and the difference is 2. The two numbers we need are 6 and 8 because their product is 48 and difference is 2. It couldn't be 12 and 4, for example, because the difference is 8. What about the signs in the factors? We look at the sign of the middle term. If the sign in front of the constant is plus, then the signs in front of the numbers in the factors are either both positive or both negative. If the sign in front of the constant is negative then the sign in the middle term could be plus or minus, but we know that the signs within each factor are going to be different, one will be plus the other minus. The sign in the middle term tells us to use the same sign in front of the larger of the two numbers in the factors. So for x^2+2x-48, the numbers are 6 and 8 and the sign in front of the larger number 8 is the same as +2x, a plus. The factors are (x+8)(x-6). If it had been -2x the factors would have been (x-8)(x+6). Let's look at a more complicated quadratic: 6x^2+5x-21. (You may also see quadratics like 6x^2+5xy-21y^2, which is dealt with in the same way.) The way to approach this type of problem is to look at the factors of the first and last terms. Just take the numbers 6 and 21 and write down their factors as pairs of numbers: 6=(1,6), (2,3) and 21=(1,21), (3,7), (7,3), (21,1). Note that I haven't included (6,1) and (3,2) as pairs of factors for 6. You'll see why in a minute. Now we make a table (see (5) below). In this table the columns A, B, C and D are the factors of 6 (A times C) and 21 (B times D). The table contains all possible arrangements of factors. Column AD is the product of the "outside factors" in columns A and D and column BC is the product of the "inside factors" B and C. The last column depends on the sign in front of the constant 21. The "twiddles" symbol (~) means positive difference if the sign is minus, and the sum if the sign in front of the constant is plus. So in our example we have -21 so twiddles means the difference, not the sum. Therefore in the table we subtract the smaller number in the columns AD and BC from the larger and write the result in the twiddles column. Now we look at the coefficient of the middle term of the quadratic, which is 5 and we look down the twiddles column for 5. We can see it in row 6 of the figures: 2 3 3 7 are the values of A, B, C and D. If the number hadn't been there we've either missed some factors, or there aren't any (the factors may be irrational). We can now write the factors leaving out the operators that join the binomial operands: (2x 3)(3x 7). One of the signs will be plus and the other minus. Which one is which? The sign of the middle term on our example is plus. We look at the AD and BC numbers and associate the sign with the larger product. We are interested in the signs between A and B and C and D. In this case plus associates with AD because 14 is bigger than 9. The plus sign goes in front of the right-hand operand D and the minus sign in front of right-hand operand B. If the sign had been minus (-5x), minus would have gone in front of D and plus in front of B. So that's it: [(Ax-B)(Cx+D)=](2x-3)(3x+7). (The solution to 6x^2+5xy-21y^2 is similar: (2x-3y)(3x+7y).) [In cases where the coefficient of the middle term of the quadratic appears more than once, as in rows 1 and 7, where 15 is in the twiddles column, then it's correct to pick either of them, because it just means that one of the binomial factors can be factorised further as in (1) above.] Quadratic factors A B C D AD BC AD~BC 1 1 6 21 21 6 15 1 3 6 7 7 18 11 1 7 6 3 3 42 39 1 21 6 1 1 126 125 2 1 3 21 42 3 39 2 3 3 7 14 9 5 2 7 3 3 6 21 15 2 21 3 1 2 63 61  
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What are greatest common factors?

All non-prime integers (composite numbers or integers) can be broken down into factors. Fractions are formed when two integers are arranged into a fraction: a/b, where a and b are different integers. If a and b are composite integers, they can each be replaced by the product of their factors. If they have a common factor, that factor can be used to reduce the fraction to a simpler form. The greatest common factor is the largest of the common factors, which is the product of all the common prime factors (prime means that the factor cannot be reduced further). The GCF is also useful in ratios as a means of simplifying the ratio. So GCF only applies if there are two or more integers. Example: 420 and 378. 420=2*2*3*5*7 and 378=2*3*3*3*7. These numbers have three factors in common: 2, 3 and 7. If we multiply them together we get 2*3*7=42. This is the GCF. If we saw the fraction: 378/420 we can reduce it by dividing top and bottom by 42 to get 9/10. Example: 24:84:144 can be reduced to 2:7:12 by dividing each number by the GCF 12.
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what do you mean exactly?

Of course, you haven't given the equation, but the rule is with equations that if a multiplication factor on one side of an equation appears on the other side you can divide both sides of the equation by that factor, provided it isn't zero. Similarly for divide, when you would multiply by the factor. If an addition or subtraction factor appears on each side of an equation you can subtract or add that factor from both sides of the equation. If the factor is called x or some multiple of x (e.g., 4x) or some other variable name the same rule applies. But you do need to make sure that the suspected factor isn't bound up in a relationship with another factor. So if your equation looks like x times something = x times something else, then dividing by x on both sides you're left with something=something else. Similarly if x were dividing, when you would multiply to cancel the common factor.  If your equation looks like something plus x or x plus something  = x plus something else or something else plus x, then subtracting x from both sides leaves something = something else. Similarly for minus, you simply add x to both sides. Does this help?
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How to find the factors of a 7th degree polynomial

If there are only positive coefficients, the zeroes (from which factors can be deduced) will all be negative. For mixed positive and negative coefficients, usually there are some simple zeroes like 1 and -1 that can be substituted for the variable that give a zero result. The size of the coefficients are also likely to be small numbers. Look out for missing powers; that may be an indication that the polynomial has a factor involving x^2 rather than x, for example. When you find a factor divide the polynomial by it using algebraic or synthetic division. This will reduce the degree. The method I find to be most effective is to use my calculator to plug in trial values for the variable between, say, -5 and +5 in steps of 1. When the result changes sign between two consecutive integers, I know there's a zero between; and occasionally an integer will itself produce zero, so I know that integer is a zero. Once a zero is located between two integers I can then home in on values in between. In a 7-degree poly you may have up to 7 factors so you do have to keep looking. Once you have found 5 factors using zeroes (if a is a zero then x-a is a factor), you will be left with a quadratic which may or may not factorise further, but you can apply the formula in any case. I hope this helps.
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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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Using the starting price of $1.964 per gallon, what is Rob’s net price after applying the 7/5/2.5 trade discount series using the net decimal equivalent?

  A lot to answer! First, the net result of applying the trade discount 7/5/2.5. A discount of 7% means a reduction to  93%; 5% a reduction to  95% of 93%; and a final reduction to 97.5% after applying the other two reductions. Net reduction is the product of these: 0.93*0.95*0.975=0.8614 or 86.14%, a discount of 13.86%. Applying this to $1.964 per gallon we get $1.692. The net result of the new supplier is 0.90*0.93*0.96=80.35% or 19.65% net discount. Apply this to $2.086=$1.676, slightly lower than $1.692, so a better prospect for Rob. Rob's customers get a reduction of 2% if they make a payment within 15 days of purchase. They must pay the full purchase price after that within 30 days of purchase or they will be penalised by the 1% service charge. Half of the monthly sales ($12,500) are paid by the customers paying within the discount period of 15 days, so $12,500 represents 98% or 0.98 of the amount that Rob would have received if there hadn't been any discount. If we call this non-discounted amount X, then 12500=0.98X, so X=12500/0.98=$12755.10. The customers therefore make a saving of $255.10. This is a little more than 2% saving (which would have been $250). The remaining $12,500 is paid by customers and 5% of this amount=$625 incurs the 1% service charge for late payment. The service charge amount is 1% of $625=$6.25. If S is the sales volume (monthly) then we know, because of the discount arrangement, Rob receives 0.98*S/4 (25%) = 0.245S from those customers paying within the discount period and S/5 (20%) paying within 30 days. This 20% doesn't include those paying within the discount period, but rather those paying within the time period 16 to 30 days. It would appear that 55% (100-25-20) of the sales volume is being paid after 30 days for which the penalty service charge applies. It also means that late payment encroaches into the next month. 0.55S delivers a total service charge of 0.01*0.55S=0.0055S. In terms of revenue then, Rob receives (0.245+0.2+0.0055)S=0.4505S. S is the non-discounted value of the stock Rob sells. He loses only what is paid in discount, that is, 0.25*0.02S=0.005S and, assuming he is eventually paid by his customers, he should receive 0.995S a month. Within that month he should receive payment from late payers from the previous month or earlier, just as in later months he receives late payments applying to the current month. The problem he has is that late payments affect his own ability to purchase fuel and other goods and commodities. The service charge penalties for late payment do not contribute very much to his revenue. Rob could encourage his customers to make prompt payment by offering a higher discount for payment within 15 days and by increasing the penalty for late payment. Although, based on current stats, he would lose more in discounts and gain more in penalties, he may be able to better ensure that he improves his purchasing ability to keep his stock at an acceptable level. He may be able in this way to reduce the percentage of late-paying customers. 
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Find m if x – 3 and x + 2 are factors of x3 + m2x2 – 11x – 15m

  36r4 + 36r3 + 3r2) / 9r3  this is normally written as (36r^4 + 36r^3 + 3r^2) / (9r^3)  this is equivalent to (36r^4 / 9r^3) + (36r^3 / 9r^3) + (3r^2 / 9r^3)  the rule for exponent arithmetic to apply is that x^a/x^b = x^(a-b).  so, basically, if the base is the same, then you subtract the exponent in the denominator from the exponent in the numerator and put the result in the numerator.  the other rule for exponent arithmetic to apply is that x^-a = 1/x^a.  you take the reciprocal and change the sign of the exponent.  x^-a = 1/x^a  x^a = 1/x^-a  normally you want to simplify by making all the exponents positive, so if you have x^-a, then show it as 1/x^a, and if you have 1/x^-a, then show it as x^a.  looking at each of the parts of (36r^4 / 9r^3) + (36r^3 / 9r^3) + (3r^2 / 9r^3) individually, you get:  36r^4 / 9r^3 becomes 36/9 * r^4/r^3 which becomes 4 * r^(4-1) which becomes 4r^1 which becomes 4r.  36r^3 / 9r^3 becomes 36/9 * r^3/r^3 which becomes 4 * r^(3-3) which becomes 4 * r^0 which becomes 4 * 1 which becomes 4.  3r^2/9r^3 becomes 3/9 * r^2/r^3 which becomes 1/3 * r^(2-3) which becomes 1/3 * r^-1 which becomes 1/3 * 1/r^1 which becomes 1/3 * 1/r which becomes 1/(3r)  your simplified expression becomes 4r + 4 + 1/(3r)  this should be your solution, but if you wanted to put everything under the same denominator, then you would get:  4r + 4 + 1/(3r) becomes (4r*3r + 4*3r + 1) / (3r)  simplify this to get:  (12r^2 + 12r + 1) / (3r)  that should be your simplified expression.  if we did this correctly, your original expression and your simplified expression should give you the same answer if you replace r with a random value.  for example, assume r = 7  your original expression of (36r^4 + 36r^3 + 3r^2) / (9r^3) becomes equal to 32.04761905  your simplified expression of (12r^2 + 12r + 1) / (3r) becomes equal to 32.04761905  since they both provide the same answer, they're equivalent and you can reasonable conclude that you simplified correctly.  if your stated solution is not this, then let me know what it is and i'll come up with a reason why it should be that and not what i just told you.  the difference, if any, is usually in what the definition of simplified is.   
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is 2=1 by applying factoring?

2=pi=3.14159 26535 89793 23846 26433 83279 5028...
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is 2=1 by applying factoring?

1=1..................
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Factorials with variables?

Notice that: 2! = 2*(1) = 2*1! 3! = 3*(2*1) = 3*2! 4! = 4*(3*2*1) = 4*3! … n! = n*(n - 1)! (n + 1)! = (n + 1)*n! A factorial can be written as the first factor times the factorial of the next factor, ie (n + 1)! = (n + 1)*n!. Applying this formula, we obtain: (n + 1)!/n! = (n + 1)*n!/n! = n + 1 For (2n + 2)!/(2n)!, just apply the formula twice: (2n + 2)!/(2n)! = (2n + 2)*(2n + 1)!/(2n)! = (2n + 2)*(2n + 1)*(2n)!/(2n)! = (2n + 2)*(2n + 1) I hope this helps!
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