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# is 2=1 by applying factoring?

prove 2=1 by applying knowledge of factoring

## Research, Knowledge and Information :

### Factoring Quadratic Form - Kuta Software LLC

Factoring Quadratic Form Date_____ Period____ Factor each completely. 1) u4 + 2u2 2) x4 + x2 − 12 3) a4 + 6a2 + 5 4) x4 − 8x2 + 15 5) u4 ...

### Applying x^2+bx+c to real life questions - StudyPug: #1 Help ...

Use your knowledge of polynomials and factoring them to solve our guided word problems related to geometric objects. ... (k 3 + 7 k 2 + 1 2 k) ...

### The Quadratic Formula Explained | Purplemath

Demonstrates the use of the Quadratic Formula and compares the QuadraticFormula to the solutions found by factoring ... the Quadratic Formula works: Solve x 2 + 3x ...

### 9-2 Multiplying and Factoring

Multiplying and Factoring Lesson 2-4 Multiply. 1. 3(302) 906 2. 41(7) 287 3. 9(504) 4536 ... The process of applying the distributive property with polynomials ...

### Special Factoring: Differences of Squares | Purplemath

Special Factoring: Differences of ... (x 2) 2 – 1 2 so, applying the formula, I get: x 4 ... because one of the factors I got — namely, the x 2 – 1 factor ...

### 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 ...

### Solve by Factoring Lessons | Wyzant Resources

Solve by Factoring Lessons. ... Steps 1 and 2. All three terms are already on the left side of the equation, so we may begin factoring. First, ...

### 2.1 Factoring out the GCF and Factoring by Grouping

2.1 Factoring out the GCF and Factoring by ... (10 5 6 3 5 2 1 3 2 1 2 1 5 3x x x x x x ... It is important to watch out for GCFs before applying bottoms-up! 8. 2 18 ...

### Lesson 4.2.5: Applying Quadratics - Algebra 1 With Mr. Eoff

Algebra 1 With Mr. Eoff. Overview Review Unit 1 > > > > Unit 2 > > ... Lesson 4.2.5: Applying Quadratics: For this lesson there are 5 steps for you to take.

## Suggested Questions And Answer :

### 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.

### 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?

### 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.

### 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.

### 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.

### is 2=1 by applying factoring?

2=pi=3.14159 26535 89793 23846 26433 83279 5028...

### is 2=1 by applying factoring?

1=1..................

### 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!