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factors of 9

factors of 9 please

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What Are the Factors of 9?


What Are the Factors of 9? The factors of 9 are 1, 3, 9. Getting 9's factors is done like this. 1 and 9 are the first two factors. Try 1. 1 x 9 = 9, so put these into ...
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What are the factors of 9 - Answers.com


9 is a composite number; it has factors other than 1 and itself. It is a square number, so it has an odd number of factors. The three factors of 9 are 1, 3, and 9.
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What are the factors of 9 [SOLVED] - Mathwarehouse.com


What are the factors of 9 [SOLVED] Factorization Calculator: The factors of 9 Answer: 1,3,9, Related Links : Is 9 a rational number? Is 9 an irrational number?
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factors of 9 | Find the Factors


When 9 is a clue in the FIND THE FACTORS puzzles, the solution will use either 1 x 9 or 3 x 3, but not both.
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What are the factors of 9? | Reference.com


The factors of 9 are 1, 3 and 9. A factor is a number that divides into another number evenly. When a person divides 9 by the numbers 9, 3 and 1, it is easy to see ...
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What are the factors and prime factors of 9 - Answers


9 is a composite number; it has factors other than 1 and itself. It is a square number, so it has an odd number of factors. The three factors of 9 are 1, 3, and 9.
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Factoring Calculator


Factoring calculator to find the factors or divisors of a number. Factor calculator finds all factors and factor pairs of any positive non-zero integer. Factors ...
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Finding factors of a number (video) | Khan Academy


Hi so we need to Find all the factors of 120 Or another way to think about it, find all of the whole Numbers that 120 is divisible by. So the first one, that's maybe ...
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Find the Prime Factors of a Number


Find the Prime Factors of a Number: A prime number is any number with no divisors other than itself and 1, such as 2 and 11. Any number can be written as a product of ...
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Factors - MCWDN


Factors . A factor is a number that can evenly be divided into another number. Any number can be divided by one and itself. Numbers that can only be divided by one ...
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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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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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what is prime factor 2430

2 x 3^5 x 5 are all the prime factors of 2430. You can discover this yourself by knowing what clues to look for.. If your number is even, i.e., ends in 0, or 2, or 4, 6, or 8, then 2 is a factor. So, divide it by 2 to get 1215.  The result can not be further divided by 2.  So, 2 is only a factor one time.  If other factors exist, they produce 1215. Thus, we will work the result, 1215, to see what other factors may be in it. If a number ends in 0 or 5 then 5 is a factor.  Clearly, 1215 does end in 5. So, divide the result 1215 by 5 to get 243. Thus far, the factors identified are 2, and 5 and 243.  But what easy test can we do on 243 to check for any other factors? If the sum of its digits, i.e., 2 + 4 + 3  which is 9 is divisible by 9 or 3, then they are a factor.  9 is not a prime, but it shows that 3 is a factor more than once. Divide 243 by 3, and you see it goes evenly 81 times. Continue working this new result.  Notice its sum of digits 8 + 1 is 9.  So, keep dividing each result by 3.as many times as it will go evenly. All in all, 3 goes this many times into 243 and each result:  3 x 3 x 3 x 3 x 3. This is 3 to the 5th power, written as 3^5.
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f(x)=x^3+5x^2-17x-21

f(x)=x^3+5x^2-17x-21 1. factors of constant: 21: 1,3,7,21 2. factors of first coefficient: 1 3. p/q is 1,-1,3,-3,7,-7,21,-21 4. use synthetic division: 1 | 1 5 -17 -21 |_____1_____4______-13____ 1 4 -13 8 not a factor -1 | 1 5 -17 -21 |____-1___-4___-13_______ 1 4 13 8 not a factor 3 | 1 5 -17 -21 |____3___6___-69____ 1 2 -23 not a factor -3 | 1 5 -17 -12 |____-3__-24__-21________ 1 8 7 9 not a factor 7 | 1 5 -17 -21 |_____7__-14_-21________ 1 -2 -3 0 this is a factor -7 | 1 5 -17 -21 |____-7___-84_________ 1 12 not a factor 21 | 1 5 -17 -21 |___21____________ 1 -16 not a factor -21 | 1 5 -17 -21 |___-21____________ 1 26 not a factor 7 is the real root.
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How do special products help us factor polynomials? Give examples.

The special products that students are usually asked to identify and remember to help in factorisation are the squares and difference of squares: (a+b)^2=a^2+2ab+b^2; (a-b)^2=(b-a)^2=a^2-2ab+b^2; (a-b)(a+b)=a^2-b^2. a and b can be composite quantities like 3x, xy, 2xyz, etc. Examples: (3xy-z)^2=9x^2y^2-6xyz+z^2; (1/x+1/y)^2=1/x^2+2/xy+1/y^2; (5-t)(5+t)=25-t^2. I include some other types of products you might find useful: The constant term in a polynomial is a strong clue to how it factors, assuming that it is meant do so. If the constant is a prime number p, then the factors will consist of p and at least one 1, because p*1*...*1= p. The degree of the polynomial tells you how many factors there are: a degree 2 (quadratic) has two, for example. Examples: x^3-7x^2-x+7=(x-1)(x+1)(x-7); x^2-12x-13=(x-13)(x+1). If the constant is a composite number (i.e., not a prime number), then you need to factorise it and group the factors according to the degree. For example, if the constant is 15 and the degree is 4, then the factors arranged in fours are (1,1,1,15), (1,1,3,5). Example: x^4+2x^3-16x^2-2x+15=(x-1)(x+1)(x-3)(x+5). If the degree is n and the constant term is p^n, where p is a prime number, then the polynomial may consist of factors +p and -p. Examples: x^3-5x^2-25x+125=(x-5)^2(x+5). But x^2-26x+25=(x-25)(x-1). The sign of the constant is significant. If the degree n is even, and the sign is plus, then the zeroes of the polynomial will consist of an even number of pluses and minuses. If the sign is minus, there is an odd number of pluses and an odd number of minuses. If the degree is odd, and the sign is plus, the factors contain an even number of minuses, or none at all, and the rest are plus. If the sign is minus the factors contain an odd number of minuses, and the rest are pluses, or there are no other factors.
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What is the common denominator of 7, 5, 4, 2

Do you mean the common factor?  In order to have a common denominator you need fractions The common factor number that goes into all of those numbers evenly without a remainder. You might also mean common multiple which is the number that all of the number can be multiplied evenly by. Common factors: Factors of 7 7 = 1 times 7 7 is a prime number which means it only has factors of itself and 1   factors of 5 5 = 1 times 5 5 is also a prime number   Factors of 4 4 = 4 times 1 but, 4 = 2 times 2 factors of 4 are: 4,2 and 1   Factors of 2 2 = 2 times 1 2 is a prime number. Its only factors are itself and 1   the only common factor of 7,5,4 and 2 is: 1 It is the only number on all of the lists. Hope this helps.  If that isn't what you needed you can ask another question.  
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64239 is divided by a certain number.while dividing the number 175,114 and,213 appears as three successive remainders.find the divisor ?

It's not clear what "successive remainders" means. There is only one remainder after a division: (dividend)=(quotient)*(divisor)+(remainder). For example, if 175 is the remainder after division then 286 divides into 64239 224 times, remainder 175. The largest remainder is 213, implying that the divisor is bigger than 213. It's clear that the divisor cannot be divided again by the same divisor to yield a remainder in excess of the divisor. In the example 286 will not divide into 224, so 224 would be a second remainder. If the divisor has to be used three times to divide successively into each quotient to obtain each remainder then the divisor cannot exceed 40 (the approximate cube root of 64239) and therefore the remainders cannot exceed 40. See explanations in square brackets below. The only other interpretation I can find to the question is that in the process of long division, we have the successive remainders as we proceed with the division. But that doesn't seem to work either. [The numbers a, b and c are integers and b=cx+213; a=bx+114; 64239=ax+175, making ax=64064. Combining these equations we get: a=64064/x=bx+114, so b=64064/x^2-114/x=cx+213, and c=64064/x^3-114/x^2-213/x. Because 64064=ax, we see that 64064 must be divisible by x. 64064 factorises: 2^6*7*11*13. Because the largest remainder is 213, x>213, because x must be bigger than any remainder it creates. But the first term 64064/x^3 can only yield an integer value if x^3<64064, that is, x<40, which contradicts the requirement x>213.] [If the successive remainders appear in the long division, and the divisor has 3 digits as expected, and three remainders suggest that the other factor also has 3 digits, then if 175 is the first remainder it must be the result of dividing the divisor into 642. Therefore the multiple of the divisor =642-175=467. Since 467 is prime then the first digit of the other factor must be 1. 64239 divides by 467 137 times with 260 as the remainder. This is not one of the three remainders, so 175 cannot be the first remainder. Try 114: 642-114=528=2^4*3*11. We need a factor >213. 264, 352. This time we appear to have partial success, because 176 (less than 213, but half of 352) goes into 64239 364 times with a remainder of 175! So we have two of the three required remainders. Unfortunately the third remainder is 87, so 114 cannot be the first remainder. Finally, try 213: 642-213=429=3*11*13. The only factor bigger than 213 is 429 itself, and the final remainder after dividing 64239 by 429 is 318, not included in the three remainders given. So the assumption that the remainders arise during long division is also false.] 64064=2^6*7*11*13. Leaving aside the powers of 2, and just looking at combinations of 7, 11 and 13, we get 7*11=77; 7*13=91; 11*13=143, and 7*11*13=1001. By progressively doubling these products we can discover factors bigger than 213: 308=4*77; 364=4*91; 286=2*143, etc. We can also take 7, 11 and 13 and multiply them by powers of 2: 224=32*7; 352=32*11; 416=32*13. We can write the factors of 64064 in pairs, starting with a factor a little bigger than 213: (224,286), (308,208), (352,182), (364,176), (416,154), (448,143), (616,104), (704,91), (1001,64), (1232,52), (2002,32), (2464,26), (4004,16), (4928,13), (8008,8), (16016,4), (32032,2). These are (a,x) or (x,a) pairs. If we take 175 as the remainder, and ignore the other two, then we have a choice of factors. The 3-digit factors are (224,286), (308,208), (352,182), (364,176), (416,154), (448,143), (616,104). The only pair in which each factor is greater than 213 is (224,286).
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how do you use break apart to divide ?

You need to find the factors of the number you're dividing. If it's an even number you can divide by 2, so do this and note the factors as you find them. Therefore you have so far 2 and another number, we'll call n. Look at n, and if it's even you have another 2, so you would write 2*2 and another n. When you no longer have an even number, or if the first number was odd, you check to see if 3 goes into it. You make a note and divide the number by 3 so you now have ...3 and another n. You may even have 2*2...*3*n. Again you try 3, until you can't divide exactly, then you move on to 5 (numbers ending in 5 or 0 are exactly divisible by 5). You carry on like this noting what factors you've found as you go along. The numbers you pick as divisors will always be prime: in order here are the first few: 2, 3, 5, 7, 11, 13. A prime number is one which can't be divided exactly by a number smaller than it (actually up to its square root). 4 isn't prime because 2 goes into it; 9 isn't prime because 3 goes into it and all primes apart from 2 are odd numbers.  Now's a good time to look at examples. First, 660. It's even so 2 goes into it. So we have 2*330. 330 is also even: 2*2*165. 165 is divisible by 3, so we divide by 3 and we have 2*2*3*55. 55 ends in 5 and it's not divisible by 3, so we divide by 5 and get 11, and now we have 2*2*3*5*11. 11 is prime so we stop. 660=2*2*3*5*11. This breakdown means we can spot the factors other than prime factors. 2*2 means that 4 is a factor; 2*3 means that 6 is a factor; 3*5 means that 15 is a factor; 2*5*11=110 means that 110 is a factor, and so on. So if you needed to divide 660 by 15 you cross out the factors that make 15 (3 and 5) leaving 2*2*11=44. Let's try another example: 2030. 2030=2*1015=2*5*203=2*5*7*29. 29 is prime. How do we know how far to divide? Take 9042=2*4521=2*3*11*137. Is 137 a prime number? We only have to try and divide by numbers up to the square root of 137. What does that mean? The square root is the number which when multiplied by itself is closest to the number we're trying to divide, but smaller than it. 11*11=121; 12*12=144, so we only have to go up to dividing by 11, because 11*11 is less than 137 while 12*12 is greater, and the square of the next prime is 13*13=169, which is bigger than 137; so we only need to go up to primes up to 11, which doesn't divide exactly into 137, so 137 must be a prime number. Now try some numbers yourself!
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what is the LCM and the GCF of 125, 235 and 270

LCM of 125, 235, 270 Factor each into primes: 125 = 5 * 5 * 5 235 = 5 * 47 270 = 2 * 3 * 3 * 3 * 5 The LCM is the product of the minimum list of primes needed to make all of those numbers (125, 235, 270). For instance, there's a factor of 2 in 270, so the LCM has to have a factor of 2. There are three 3's as factors in 270, so the LCM has to have a factor of three 3's. There are three 5's as factors in 125, so the LCM has to have a factor of three 5's.  There's a 5 in 235, but that's already covered by one of the three 5's. There's a 47 in 235, so the LCM has to have a factor of 47. Basically you start with this: 125 = 5 * 5 * 5 235 = 5 * 47 270 = 2 * 3 * 3 * 3 * 5 Then make a list like this: 2 * 3 * 3 * 3 * 5 * 5 * 5 * 27 Then multiply it together like this: 2 * 3 * 3 * 3 * 5 * 5 * 5 * 27 = 182250 Answer:  The LCM of 125, 235, and 270 is 182250 . GCF of 125, 235, and 270 Start with a list of primes: 125 = 5 * 5 * 5 235 = 5 * 47 270 = 2 * 3 * 3 * 3 * 5 The GCF is the product of the list of primes that appear in all of those numbers (125, 235, 270). 47 only appears as a factor in 235, so 47 does not appear in the GCF. The only number that appears in all three numbers (125, 235, 270) is 5. 5 appears in 125 three times, but it only appears once in all three numbers (125, 235, 270), so the GCF only includes one 5. Answer:  The GCF of 125, 235, and 270 is 5.
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What is the greatest common factor? How do you know when you have found the greatest one?

Example: The common factors of 15 and 30 Factors of 15 are 1, 3, 5, and 15 Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30 The factors that are common  are 1, 3, 5 and 15 What is the "Greatest Common Factor" ? It is simply the largest of the common factors. The largest of the common factors is 15, so the Greatest Common Factor of 15 and 30  is 15
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