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what is 250000000 divisable by?

if 25 is the divisible by 1,5,25. Then what is 250,000,000 divisable by?

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N is a number which is divisible by 19. if (n+1) (n+8) is ...


N is a number which is divisible by 19. if (n+1) (n+8) is divided by 19, the remainder will be? Find answers now! No. 1 Questions & Answers ... What is 250000000 ...
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What is 250000000 divided by 4 - answers.com


What is 250000000 divided by 4? SAVE CANCEL. already exists. Would you like to merge this question into it? MERGE CANCEL. already exists as an ...
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What is 0.4 divided by 6? - weknowtheanswer.com


What is 250000000 divided by 4? ... Are all even numbers divisible by 10? What is the sum of the first 100 natural number? How many days in week-?
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Math Forum - Ask Dr. Math


You need to figure out how many times your number is divisible by 10. Any number divisible by 10 ends in a zero ... we get 250,000,000 which does end in seven zeros ...
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What is 14.7 percent of 250000000 - step by step solution


Simple and best practice solution for 14.7% of 250000000. Check how easy it is, and learn it for the future. ... What is 14.7 percent of 250000000 ...
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Integer Number 1000000000


Integer Number 1000000000 ... , 12500000, 15625000, 20000000, 25000000, 31250000, 40000000, 50000000, 62500000, 100000000, 125000000, 200000000, 250000000, ...
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What is the sum of 1000000000 divided by 4 - Answers.com


What is the sum of 1000000000 divided by 4? ... 250000000 Minor edit ... 64 is the only number divisible by 5 with a remainder of 4 ...
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Engineering notation - Wikipedia


Engineering notation or engineering form is a version of scientific notation in which the exponent of ten must be divisible by ... in engineering notation is ...
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Suggested Questions And Answer :


10 vedic maths rules for class IX

2 instead of 5: 34/5 can be calculated by multiplying 34 by 2 instead of dividing by 5. 34*2=68. We just move the decimal point one place to the left: 34/5=6.8. 124/5=24.8 because 124*2=248. Move the decimal point: 248 becomes 24.8. 34*5 is the same as 34/2=17 but we add a zero to make 17 into 170. 73*5 is the same as 73/2=36.5 then move the decimal point one place to the right (or add zero): 36.5 becomes 365=73*5. So we only need to know how to multiply and divide by 2 to divide and multiply by 5. We just move the decimal point. Divisibility by 9 or remainder after dividing by 9. All multiples of 9 contain digits which added together give 9. As we add the digits together, each time the result goes over 9 we add the digits of the result together and use that result and continue in this way up to the last digit. Is 12345 divisible by 9? Add the digits together 1+2+3=6. When we add 4 we get 10, so we add 1 and zero=1 then we add 5 to get 6. The number is not exactly divisible by 9, but the remainder is 6. We can also ignore any 9's in the number. Now try 67959. We can ignore the two 9's. 6+7=13, and 1+3=4; 4+5=9, so 67959 is divisible by 9. Multiplying by 11. Example: 132435*11. We write down the first and last digits 1 ... 5. Now we add the digits in pairs from the left a digit step at a time. So 1+3=4; 3+2=5: 2+4=6; 4+3=7; 3+5=8. Write these new digits between 1 and 5 and we get 1456785=132435*11. But we had no carryovers here. Now try 864753*11. Write down the first and last digits: 8 ... 3. 8+6=14, so we cross out the 8 and replace it with 8+1=9, giving us 94 ... 3. Next pair: 6+4=10. Again we go over 10 so we cross out 4 and make it 5. Now we have 950 ... 3. 4+7=11, so we have 9511 ... 3. 7+5=12, giving us 95122 ... 3; 5+3=8, giving us the final result 9512283.  Divisibility by 11. We add alternate digits and then we add the digits we missed. Subtract one sum from the other and if the result is zero the original number was divisible by 11. Example: 1456785. 1 5 7 5 make up one set of alternate digits and the other set is 4 6 8. 1+5+7=13. We drop the ten and keep 3 in mind to add to 5 to give us 8. Now 4 6 8: 4+6=10, drop the ten and add 0 to 8 to give us 8 (or ignore the zero). 8-8=0 so 11 divides into 1456785. Now 9512283: set 1 is 9 1 2 3 and set 2 is 5 2 8; 9+1=0 (when we drop the ten); 2+3=5; set 1 result is 5; 5+2+8=5 after dropping the ten, and 5-5=0 so 9512283 is divisible by 11. Nines remainder for checking arithmetic. We can check the result of addition, subtraction, multiplication and (carefully) division. Using Method 2 above we can reduce operands to a single digit. Take the following piece of arithmetic: 17*56-19*45+27*84. We'll assume we have carried out this sum and arrived at an answer 2365. We reduce each number to a single digit using Method 2: 8*2-1*9+9*3. 9's have no effect so we can replace 9's by 0's: 8*2 is all that remains. 8*2=16 and 1+6=7. This tells us that the result must reduce to 7 when we apply Method 2: 2+3+6=11; 1+1=2 and 2+5=7. So, although we can't be sure we have the right answer we certainly don't have the wrong answer because we arrived at the number 7 for the operands and the result. For division we simply use the fact that a/b=c+r where c is the quotient and r is the remainder. We can write this as a=b*c+r and then apply Method 2, as long as we have an actual remainder and not a decimal or fraction. Divisibility by 3. This is similar to Method 2. We reduce a number to a single digit. If this digit is 3, 6 or 9 (in other words, divisible by 3) then the whole number is divisible by 3. Divisibility by 6. This is similar to Method 6 but we also need the last digit of the original number to be even (0, 2, 4, 6 or 8). Divisibility by 4. If 4 divides into the last two digits of a number then the whole number is divisible by 4. Using 4 or 2 times 2 instead of 25 for multiplication and division. 469/25=469*4/100=1876/100=18.76. 538*25=538*100/4=134.5*100=13450. We could also double twice: 469*2=938, 938*2=1876, then divide by 100 (shift the decimal point two places to the left). And we can divide by 2 twice: 538/2=269, 269/2=134.5 then multiply by 100 (shift the decimal point two places left or add zeroes). Divisibility by 8. If 8 divides into the last three digits of a number then the whole number is divisible by 8. Using 8 or 2 times 2 times 2 instead of 125 for multiplication and division. Similar to Method 9, using 125=1000/8. Using addition instead of subtraction. 457-178. Complement 178: 821 and add: 457+821=1278, now reduce the thousands digit by 1 and add it to the units: 278+1=279; 457-178=279. Example: 1792-897. First match the length of 897 to 1792 be prefixing a zero: 0897; complement this: 9102. 1792+9102=1894. Reduce the thousands digit by 1 and add to the result: 894+1=895. Example: 14703-2849. 2849 becomes 02849, then complements to 97150. 14703+97150=111853; reduce the ten-thousands digit by 1 and and add to the result: 11854. Squaring numbers ending in 5. Example: 75^2. Start by writing the last two digits, which are always 25. Take the 7 and multiply by 1 more than 7, which is 8, so we get 56. Place this before the 25: 5625 is the square of 75. The square of 25 is ...25, preceded by 2*3=6, so we get 625. All numbers ending in 0 or 5 are exactly divisible by 5 (see also Method 1). All numbers ending in zero are exactly divisible by 10. All numbers ending in 00, 25, 50 or 75 are divisible by 25. Divisibility by 7. Example: is 2401 divisible by 7? Starting from the left with a pair of digits we multiply the first digit by 3 and add the second to it: 24: 3*2+4=10; now we repeat the process because we have 2 digits: 3*1+0=3. We take this single digit and the one following 24, which is a zero: 3*3+0=9. When we get a single digit 7, 8 or 9 we simply subtract 7 from it: in this case we had 9 so 9-7=2 and the single digit is now 2. Finally in this example we bring in the last digit: 3*2+1=7, but 7 is reduced to 0. This tells us the remainder after dividing 2401 by 7 is zero, so 2401 is divisible by 7. Another example: 1378. 3*1+3=6; 3*6=18 before adding the next digit, 7 (we can reduce this to a single digit first): 3*1+8=3*1+1=4; now add the 7: 4+7=4+0=4;  3*4=12; 3*1+2+8=5+1=6, so 6 is the remainder after dividing 1378 by 7.  See also my solution to: http://www.mathhomeworkanswers.org/72132/addition-using-vedic-maths?show=72132#q72132
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please help with describing and ordering fractions

The divisions between 0 and 1 show how a fraction is made up. Let's take the case where the interval is divided up into 12 equal divisions. One division represents 1/12. Two divisions represent 2/12 which together represent 1/6 because 6*2=12 and the interval between 0 and 1 contains 6 sixths. Three divisions represent 3/12 which represent 1/4 because 4*3=12 and the interval between 0 and 1 contains 4 fourths or quarters, just like there are four quarters to $1. Four divisions represent 4/12 or one third. Five divisions just represent 5/12 and 7 divisions represent 7/12; but 6 divisions represent 6/12 or one half and two lots of these make 1. So 2*6/12 is the same as 2*1/2. 11 divisions make 11/12. Eight divisions is 8/12. Since 4/12 is a third, 8/12 must be two thirds because 2*4=8 or 2*4/12=8/12. Nine divisions are 3*3/12=9/12, and since 3/12 is the same as a quarter, 9/12 must be three quarters. Ten divisions make 10/12, but since 5*2/12=10/12 and 2/12 is one sixth, 10/12 must be 5 sixths or 5/6. If the interval 0 to 1 is divided into 60 divisions we get more fractions. The divisions help us to add and subtract. The number of divisions is associated with the least common multiple (LCM). So the LCM of 2, 3, 4 and 6 is 12. For 60 as LCM we have 2, 3, 4, 5, 6, 10, 12, 15, 20, 30. These numbers of divisions give us the fractions 1/30, 1/20, 1/15, 1/12, 1/10, 1/6, 1/5, 1/4, 1/3, 1/2. So to add 1/5 and 1/6 we use the divisions 12/60+10/60=22/60=11/30 because 22=2*11 and 22/60=2*11/60=11/30. We can also see that 11/30-1/5 is the same as 22/60-12/60=10/60=1/6. This should give you some idea how dividing 0 to 1 into different divisions helps you see how fractions add and subtract.
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n^3-9n^2+20n prove that it is divisible by 6 for all integers n greater or equal to 1

Factor the given expression.   We have: n³-9n²+20n=n(n-4)(n-5) ··· Ex.1 If Ex.1 is divisible by 6, Ex.1 is also divisible by two prime factors of 6, 2 and 3. (6=2x3) A. If n is odd, (n-5) is even.   If n is even, (n-4) is also even.   Therefore, n(n-4)(n-5) is always a multiple of 2. B. If Exp.1 is divisible by 3, the remainder is 0,1,or 2. If n ≡ 0 (mod 3), n-4 ≡ -4 ≡ -1 (mod 3), and n-5 ≡ -5 ≡ -2 (mod 3), that is: n is a multiple of 3. If n ≡ 1 (mod 3), n-4 ≡ -3 ≡ 0 (mod 3), and n-5 ≡ -4 ≡ -1 (mod 3), that is: (n-4) is a multiple of 3. If n ≡ 2 (mod 3), n-4 ≡ -2 (mod 3), and n-5 ≡ -3 ≡ 0 (mod 3), that is: (n-5) is a multiple of 3. Therefore, n(n-4)(n-5) is always a multiple of 3. CK: If n=1, n(n-4)(n-5)=1(-3)(-4)=12, 12(n=2), 6(n=3), 0(n=4), 0(n=5), 12, 42, 96, 180 … CKD.  Therefore, n³-9n²+20n is divisible by 6 for all integers greater than or equal to 1.  
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Divide x4 -2 by x+1

Well!   I think you meant (x^4 - 2) / (x+1)   (x^4 - 2) / (x+1) Writing it in polynomial form with co-efficients x^4 - 2 = 1 x^4 + 0 x^3 + 0 x^2 + 0 x - 2   Now, writing the co-efficients alone, for synthetic division Synthetic Division x c   x^4 x^3 x^2 x c   Value Poly 1 1   1 0 0 0 -2             1 1         1 x^3       0 -1 0 0 -2     REM         -1 -1       -1 x^2         0 +1 0 -2     REM           1 1     1 x           0 -1 -2     REM             -1 -1   -1 c             0 -1     REM Further Division is not Possible. Thus, (x^4 - 2) / (x+1) Quotient: x^3 - x^2 + x -1 Remainder: -1
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find HCF of3x^4+17x^3+27x^2+7x-6;6x^4+7x^3-27x^2+17x-3 by division method.

3x^4+17x^3+27x^2+7x-6=(x+1)(3x^3+14x^2+13x-6)=(x+1)(x+3)(3x^2+5x-2)= (x+1)(x+3)(3x-1)(x+2) CLUE: Write coefficients: 3 17 27 7 -6. If we add them together, do we get zero? No, so 1 is not a zero and x-1 is not a factor. Look at the odd powers of x (17x^3 and 7x). Change the sign in front of them and do the arithmetic: 3-17+27-7-6=0. Did we get zero? Yes, so -1 is a zero and x+1 is a factor. Now use synthetic division: -1 | 3 17..27...7..-6 ......3 -3 -14 -13...6 ......3 14..13..-6 | 0 gives us 3x^3+14x^2+13x-6. Neither 1 nor -1 is a zero. The factors of the constant 6 include 2 and 3. So the zeroes may be 2 or -2, 3 or -3. We find that -3 is a zero, so x+3 is a factor: -3 | 3 14..13..-6 ......3..-9 -15...6  .....3...5...-2 | 0 gives us 3x^2+5x-2, which we can easily factorise (see above). 6x^4+7x^3-27x^2+17x-3=(x-1)(6x^3+13x^2-14x+3)=(x-1)(x+3)(6x^2-5x+1)= (x-1)(x+3)(3x-1)(2x-1) CLUE: Coefficients sum to zero, so 1 is a zero and x-1 a factor. 1 | 6...7 -27..17..-3 .....6...6..13 -14...3 .....6 13 -14....3 | 0 gives us 6x^3+13x^2-14x+3. Try -3 as a factor because the constant is 3 and +3 does not give us zero: -6*27+13*9+14*3+3=-162+117+42+3=0, so x+3 is a factor. -3 | 6..13 -14...3 ......6 -18..15..-3 ......6..-5.....1 | 0 gives us 6x^2-5x+1, which is easy to factorise (see above).   By inspecting the bracketed factors we can see those in common between the two polynomials. There are two, so we take both of them: HCF=(x+3)(3x-1)=3x^2+8x-3
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find HCF and LCM of3x^4+17x^3+27x^2+7x-6;6x^4+7x^3-27x^2+17x-3 by division method.

3x^4+17x^3+27x^2+7x-6=(x+1)(3x^3+14x^2+13x-6)=(x+1)(x+3)(3x^2+5x-2)= (x+1)(x+3)(3x-1)(x+2) CLUE: Write coefficients: 3 17 27 7 -6. If we add them together, do we get zero? No, so 1 is not a zero and x-1 is not a factor. Look at the odd powers of x (17x^3 and 7x). Change the sign in front of them and do the arithmetic: 3-17+27-7-6=0. Did we get zero? Yes, so -1 is a zero and x+1 is a factor. Now use synthetic division: -1 | 3 17..27...7..-6 ......3 -3 -14 -13...6 ......3 14..13..-6 | 0 gives us 3x^3+14x^2+13x-6. Neither 1 nor -1 is a zero. The factors of the constant 6 include 2 and 3. So the zeroes may be 2 or -2, 3 or -3. We find that -3 is a zero, so x+3 is a factor: -3 | 3 14..13..-6 ......3..-9 -15...6  .....3...5...-2 | 0 gives us 3x^2+5x-2, which we can easily factorise (see above). 6x^4+7x^3-27x^2+17x-3=(x-1)(6x^3+13x^2-14x+3)=(x-1)(x+3)(6x^2-5x+1)= (x-1)(x+3)(3x-1)(2x-1) CLUE: Coefficients sum to zero, so 1 is a zero and x-1 a factor. 1 | 6...7 -27..17..-3 .....6...6..13 -14...3 .....6 13 -14....3 | 0 gives us 6x^3+13x^2-14x+3. Try -3 as a factor because the constant is 3 and +3 does not give us zero: -6*27+13*9+14*3+3=-162+117+42+3=0, so x+3 is a factor. -3 | 6..13 -14...3 ......6 -18..15..-3 ......6..-5.....1 | 0 gives us 6x^2-5x+1, which is easy to factorise (see above).   LCM=(x-1)(x+1)(x+3)(3x-1)(x+2)(2x-1) contains a combination of the factors of both numbers. By inspecting the bracketed factors we can see those in common between the two polynomials. There are two, so we take both of them: HCF=(x+3)(3x-1)=3x^2+8x-3
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Algebra 2 homework, please help.

  First set of questions is all roots, or solutions, of the equations and second set is to find where the functions are zero, the rational roots only. The methods for roots and zeroes are basically the same. 1. 3x^2+10x+3=3x^2+9x+x+3=3x(x+3)+(x+3)=(3x+1)(x+3)=0; x=-1/3, -3 2. 5x^2+5x-x-1=5x(x+1)-(x+1)=(5x-1)(x+1)=0; x=1/5, -1 3. Cubic equation. Trial shows that x=-1 is a solution, so divide cubic by factor (x+1). Use synthetic division: -1 | 3 11  5 -3      | 3 -3 -8  3        3  8 -3 | 0 and  (x+1)(3x^2+8x-3)= (x+1)(3x^2+9x-x-3)=(x+1)(3x(x+3)-(x+3))=(x+1)(3x-1)(x+3)=0; x=-1, 1/3, -3 4. 3x^4-2x^2-5=(3x^2-5)(x^2+1)=0; 3x^2=5, so x=+/-sqrt(5/3) 5. Trial shows that 3x=-1 is a solution, i.e., x=-1/3. Use synthetic division: -1/3 | 3 10 30   9         | 3  -1 -3  -9           3   9 27 | 0      (3x+1)(x^2+3x+9)=0; x=-1/3 is the only real solution 6. (5x^2+7)(x^2-2)=0; x=+/-sqrt(2) 7. No real roots 8. (5x^2+2)(x^2-8)=0; x=+/-sqrt(8) or +/-2sqrt(2) 9. (5x^2+9)(x^2-4)=(5x^2+9)(x+2)(x-2)=0; x=+/-2 10. (5x^2+3)(x^2-8)=0; x=+/-sqrt(8) or +/-2sqrt(2) 11. f(x)=5x(x^2-1)-(x^2-1)=(5x-1)(x-1)(x+1); x=1/5, 1, -1 12. Try x=1/2, f(x)=0. So f(x)=(2x-1)(x^2-11x-1) has no more rational roots 13. f(x)=3x(x^2-1)-(x^2-1)=(3x-1)(x-1)(x+1); zeroes at x=1/3, 1, -1 14. f(x)=(x+2)(x^2+7x+3); only rational zero at x=-2 15. (x-3)(2x^2+7x-3); rational zero at x=3 16. (x-1)^2(4x-1); x=1, 1/4 17. (x+1)(2x-1)(2x+1); x=-1, +/-1/2 18. (x+1)^2(3x-1); x=-1, 1/3 19. (x-5)(3x^2-7x-75); x=5, no other rational roots 20. (x-3)(2x^2+11x-4); x=3, other roots irrational
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.ABXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXQW x 4 = WQXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXBA

219999...9978*4=879999...9912, making A=2, B=1, Q=7, W=8, X=9. SOLUTION DETAILS  Let's find X first by taking an X within the row of Xs. 4X+c=10c+X, where c is the carryover in the result of multiplying the following X by 4. 3X=9c, X=3c, and c must be between 0 and 3. Therefore X=0, 3, 6, or 9. Now look at W. 4W=10p+A, where p is a carryover. And 4A+q=W where q is a carryover. Both p and q are less than 4. Multiply the last equation by 4: 16A+4q=4W=10p+A, so 15A=10p-4q. Therefore, we need to find values of p and q that are multiples of 15.   0 4 8 12 0 0 -4 -8 -12 10 10 6 2 -2 20 20 16 12 8 30 30 26 22 18 The first column contains 10p values and the first row 4q values. The other cells are the result of subtracting the column header from the row header. Only two cells are divisible by 15: 0 and 30, implying (p,q)=(0,0) or (3,0). We can reject (0,0) because it's likely to lead to the trivial solution where the numbers are zero. So we go for p=3, q=0, making A=2. This tells us that 4W=32 and W=8. And this means that there is no carryover for 4B, therefore B must be less than 3. We can also see that 4Q+3=10c+B, because 4W=32 making the carryover 3. We know B can only be 0, 1 or 2. So 4Q=10c-3, 10c-2, or 10c-1. This has to be divisible by 4, so we can eliminate 10c-3 and 10c-1 which are odd numbers. So B=1. That leaves 4Q=10c-2, and Q=2, c=1 or Q=7, c=3. X is therefore 3 or 9. To recap, we are left with 2133...328*4?=?8233...312 or 2199...978*4?=?8799...912. Only the latter satisfies the arithmetic.
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Use principle of mathematical induction to 3 prove that n3 —n is divisible by 3

Let p(n) = n3 - n i)If n=1 , p(1 )  = 1^3 - 1                  [ 1^3 means 1 cube ]                        = 1 - 1                        = 0                       [ 0 divisible by 3] If n =1 , p(1) is true--------------------( 1) ii)  Assume  that p( m ) is true.--------------------------(2 ) iii) we have to prove that p(m+ 1) is true.   p(m + 1) = (m + 1) ^3 - (m + 1)                      = (m + 1 ) [ (m + 1)^2  -1]                [ (m + 1) is the common factor ]                     = ( m + 1) [ m 2 + 2m + 1 - 1 ]         [ expansion of (m + 1 ) ^2 ]                     = ( m + 1 ) [ m2 + 2m ]                    = ( m + 1 ) m ( m + 2 )                      [ m is the common factor ]  {   divisibility rule of 3 is  If sum of the terms divisible by 3 then it is divisible by 3} sum = m + 1 + m + m + 2            = 3m + 3         = 3 ( m + 1)      it is divisible by 3. Therefore ( m+1) m (m+2 ) is divisible by 3 p(m+1) is divisible by 3 p(m+1) is true---------------------------(3) From (1), (2) and (3) for all values of n , p(n)  = n3 - 3 is divisible by 3
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using synthetic division what it x to the fourth power minus 1 divided x-1

x^4 - 1 = x^4 + 0 x^3 + 0 x^2 + 0 x - 1 1 . | . 1 . 0 . 0 . 0 . -1 .... | ...... 1 . 1 . 1 .. 1 .... ----------------------- ....... 1 . 1 . 1 . 1 .. 0 (x^4 - 1) / (x - 1) = (x^3 + x^2 + x + 1) ========================= NOTE: -1 is a zero of (x^3 + x^2 + x + 1), so you can use synthetic division again to find (x^3 + x^2 + x + 1) / (x + 1) -1 . | . 1 . 1 . 1 . 1 ..... | ..... -1 . 0 . -1 ..... ------------------ ....... 1 .. 0 .. 1 .. 0 (x^3 + x^2 + x + 1) / (x + 1) = x^2 + 1
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