Is the result of multiplication of a polynomial by a single term the same as that of the number of terms

Is the result of multiplication of a polynomial by a single term the same as that of the number of terms

dissimilarity
If the monomial is 0, then it will be 0 after multiplication;
If it is an integer that is not zero, then after multiplication, it will be a polynomial with the same term as the polynomial, and the highest power will not change;
If it is a monomial of x power, then after multiplication, it is a polynomial with the same term as the polynomial, but the highest power has changed
A formula for finding the number of terms of a polynomial
For example, how many terms of (a + B + C + D) to the 10th power? Is there a formula for direct calculation?
How to find the coefficients of polynomial terms? There should be certain rules.
For example, the power of (a + b) is 10. The binomial theorem can be used. Find the coefficients of (a + B + C + D + e) to the 15th power?
The formula has not been summed up. The method is as follows: divide 10 into the sum of four non negative integers, and the four numbers are not exchangeable. For example, 10 + 0 + 0 + 0,0 + 10 + 0 + 0. Keep writing. After writing, there are as many terms in the formula as there are. Your (a + B + C + D) ^ 10 is too much, for example, a simple (a
I have found such a formula. LZ can try it:
Let the degree be n, and the expanded form has m terms, that is, (a1 + A2 +...) ＋am）^n
After expansion, there is (n + m-1)! /（n！ (m-1)！） Item.
For example, (a + B + C + D) ^ 10, n = 10, M = 4, then 13! ÷10！ ÷3！ = 286 (items)
(a + B + C + D + e) ^ 100, n = 100, M = 5, then 104! ÷100！ ÷4！ = 4598126 (items)... Expanded
I have found such a formula. LZ can try it:
Let the degree be n, and the expanded form has m terms, that is, (a1 + A2 +...) ＋am）^n
After expansion, there is (n + m-1)! /（n！ (m-1)！） Item.
For example, (a + B + C + D) ^ 10, n = 10, M = 4, then 13! ÷10！ ÷3！ = 286 (items)
(a + B + C + D + e) ^ 100, n = 100, M = 5, then 104! ÷100！ ÷4！ = 4598126 (item)
Find the greatest common divisor and the least common multiple of each group of numbers below
42 and 36 91 and 39 48 and 32
42 and 36: greatest common divisor: 6, least common multiple: 252
91 and 39: greatest common divisor: 13, least common multiple: 273
48 and 32: greatest common divisor: 16, least common multiple: 96
If the result of division by the same base power is - 2 power of X, then the value of X cannot be
The - 2 power of x = 1 / X & # 178;
So x ≠ 0
Find the greatest common divisor and the least common multiple of the following groups of numbers. 24 and 36; 18, 24 and 40 (only the least common multiple)
(1) 24 and 36; 24 = 2 × 2 × 2 × 3; 36 = 2 × 2 × 3 × 3; their greatest common divisor is 2 × 2 × 3 = 12; their least common multiple is 2 × 2 × 3 × 2 × 3 = 72; (2) 18, 24 and 40; 24 = 2 × 2 × 2 × 3; 18 = 2 × 3 × 3; 40 = 2 × 2 × 2 × 5; their least common multiple is 2 × 3 × 2 × 3 × 5 = 360
Given the complex number Z = 3 + AI and | Z-2 | 2, find the value range of real number a
Solution 1: using the definition of module, eliminate Z. ∵ z = 3 + AI (a ∈ R) from two known conditions. From | Z-2 | 2, get | 3 + AI-2 | 2, that is | 1 + AI | 2, and get − 3 | a | 3. Solution 2: using the geometric meaning of complex number, from condition | Z-2 | 2, we can see that the corresponding point Z in the complex plane is in the circle (excluding boundary) with (2,0) as the center and 2 as the radius The corresponding point Z is on the straight line x = 3, so the line AB (except the end point) is the set of moving points Z. it can be seen from the figure that − 3 < a < 3
How to calculate the greatest common divisor and the least common multiple!
Ball algorithm
The greatest common divisor (GCD) or the highest common factor (HCF) is the largest common factor among several integers
For example, the common divisor of 12 and 30 are: 1, 2, 3, 6, where 6 is the greatest common divisor of 12 and 30
There are two ways to find the greatest common divisor of two integers
*The two numbers decompose the prime factors, then take out the same terms and multiply them
*Division by rotation (Extended Version)
The relation between GCD (a, b) × LCM (a, b) = ab
The greatest common factor of two integers can be used to calculate the least common multiple of two integers, or the fraction can be reduced to the simplest fraction
There is a distribution law in the greatest common factor and the least common multiple of two integers
* gcd(a,lcm(b,c)) = lcm(gcd(a,b),gcd(a,c))
* lcm(a,gcd(b,c)) = gcd(lcm(a,b),lcm(a,c))
In coordinates, connect points (0,0) and (a, b), and the number of points passing through integer coordinates (except (0,0) is GCD (a, b)
The common multiple of several numbers is called the common multiple of these numbers, and the smallest one is called the least common multiple of these numbers
The expression of the least common multiple:
In mathematics, it is often expressed in square brackets. For example, [12,18,20] is the least common multiple of 12, 18 and 20
The method of finding the least common multiple is as follows
There are two ways to find the least common multiple of several natural numbers
(1) The method of decomposing prime factors. First, decompose these numbers into prime factors, and then multiply all the common prime factors, the common prime factors of several numbers and the unique prime factors of each number. The product is their least common multiple
For example, find [12,18,20], because 12 = 22 × 3,18 = 2 × 32,20 = 22 × 5, in which the common prime factor of three numbers is 2, the common prime factor of two numbers is 2 and 3, and the unique prime factor of each number is 5 and 3. Therefore, [12,18,20] = 2 ^ 2 × 3 ^ 2 × 5 = 180
(2) Formula method. Because the product of two numbers is equal to the product of the greatest common divisor and the least common multiple of the two numbers, that is, (a, b) × [a, b] = a × B. therefore, to find the least common multiple of two numbers, we can first find their greatest common divisor, and then use the above formula to find their least common multiple
For example, if we find [18,20], we can get [18,20] = 18 × 20 ^ (18,20) = 18 × 20 ^ 2 = 180. To find the least common multiple of several natural numbers, we can first find the least common multiple of two of them, then find the least common multiple of this number and the third number, and then find them in turn until the last one. The last least common multiple is the least common multiple of several numbers
The problem of modulus of complex number
Suppose Z, Z0, Z1, Z2 are complex numbers, and Z2 = (z-z1) * Z0 + Z1
Where: | Z2 | = 1, Z1 = - 2, Z0 = cos (- 60 degrees) + isin (- 60 degrees)
For the two sides of the above formula, the left side = 1. How to simplify the right side?
Z1 is not a complex number, but this problem is solved from the complex plane
It should be directly calculated that z = a + bi (z-z1) = (a + 2) + bi (z-z1) * Z0 = 1 + A / 2 + (sqrt [3] b) / 2 + (- sqrt [3] - (sqrt [3] a) / 2 + B / 2) I (z-z1) * z0-z1 = - 1 + A / 2 + (sqrt [3] b) / 2 + (- sqrt [3] - (sqrt [3] a) / 2 + B / 2) I | (z-z1) * z0-z1 | = SQR
16. What is the least common multiple of 12 and 15______ .
16 = 2 × 2 × 2, 12 = 2 × 2 × 3, 15 = 3 × 5, so the least common multiple of 16, 12 and 15 is 2 × 2 × 2 × 2 × 3 × 5 = 240
Modulus of complex number
complex
（1-i）^10（3-4i）^4
What is the module of ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ ￣ can we calculate each module separately and then
(radical 3-I) ^ 5
Proposition 1: if Z1 and Z2 are complex numbers, the modules of their products are equal to the products of their modules
Z1 = x + iy, Z2 = a + IB, then | Z1 | = x ^ 2 + y ^ 2 under the root sign; | Z2 | = a ^ 2 + B ^ 2 under the root sign
Z1 * Z2 = (x + iy) (a + IB) = XA + IYA + IXB + I ^ 2by = (because I ^ 2 = - 1) XA by + I (Ya + BX)
So | Z1 * Z2 | ^ 2 = (XA by) ^ 2 + (Ya + BX) ^ 2 = (XA) ^ 2-2abxy + (by) ^ 2 + (ya) ^ 2 + 2abxy + (BX) ^ 2
=(XA) ^ 2 + (by) ^ 2 + (ya) ^ 2 + (BX) ^ 2 | Z1 * Z2 | = under the root sign (XA) ^ 2 + (by) ^ 2 + (ya) ^ 2 + (BX) ^ 2
And | Z1 | Z2 | = under the root sign (x ^ 2 + y ^ 2) (a ^ 2 + B ^ 2) = under the root sign (XA) ^ 2 + (BX) ^ 2 + (ya) ^ 2 + (by) ^ 2
It is the same as | Z1 * Z2 |
So it doesn't matter to find the module separately and then multiply it. Finding the module is almost the same as the absolute value of the ball
Proposition 2: |1 / w| = 1 / |w|
The proof is the same as the above. It's pure verification. It's too flattering to say that it's proof. There's no skill or suspense
From the combination of proposition 1 and Proposition 2, we can know that the module of multiplication and division can be calculated first and then multiplied and divided
But addition and subtraction don't work
But the module of addition and subtraction is definitely not equal to the module of addition and subtraction, and the absolute value after addition and subtraction is not necessarily equal to the absolute value of addition and subtraction
|1+(-1)|=0 ≠ |1|+|-1|=2
OK, I remember that the modulus of addition, subtraction, multiplication and division is equal to the modulus of addition, subtraction, multiplication and division
We can calculate each module separately and then perform the operation | UV / w | = | u | V | / | W|
The module is | (1-I) | ^ 10 * | (3-4i) | ^ 4 / | (3 ^ 0.5-i) | ^ 5 = 2 ^ 5 * 5 ^ 4 / 2 ^ 5 = 5 ^ 4 = 625
sure
=|1-i|^10*|3-4i|^4/(√3-i|^5
=(√2)^10*5^4/2^5
=625