Given the set a = {x | x2-2x-3 ＞ 0}, then the number of elements in the set Z ∩ ∁ RA is () A. 5B. 4C. 3D. 2

Given the set a = {x | x2-2x-3 ＞ 0}, then the number of elements in the set Z ∩ ∁ RA is () A. 5B. 4C. 3D. 2

A = {x | x2-2x-3 ＞ 0} = {x | x ＞ 3 or X ＜ - 1}, ∁ RA = {x | - 1 ≤ x ≤ 3}, then Z ∩ ∁ RA = {- 1, 0, 1, 2, 3}, there are five elements, so a

A question about factorization
x^2+xy-2y^2=7
Finding integers x and Y

x^2+xy-2y^2=7
(x-y)(x+2y)=7×1.
Because X and y are integers (including positive and negative integers),
So {X-Y = 1, x + 2Y = 7} -- (1);
Or {X-Y = 7, x + 2Y = 1} -- (2)
From the equations (1), it is concluded that
x=2,y=3;
From the equations (2), it is concluded that
x=5,y=-2;
Both cases fit the meaning of the question

The problem of factorization (1) is urgent
1.(x+y-2xy)(x+y-2)+(xy-1)^2
2.(2x-3y)^3+(3x-2y)^3-125(x-y)^3
3. In order to factorize the quadratic trinomial x ^ 2-5x + P in the range of integers, what is the value of integer p?
The answers are numerous, but I don't quite understand

(1),(x+y-2xy)(x+y-2)+(xy-1)²
=(x+y)²-2(1+xy)(x+y)+4xy+(xy-1)²
=(x+y)²-2(1+xy)(x+y)+(xy+1)²
=(x+y-xy-1)²=(x-1)²(y-1)²
(2) Let 2x-3y = a, 3x-2y = B, 5x-5y = a + B;
(2x-3y)³+(3x-2y)³-125(x-y)³
=a³+b³-(a+b)³=-3a²b-3ab²=-3ab(a+b)=15(y-x)(2x-3y)(3x-2y);
(3) Suppose X & sup2; - 5x + P = (x + a) (x + b)
=X & sup2; + (a + b) x + AB, then: a + B = 5, ab = p;
A + B = 5 has innumerable integer solutions, a = n, B = 5-N (n is an integer), P = n (5-N);
N = 1,2,3,4,5,6,... - 1, - 2, - 3, - 4, - 5,... In this way, we can get the value of countless integers P

It is known that 2x-3 and 3x + 1 are the factors of F (x) = AX3 + bx2 + 32x + 15

If (2x-3) and (3x + 1) are factors of F (x) = AX2 + bx2 + 32x + 15, then (2x-3) (3x + 1) = 6x2-7x-3 can divide f (x). Solution 1: by using the large division of polynomials and Polynomials: ∧ B + 7ab = − 30 and 32 + A2 = 35, ∧ a = 6 and B = - 37, that is, f (x) = bx3-37x2 + 32x + 15 = (2x-3) (3x + 1) (MX + n) = (6x2 − 7x − 3) (MX + n) +n) = 6mx3 + (6N − 7m) x2 − (3m + 7n) x − 3N = AX3 + bx2 + 32x + 15 〉 a = BMB = 6N − 7m32 = − (3m + 7n) 15 = − 3N 〉 n = - 5, M = 1, B = - 37, a = 6, that is, f (x) = (2x-3) (3x + 1) (X-5) = 6x3-37x2 + 32x + 15

Can you ask some factorization questions in grade one?
5a^2(x-y)+10a(y-x)
a(x-3)(y-2)+b(3-x)(2-y)
x^2+y^2/9-2/3 xy
(m^2+1)^2+4m(m^+1)+4m^2
8ax+12ay-10bx-15by
2ab-a^2-b^2+4

5a^2(x-y)+10a(y-x)
=5a^2(x-y)-10a(x-y)
=5a(x-y)(a-2)
a(x-3)(y-2)+b(3-x)(2-y)
=a(x-3)(y-2)+b(x-3)(y-2)
=(x-3)(y-2)(a+b)
x^2+y^2/9-2/3 xy
=(x-y/3)^2
(m^2+1)^2+4m(m^+1)+4m^2
=(m^2+1+2m)^2
=(m+1)^4
8ax+12ay-10bx-15by
=(8ax+12ay)-(10bx+15by)
=4a(2x+3y)-5b(2x+3y)
=(2x+3y)(4a-5b)
2ab-a^2-b^2+4
=4-(a^2-2ab+b^2)
=2^2-(a-b)^2
=(2+a-b)(2-a+b)

Five ways of factoring mathematical problems
Factorization: 5 questions, * = multiply by
Quadratic power of X * (quadratic power of Y-9) + 4 (quadratic power of 9-y)=
The second power of (x-2y) - 3 * (x-2y) - 10=
6X's N + 1 power + 5x's n power Y - 4x's n-1 power Y's quadratic power=
The second power of 9-a-the second power of B + 2Ab=
The quadratic power of x-2xy + the quadratic power of Y + 4x-4y + 4=

X ^ 2 is the square of X
1. The original formula = (y ^ 2-9) (x ^ 2-4) = (y + 3) (Y-3) (x + 2) (X-2)
2. Multiplication by cross = [(x-2y) + 2] [(x-2y) - 5]
Please check again
4. The original formula = 9 - (a-b) ^ 2 = [3 + (a-b)] [3 - (a-b)] = (3 + a-b) (3-A + b)
5. The original formula = (X-Y) ^ 2 + 4 (X-Y) + 4 = (X-Y + 2) ^ 2

1. Given x + 3Y = 10, find the value of (x ^ 2-9y ^ 2) / (x-3y)
2.a^2(a-3)-(a-3)
3. If x ^ 2-5x-6 = (x + 1) (x-m), then the value of M is
4. If 4A ^ 2 + 4A + B ^ 2-6b + 10 = 0, then a ^ 3b-ab ^ 3=
5. If m ^ 2-4mn + 4N ^ 2 = 0, find the value of (m ^ 2 + n ^ 2) / Mn

1. (x ^ 2-9y ^ 2) / (x-3y) = (x-3y) (x + 3Y) / (x-3y) = x + 3Y = 102. A ^ 2 (A-3) - (A-3) = (a ^ 2-1) (A-3) = (A-1) (a + 1) (A-3) 3. According to the meaning of the question: x ^ 2-5x-6 = (x + 1) (X-6), that is: M = 64.4a ^ 2 + 4A + B ^ 2-6b + 10 = 0 (4a ^ 2 + 4A + 1) - 1 + (b ^ 2-6b + 9) - 9 + 10 = 0 (2a + 1) ^

How to factorize 3x ^ 2-6x + 2 = 0

3(x-1+√3/3)(x-1-√3/3)=0

Factorization of 3x ^ 3 + 6x ^ 4

3x^3+6x^4
=2x³(1+3x)

Three mathematical problems of factorization
1. (y + 1) (Y squared-1) - (y + 1) cubic
2. (xsquare-x) square + 0.5 (xsquare-x) + 1 / 16
3. A square - b square + a-b

1.-2(y+1)^2 2.(x^2-x+1/4)^2 3.(a+b+1)(a-b)