Square of factorization factor (x's square + 3x) - 8 (x's square + 3x) - 20

Square of factorization factor (x's square + 3x) - 8 (x's square + 3x) - 20

Square of (x + 3x) - 8 (x + 3x) - 20
=(square of X + 3x-10) (square of X + 3x + 2)
=（x+5)(x-2)(X+1)(X+2)

(x squared + 3x) squared - 8 (x squared + 3x) - 20, factorization is required
I've already done it. I don't need it,

Consider x square + 3x as a variable t
Namely
T square-8t-20
=(t-10)(t+2)
=(xsquare + 3x-10) (xsquare + 3x + 2)
=(x-2)(x+5)(x+1)(x+2)

The number of proper subsets of set {x ∈ n | 0 ＜ X-1 | 4} is ()
A. 32B. 31C. 16D. 15

Set {x ∈ n | 0 ＜ X-1 ＜ 4} = {x ∈ n | 0 ＜ X-1 ＜ 4, or - 4 ＜ X-1 ＜ 0} = {0, 2, 3, 4}. The number of all subsets is 24 = 16, so the number of proper subsets is 15, so the answer is d

2b²c²+2c²a²+2a²b²-a^4-b^4-c^4
2x²+5xy-4y²
（y+1)^4+(y+3)^4-272
(xy-1)²+（x+y-2)(x+y-2xy)
x^4-x^3+4x²+3x+5

2b²c²+2c²a²+2a²b²-a^4-b^4-c^4
=-（a-b-c）（a+b-c）（a-b+c）（a+b+c）
2x²+5xy-4y²
This expression cannot be decomposed
（y+1)^4+(y+3)^4-272
=2（y-1）（y+5）（y^2+4y+19）
(xy-1)²+（x+y-2)(x+y-2xy)
=(x-1)^2(y-1)^2
x^4-x^3+4x²+3x+5
(x^2-2x+5)(x^2+x+1)
~~~~~~~~~~~Please accept~~~~~~~~~~~

About factorization
This is what the book says
(a+b)^2+2(a+b)+1
=[(a+b)+1]^2
=(a+b+1)^2
Especially from the topic to the explanation of the first step

Consider (a + b) as X, (a + b) = X
(a+b)^2+2(a+b)+1
=x^2+2x+1
=(x+1)^2
Complete square formula
（a+b)^2+2(a+b)+1
=(a+b+1)^2

On factorization
1. X ^ n + 2x ^ (n + 1) + x ^ (n + 2) decompose it into factors
2. (x + 1) (x + 2) (x + 3) (x + 4) + 1 decompose it into a factor. Pay attention to the processing of the result. My result is (x ^ 2 + 5x + 5) ^ 2, but it's not right. It seems that I have to calculate the result in brackets according to the discriminant B ^ 2-4ac. What should the final result be?
Ax ^ 2 + BX + C = a (x-m) (x-n) why?

1. X ^ n + 2x ^ (n + 1) + x ^ (n + 2) = x ^ n (1 + 2x + x ^ 2) = x ^ n (1 + x) ^ 22.. (x + 1) (x + 2) (x + 3) (x + 4) + 1 = (x ^ 2 + 5x + 4) (x ^ 2 + 5x + 6) + 1 = (x ^ 2 + 5x + 4) ^ 2 + 2 (x ^ 2 + 5x + 4) + 1 = (x ^ 2 + 5x + 5) ^ 2 yes, it has nothing to do with the discriminant B ^ 2-4ac, if the solution in the range of real number is also equal to [x - (- 5 +... "

Problems related to factorization
After the factorization of 2x ＾ 3-x ^ 2 + MX + N, there is a factorization of x ^ 2 + X + 6 and the value of M, n

Let (x ^ 2 + X + 6) (AX + b) = 2x ^ 3-x ^ 2 + MX + n
Then:
ax^3 + (a+b)x^2 +(6a+b)x + 6b = 2x^3 -x^2 +mx+n
The corresponding coefficients on both sides are equal
From the cubic coefficient, a = 2
At this point,
2+b=-1,12+b=m,6b=n
b=-3,
m=9.n=-18

Factorization
It is known that m and N are positive integers, and m (m-n) - (n-m) = 12 to find the value of M and n

M (m-n) - (n-m) = 12, then (M + 1) (m-n) = 12, because 12 = 1 × 12 = 3 × 4 = 2 × 6 and M + 1 ＞ M-N, there are three cases: ① m + 1 = 12, M-N = 1, M = 11, n = 10; ② m + 1 = 4, M-N = 3, M = 3, n = 0; ③ m + 1 = 6, M-N = 2, M = 5, n = 3, M = 11, n = 10 or M = 5, n = 3

On factorization
(1) (19x-31) (13x-17) - (13x-17) (11x-23) can be factorized into (AX + b) (8x + C), where ABC is an integer and a + B + C =?
(2) What is the result of factoring the polynomial ax squared-ax-2a?
(3) If M + n = 3, what is the value of 2m square + 4Mn + 2n square - 6?

(19X-31）（13X-17)-(13X-17）（11X-23)
=（13X-17)（19x-31-11x+23）
=(13x-17)(8x-8)
So a = 13, B = - 17, C = - 8
abc =1768
ax²-ax-2a
=a(x²-x-2)
=a(x-2)(x+1)
2m²+4mn+2n²-6
=2(m²+2mn+n²)-6
=2(m+n)²-6
=2×3²-6
=12

On factorization
Find the factorization result of [(x ^ 2 + y ^ 2) (a ^ 2 + B ^ 2) + 4abxy] ^ 2-4 [XY (a ^ 2 + B ^ 2) + AB (x ^ 2 + y ^ 2)] ^ 2

[(x ^ 2 + y ^ 2) (a ^ 2 + B ^ 2) + 4abxy] ^ 2-4 [XY (a ^ 2 + B ^ 2) + AB (x ^ 2 + y ^ 2)] ^ 2 let X & sup2; + Y & sup2; = m, a & sup2; + B & sup2; = n, xy = P, ab = q, then the original formula is (nm + 4pq) & sup2; - 4 (PN + QM) & sup2; = (nm + 4pq + 2pn + 2qm) (nm + 4pq-2pn-2qm) = [n (M + 2P) + 2q (M + 2P)] [n (