Factorization 2003³-2×2003²-2001 ——————————— 2003³+2003²-2004 Yes + points·· Amen. You're all wrong. I don't think we've missed a factor. What's more, there's no 2001 / 2004 in the multiple-choice answers-

Factorization 2003³-2×2003²-2001 ——————————— 2003³+2003²-2004 Yes + points·· Amen. You're all wrong. I don't think we've missed a factor. What's more, there's no 2001 / 2004 in the multiple-choice answers-

Let 2003 = a primitive = [a ^ 3-2 * a ^ 2 - (A-2)] / [a ^ 3 + A ^ 2 - (a + 1)] = [a ^ 2 (A-2) - (A-2)] / [a ^ 2 (a + 1) - (a + 1)] = [(a ^ 2-1) (A-2)] / [(a ^ 2-1) (a + 1)] = (a + 1) (A-2) / (a + 1) (A-1) (a + 1) = (A-2) / (a + 1) = 2001 / 2004

A mathematical problem about factorization in grade two of junior high school
No matter what the value of X, y is, the value of polynomial x + y + 2x-4y + 5 is nonnegative

x^2+y^2+2x+4y+5
=x^2+2x+1+y^2+4y+4
=(x+1)^2+(y+2)^2
Therefore, the polynomial value is nonnegative
Note: ^ 2 means square

Factorization of 3N [x + y] - X-Y

3n【x+y】-x-y
=3n【x+y】-(x+y)
=(3n-1)(x+y)
If you don't understand this question, you can ask,

Factorization 1:2an-50a ^ 3N 2:4x (Y-X) - y ^ 2

2an-50a^3n
=2an(1-25a^2)
=2an(1-5a)(1+5a)
4x(y-x)-y^2
=4xy-4x^2-y^2
=-(4x^2+y^2-4xy)
=-(2x-y)^2

The 3nth power of 3 minus the nth power of a third of X (using factorization)

The title is like this
3^(3n)-(1/3)^n
=3^(3n)-3^(-n)
=3^n[3^(2n)-3^(-2n)]
=3^n[3^n+3^(-n)]*[3^n-3^(-n)]

①)a（x-1）+b（x-1）-c（1-x）.②10a（x-y)²-5b（y-x).③-ab（a-b)²+a（b-a)²

① (a) a (x-1) + B (x-1) - C (1-x) = (a + B + C) (x-1) ② 10A (X-Y) & # 178; - 5B (Y-X) = 5 (X-Y) (2A (X-Y) + b) = 5 (X-Y) (2ax-2ay + B) ③ - AB (a-b) & # 178; + a (B-A) & # 178; = - A (a-b) & # 178; (B-1); if you don't understand this question, you can ask

The square of Y - the square of 9 (x + y) is the problem of extracting the common factor

The original formula = y & # 178; - [3 (x + y)] &# 178;
=y²-(3x+3y)²
=(y+3x+3y)(y-3x-3y)
=-(3x+4y)(3x+2y)

(a-b) ^ + 10a-10b + 25 do it by factoring

(a-b)^+10a-10b+25
=(a-b)^2+2*5(a-b)+5^2
=(a-b+5)^2

Given the polynomial x-2x-3, when what is the value of X, the value of the polynomial is 12?

X2-2x-3 = 12 (X-5) (x + 3) = 0, x = 5 or - 3

Given that the value of polynomial (x + 5) / 2 is 2 greater than that of polynomial (2x - 1) / 3, find the value of X

(x+5)/2=(2x-1)/3+2
3(x+5)=2(2x-1)+12
3x+15=4x-2+12
-x=-5
x=5