All by factorization 3x [X-1] = 2 [1-x] [x + 1] ^ 2-25 = 0 16 [X-2] ^ 2 = 9 [x + 3] ^ 2

All by factorization 3x [X-1] = 2 [1-x] [x + 1] ^ 2-25 = 0 16 [X-2] ^ 2 = 9 [x + 3] ^ 2

3x[x-1]=2[1-x] 3x(x-1)=-2(x-1) 3x(x-1)+2(x-1)=0 ( 3x+2)(x-1)=0 x1=-3/2,x2=1 [x+1]^2-25=0 (x+1+5)(x+1-5)=0 (x+6)(x-4)=0 x1=-6,x2=4 ...

Factorization of x ^ 2-9 = 3 (x + 3)

(x+3)(x-3)=3(x+3)
(x+3)(x-3)-3(x+3)=0
(x+3)(x-3-3)=0
(x+3)(x-6)=0
x=-3 x=6

The factorization shows that 3 ^ 10-4 * 3 ^ 9 + 10 * 3 ^ 8 must be divisible by 7

3^10 - 4×3^9 + 10×3^8
= 3^8×(3^2 - 3 + 1)
= 3^8×7
Because 3 ^ 8 is an integer
So 3 ^ 8 × 7 must be divisible by 7
So 3 ^ 10 - 4 × 3 ^ 9 + 10 × 3 ^ 8 must be divisible by 7