The square of (x-4) = (5-2x) (by factorization) And explain

(x-4)^2 =（5-2x)^2
(x-4) ^ 2 - (5-2x) ^ 2 = 0
(x-4 + 5-2x) (x-4-5 + 2x) = 0 square difference formula
(3x-1) = same class
3 (x-3) (1-x) = 0 extract the greatest common factor "3"
X-3 = 0 or 1-x = 0
So: X1 = 3, X2 = 1

Factorization: (2x-7) (2x + 5) (x ^ 2-9) - 91 Please answer as soon as possible

(2x-7)(2x+5)(x^2-9)-91 =(2x-7)(x+3)(2x+5)(x-3)-91=(2x^2-x-21)(2x^2-x-15)-91=(2x^2-x)^2-36(2x^2-x)+315-91 =(2x^2-x)^2-36(2x^2-x)+224 =(2x^2-x-8)(2x^2-x-28) =(2x+7)(x-4)(2x^2-x-8)

Square factorization of square-4 (x + 2Y) of 9 (2x-y)

9(2x - y)^2 - 4(x + 2y)^2
=(6x - 3y)^2 - (2x + 4y)^2
=(6x - 3y + 2x + 4y)(6x - 3y - 2x - 4y)
=(8x + y)(4x - 7y)
Hope to help you!

How to use factorization in 2x-5 Square-6 + 9 = 0

"2x-5" square - 6 + 9 = 0
（2x-5-3）²=0
（2x-8）²=0
X=4
If there is anything you don't understand, you can ask,

(xsquare-2x + 1 / 9) greater than 0

X^2-2X+1/9>0
(x-1)^2-8/9>0
|x-1|>2√2/3
10> 2 √ 2 / 3 + 1 or X

If the real number x satisfies x square = 9 2x-3 = 3-2x, find the value of X

x²=9
So x = - 3 or x = 3
|2x-3|=3-2x
- 2x3
The absolute value is the opposite of itself
So 2x-3 ≤ 0
x≤3/2
So x = - 3

Given 2x-3 = 0, find the value of the algebraic expression x (x 2-x) + x 2 (5-x) - 9

x（x2-x）+x2（5-x）-9，
=x（x2-x）+x2（5-x）-9，
=x3-x2+5x2-x3-9，
=4x2-9，
=（2x+3）（2x-3）．
When 2x-3 = 0, the original formula = (2x + 3) (2x-3) = 0

The square of X + 2x + 3 is 8. Find the square of 9 minus 2x minus 4x

x^2+2x=5
9-2x^2-4x
=9-2*（x^2+2x）
=9-10
=-1

If the square of X + X-1 = 0, then the value of the algebraic formula x to the third power + the square of 2x-7

x^3+2x^2-7
=x^3+x^2-x+x^2+x-1-6
=x(x^2+x-1)+(x^2+x-1)-6
=-6

If the square root of a positive number is 2x + 1 and X-7, what is the square root of x2-2x + 3?

From the meaning of the title: 2x + 1 + X-7 = 0,
The solution is: x = 2,
∴x2-2x+3=3，
The square root is ±
3．

The square of (x-4) = (5-2x) (by factorization) And explain