How to use factorization to solve the square of 9x - 25 = 0?

How to use factorization to solve the square of 9x - 25 = 0?

9x²-25=0
(3x + 5) (3x-5) = 0 using the square difference formula
The solution is as follows
X = - 5 / 3 or x = 5 / 3

9x square - (x-1) square = 0

9x ^ 2-12x + 4 = 0 (3x-2) ^ 2 = 0 X1 = x2 = 2 / 3 because it is a quadratic equation with one variable, there will be two equal real roots, so we can't write only one solution. 9x square-12x + 4 = 0（

How to factorize the square of x minus 3 in the real range?

x²-﹙√3﹚²=﹙x+√3﹚﹙x-√3﹚

If the polynomial 8x ^ 2 + mxy-5y ^ 2 + XY-8 contains no XY term, then the value of M is?

According to the meaning of the problem, the coefficient of the term of XY is equal to 0, so m + 1 = 0, and the solution is m = - 1

(1) . x ^ 2-4y ^ 2 + X + 2Y (2). (X-Y) ^ 2-4 (x + Y-1) (3). X ^ n ^ + ^ 1-2x ^ n + x ^ n-1 (n is a positive number greater than 1) polynomial factorization

(1) . x ^ 2-4y ^ 2 + X + 2Y = (x + 2Y) (x-2y) + (x + 2Y) = (x + 2Y) (x-2y + 1) (2). (X-Y) ^ 2-4 (x + Y-1) = (X-Y) & # 178; - 4 (x + y) + 4 = (x-y-2) & # 178; (3). X ^ n ^ + ^ 1-2x ^ n + x ^ n-1 = x ^ n-1 (X & # 178; - 2x + 1) = x ^ n-1 (x-1) & # 178; if there is any ambiguity in this question, you can trace it

Factorization of 2x ^ 2-10. And x ^ 4-6x ^ 2 + 9

First question:
Original formula = 2 (x ^ 2-5)
=2（x+√5）（x-√5）
Second question:
Original formula = (x ^ 2-3) ^ 2
((x+√3)(x-√3))^2

Factorization: (x ^ 2-2x) ^ 2-11 (x ^ 2-2x) + 33-9

Original = (X & # 178; - 2x) & # 178; - 11 (X & # 178; - 2x) + 24
=(x²-2x)²-3(x²-2x)-8(x²-2x)+24
=(x²-2x-3)(x²-2x)-8(x²-2x-3)
=(x²-2x-8)(x²-2x-3)
=(x²-4x+2x-8)(x²+x-3x-3)
=[x(x-4)+2(x-4)][x(x+1)-3(x+1)]
=(x+2)(x-4)(x-3)(x+1)
Believe me, I'm right
If you learn cross multiplication, you don't have to be so troublesome!

Multiplication and division of integers,
（1）(x-y)(x+y)(x²+y²)
（2）（2x-3）²（2x+3）（2x-3）
（3）（6m²n-6m²n²-3m²）÷(-3m²)
(4)(5a²b-3ab-1)(-3a²)
（5）（-3x²y）³(-2xy³z)²

（1）(x-y)(x+y)(x²+y²)=(x²-y²)(x²+y²)=x⁴-y⁴（2）(2x-3)²(2x+3)(2x-3)=(2x+3)(2x-3)(2x-3)²=(4x²-9)(2x-3)²=(4x²-9)(4x²+9-12x)=(4x
...

1. - [(2 / 3x) tripartite]
2. (2x two times y three) square / (2x two times y three) square
Just write the final answer

1.64/729 * x hexagonal
The square of 2.4x times the hexagonal of Y

Multiplication and division of integers
If it is a real number, the value of the algebraic formula - 2x ^ 2 + 3x-2 is less than 0

-2x^2+3x-2
= -2[x^2-(3/2)x+1]
= -2[x^2-(3/2)x+9/16+7/16]
= -2[(x-3/4)^2+7/16]
Because the square number plus the positive number is always greater than 0, so the number (x-3 / 4) ^ 2 + 7 / 16 is always greater than 0. After multiplying by - 2, it is always less than 0, so the value of - 2x ^ 2 + 3x-2 is always less than 0