Other methods of factoring the first 2

4m squared minus 9n squared minus 4m plus 1

4m(m-1)-(3n+1)(3n-1)2a squared plus 2b squared + 2c squared minus 2ab minus 2bc minus 2ac = (a-b) 2+(b-c) 2+(a-c) 2

=2012²-2002×2012+2000×2002-2000×2012

2012 (2012-2002) + 2000 (2002-2012) = (2012-2002) (2012-2000) The factorization ends here.

1)2x(y-z)+3(z-y)

2x-3)(y-z)

2) 9 (x+y) squared - 4 (x-y) squared.

3(x+y)+2(x-y))(3(x+y)-2(x-y))(5x+y)(x+5y)

3) (a-b) squared - 10 (a-b) + 25 (a-b-5) 2

4) 2nd power of AB + 2nd power of 6a B + 3rd power of 9a.

a(b^2+6ab+9a^2)

a(b+3a)^2

5) x to the 2nd power - 5x-14

x-7)(x+2)

6) x to the 2nd power - 1 + 2 x y + y squared.

x+y)^2-1

x+y+1)(x+y-1)

1.Knowing that the excitation difference of a to the power of m is equal to 3, and the power of a is equal to 2, then the value of m+n+2 of a is the value of lead auspicious?

a^(m+n+2)=a^m*a^n*a^2=3*2*a^2=6a^22.Knowing that m+m 1=5, then the value of the 2nd power of m + the 2nd power of m is the value of 1 of the wax skin?

m^2+(1/m)^2=(m+1/m)^2-2=5^2-2=23x(3x+1)^2

xy(25x^2-20xy+4y^2)

xy(5x-2y)^2

2(x+4)(x-4)

4x+3y)(4x-3y)

a+4b-2c)(a-4b+2c)

a^2-(4b-2c)^2

a^2-16b^2+16bc-4c^2

Multiply them out, a person has the right coefficient of the first term, and a person has the right coefficient of the second term, understand?

The first one, ab-bc-ac is already the simplest, and it can't be simplified anymore, and it can't be factored anymore.

=a²(x-y)-b²(x-y)

x-y)(a²+n²

4m²-b²-(3b)²

a+3b)(a-3b)

Solution (2) -16m + (m-n).

m-n)²-16m²

m-n)²-4m)²)4m²-n²-4(2x-y)²=[(m-n)+4m][(m-n)-4m]=(5m-n)(-3m-n)

(5m-n)(3m+n)

Solution (3) (x-2y).

x-y)(a+b)(a-b)

Solution (5) 16m 4-n 4 means power = (4m )-n )

x-2y)²-2(2x-y)]²

(x-2y)+2(2x-y)][x-2y)-2(2x-y)]=(x-2y+4x-2y)(x-2y-4x+2y)=-3x(5x-4y)

Solution (4) A Solution (1) A -9b