Factorization Practice Questions Put the questions in front and put the answers behind

1.((m+3n)square-12nm) divided by (m-3n)2.If the square of the polynomial 3x +7x-k has a factor of (3x+4) where k is constant, then k = hour.

3.After learning to use the square difference formula to decompose the factor, when completing the exercise assigned by the teacher, the nickname memorized a problem with the wrong symbol, he wrote down the square of -4x and the square of y

The square of the cubic - 2 * 2010 square - 2008

- semicolon) = cubic of 2010 + square of 2010 - square of 2011 = square of (x-2).

6.If the square of x + mx-n can be decomposed into (x-2) (x-5), then m= , n= .

1,a2-b2-2b-1 =a2-(b2+2b+1)=a2-(b+1)2

a+b+1)(a-b-1)

2,（xy+1）2-（x+y）2=(xy+1+x+y)(xy+1-x-y)

(xy+x)+(y+1)][xy-x)-(y-1)]=[x(y+1)+y+1)][x(y-1)-(y-1)]=(x+1)(y+1)(x-1)(y-1)3,x2-2x-4y2-4y= x2-4y2-2x-4y=(x+2y)(x-2y)-2(x+2y)=(x-2y-2)(x+2y)

4,（x2+2x）2-2（x2+2x）-3=(x2+2x+1)(x2+2x-3)

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

Landlord,The above process is quite detailed.,I don't believe you can't understand it.。。。

Don't forget to give points

Got it wrong! It should be (a+b) 2-4*b 2*c 2, right?

a+b)^2-4*b^2*c^2

a-b)^2

The title should be: (b + c) 4b c (otherwise the first bracket is redundant) = b 4 + c 4 + 2b c -4b c

b^4+c^4-2b²c²

b²-c²)²

(b+c)(b-c)]²

b+c)²(b-c)²

If the title is correct.

b-c+√2bc)(b-c-√2bc)

It's the root number.

The first: (x-y)[xy+z(x+y)].

The second: (a+2*b-1)*(3*a-b+1) the third: (x-2+y)*(x+2-y).

Fourth: (b+c)*(c+a)*(a+b) Is the landlord satisfied?