Middle School 8th Grade Math Olympiad about factorization

4(x+5)(x+6)(x+10)(x+12)－3x^2

4*[(x+5)(x+12)][x+6)(x+10)] 3x^2

4*(x^2 + 60 + 17x)(x^2 + 60 + 16x) -3x^2

4*[(x^2+60)^2 + 33x(x^2+60) +272x^2)] 3x^2

4(x^2+60)^2 + 132x(x^2+60) +1085x^2

2(x^2 +60) +35x][2(x^2 +60) +31x]

2x^2 + 35x + 120)(2x^2 + 31x + 120)

2x^2 + 35x + 120)(2x + 15)(x+8)

xy(x^2-y^2)+yz(y^2-z^2)+zx(z^2-x^2)

x^3y-xy^3+y^3z-yz^3+z^3x-zx^3

x^3(y-z)+y^3(z-x)+z^3(x-y)

Because x-y = (x-z) + (z-y), so:

x^3(y-z)+y^3(z-x)+z^3[(x-z)+(z-y)]

x^3-z^3)(y-z)+(y^3-z^3)(z-x)

x-z)(x^2+xz+z^2)(y-z)+(y-z)(y^2+yz+z^2)(z-x)

x-z)(y-z)(x^2+zx+z^2-y^2-yz-z^2)

x-z)(y-z)(x-y)(x+y+z)

x^2+3xy-2y^2-x+8y-6

x+2y)(2x-y)-x+8y-6

x+2y)(2x-y)-(x-8y)-6

x+2y-2)(2x-y+3) (cross multiplication).

The pending coefficient decree (x+2y)(2x-y)-(x-8y)-6=(x+2y+a)(2x-y+b) } can also be used

(y-2)(y-15)=0 (y-12)(y+5)=0 (x-14)(x+2)=0

The first three are cross multiplication.

The fourth one opens 5x2-44x+32=0 and then crosses multiplication (x-8)(5x-4)=0

The fifth one is the same (x+7)(7x+5)=0

Actually, I won't either, hahahahahaha.

1²-0²=1+0=1；

.(1) Please express the observed law in the formula containing n:

n^2-(n-1)^2 = n + n-1) = 2n-1

2) Is your conjecture correct? Use the relevant knowledge to illustrate.

According to the squared difference formula:

n 2-(n-1) 2 = =(2n-1)*1=2n-1, which is consistent with the above conjecture.

3) According to the formula of square difference, multiply (2-1) in front of the equation to obtain:

2-1) (2 + 1) (2 + 1) (2 to the fourth power + 1) (2 to the eighth power + 1) (2 to the sixteenth power + 1).

2 -1) (2 + 1) (2 to the fourth power + 1) (2 to the eighth power + 1) (2 to the sixteenth power + 1).

2 to the fourth power - 1) (2 to the fourth power + 1) (2 to the eighth power + 1) (2 to the sixteenth power + 1).

2 to the eighth power - 1) (2 to the eighth power + 1) (2 to the sixteenth power + 1).

2 to the sixteenth power - 1) (2 to the sixteenth power + 1).

2 of the thirty-second - 1

1.The square of n - the square of (n-1) = 2n-12The square of n - the square of (n-1).

n+n-1)(n-n+1) – square difference formula = 2n-1

Your little number is 、、、

n+1)^2 - n^2 = (n+1) +n =2n+1=(n+1+n)(n+1-n)

The conjecture is obviously correct, because (n+1) 2 = (n+1)*(n+1)=n 2+2n+1

Actually, there is such a formula.

a^2-b^2=(a+b)(a-b)

n^4-1=(n^2+1)(n^2-1)=(n^2+1)(n+1)(n-1)

1.Extracting the common factor x-y then the above equation is equal to x-y (x-y+y) =x* x-y

2.b 2-2b+1 is a perfect square (b-1) 2, which is connected to a-(b-1))*a+(b-1))=(a-b+1)*(a+b-1).

3.is a perfect square (a-1-2) 2=(a-3) 2

1.Original = (x+y)(x+y+y)=(x+y)(x+2y)2Original formula = a - (b-1) = (a + b - 1) (a - b + 1) 3

Original = [(a-1)-2] = (a-3) Pay attention to the common factor and make use of the perfect flat method.

You won't be able to! Kid us too! You just. Just put one of the same.

First of all, the question you typed should not be x 2 + 2 x y + y 2, x 2-y 2, if so, then the solution is as follows:

Because x= 3+1 and y= 3-1, x+y=2 3

x-y=2, so (1) x 2+2xy+y 2

x+y)^2

12(2) x^2-y^2

x+y)(x-y)

And C got stuck, indigestion must be considered, and there was a panic in the southwest. Look, it's not good for the child not to cry, f

Decomposition factor m -n +5m+5n = (m+n)(m-n+5) m -n = n+3-m-3

m+n)(m-n)=-(m-n)

m+n = -1

m cubic - 2mn + n cubic = m * m - 2mn + n * n = m (n + 3) - 2 mn + n (m + 3).

mn+3m-2mn+mn+3n

3m+3n3(m+n)=-3

Decompose the factor m -n +5m+5n=

m²-n²)+5m+5n)

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

Finally (m+n)(m-n+5).

Knowing m = n+3, n = m+3, and m ≠n asks: find the value of m+n process.

m²-n²= (n+3-m-3)

m+n)(m-n)=-(m-n)

m+n = -1

Find the value of m to the third power - 2mn + n to the third power process.

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

2, m^2-n^2=n+3-(m+3)

m-n)(m+n)=n-m

m and n are not equal.

So you can make an appointment with m-n

m+n= -1

m^3-2mn+n^3

m+n)^3-2m^2n-2mn^2-2mn= (m+n)^3-2mn(m+n+1)

Got it that m+n= -1

So the end result is 1

I'll give you two of these questions first.

2(x+y)^3-8(x+y)=2（x+y)（x+y+2)（x+y-2)

16x 4-8x 2+1=(2x+1) 2(2x-1) 2 The first question seems to be wrong, right, do you still need answers for the rest of the questions??

a-b or b-a differs only by one symbol, turning b-a into -(a-b).

Extract the common factors (a-b) as.

a-b)(a^2-b^2)

Look at the square difference fraction to decompose a 2-b 2 = (a + b) (a - b) i.e. a 2 (a - b) + b 2 (b - a).

a^2(a-b)-b^2(a-b)

a-b)(a^2-b^2)

a-b)[(a+b)(a-b)]

a+b)(a-b)^2