Factorization of a ^ 2 (a + 2b) ^ 2-9 (x + y) ^ 2 Q (P + Q) ^ 2-6 (P-Q) + 1 a^2（a+2b）^2-9（x+y）^2 q（p+q）^2-6（p-q）+1 Factorization

Factorization of a ^ 2 (a + 2b) ^ 2-9 (x + y) ^ 2 Q (P + Q) ^ 2-6 (P-Q) + 1 a^2（a+2b）^2-9（x+y）^2 q（p+q）^2-6（p-q）+1 Factorization

a^2（a+2b）^2-9（x+y）^2
=[a(a+2b)+3(x+y)][a(a+2b)-3(x+y)]
=(a^2+2ab+3x+3y)(a^2+2ab-3x-3y)
9（p+q）^2-6（p+q）+1
=[3(p+q)-1]^2
=(3p+3q-1)^2

Factorization p ^ 2 (P + Q) ^ 2-Q ^ 2 (P-Q) ^ 2 x ^ 2 (Y-1) + (1-y)
Factorization p ^ 2 (P + Q) ^ 2-Q ^ 2 (P-Q) ^ 2 x ^ 2 (Y-1) + (1-y)

p^2(p+q)^2-q^2(p-q)^2
=[p(p+q)+q(p-q)][p(p+q)-q(p-q)]
=(p^2+2pq-q^2)(p^2+q^2)
x^2(y-1)+(1-y)
=(y-1)(x^2-1)
=(x+1)(x-1)(y-1)

Factorization of [(x + P) - (x + Q)]

(x+p)-(x+q) =x+p-x-q =p-q

If the root P-2 and Q ^ 2-8q + 16 are opposite to each other, the result of factoring x ^ 2 + y ^ 2 - (PXY + Q) is?

Solution
√ P-2 and Q ^ 2-8q + 16 are opposite to each other
∴√p-2+(q^2-8q+16)=0
∴√p-2+(q-4)^2=0
∴p-2=0,q-4=0
∴p=2,q=4
∴x^2+y^2-(pxy+q)
=x^2+y^2-(2xy+4)
=(x^2+y^2-2xy)-4
=(x-y)^2-2^2
=(x-y+2)(x-y-2)

Factorization (2x + 3y-3) (2x + 3y-7) - 11
How does 15 in = (2x + 3y-5) & sup2; - 15 come from? And don't open a root!

You can make 2x + 3Y as a number
Original formula = (2x + 3Y) ^ 2-11 (2x + 3Y) + 21-11
= (2x+3y)^2-11(2x+3y)+10
=(2x+3y-1)(2x+3y-10)

2X ^ 2 + 5x-3 factorization and 2x ^ 2-5xy-3y ^ 2 process!

Cross multiplication
2x -1
x +3
2x^2+5x-3
=(2x-1)(x+3)
2x ^2-5xy-3y^2
(2x-y)(x+3y)
2x -y
x +3y

If the polynomial x ^ - ax + 1-A about X is a complete square, a =? (^ denotes Square)
If the polynomial x ^ - ax + 1 - a about X is a complete square, a =? (^ denotes square, please use this expression as well)

According to the meaning of the title, it can be set as follows:
x^2-ax+1-a=(x+t)^2
That is: x ^ 2-ax + 1-A = x ^ 2 + 2tx + T ^ 2
The comparison coefficient is: - a = 2T, 1-A = T ^ 2
Then: 1-A = (- A / 2) ^ 2
That is: A ^ 2 + 4a-4 = 0
The solution is: a = - 2 ± 2 √ 2

If the square + ax + 9 of polynomial x is a complete square, then the constant a =?
Solution! Only the answer is OK!

Let (x + m) ² = x & #178; + ax + 9
x²+2mx+m²=x²+ax+9
So 5m = a
m²=9
Then M = ± 3
So a = 2m = ± 6

Given the cubic power of the polynomial ax + BX + 4, when x = 2012, the value is 8, and when x = - 2012, what is the value?

f(x)=Ax^3+Bx+4
f(2012)-4=Ax^3+Bx=8
f(-2012)=-8+4=-4

When the value of X is - 1,1 and the value of polynomial AX2 + BX + 3 is 2,6 respectively, then a=______ ，b=______ ．

When x = - 1, A-B + 3 = 2, when x = 1, a + B + 3 = 6, the solution is a = 1, B = 2, so the answer is 1, 2