If the two real roots of square + BX + C = 0 of ax are x1x2, then the square + BX + C of quadratic trinomial ax can be decomposed into

If the two real roots of square + BX + C = 0 of ax are x1x2, then the square + BX + C of quadratic trinomial ax can be decomposed into

ax^2＋bx＋c＝a(x－x1)(x－x2)

If the quadratic trinomial AX2 + BX + C is a monomial with respect to X ()
A. a≠0，b=0，c=0B. a=0，b≠0，c=0C. a=0，b=0，c≠0D. a=0，b=0，c=0

One time monomial is the one with the number of times 1, so the conditions that meet the meaning of the question should be a = 0, B ≠ 0, C = 0

Given a quadratic monomial x square + 2x + 3, multiply it with a quadratic term ax + B, the product does not appear a quadratic term, and the coefficient of the quadratic term is 1, find the value of a, B

The n-th power of the square * x of (M + 1) is the binomial of X. find m and n

(M + 1) & sup2; * x ^ n is the binomial of X, and m and N are obtained
From the meaning of the title, it is concluded that:
(m+1)²≠0
n=2
The solution is as follows:
m≠-1
n=2
Note: this problem can only be solved here, or the condition is missing;

Factorization-4a ^ 2B + 2Ab ^ 2-12ab

-2ab(2a-b+6)

-4a^3+16a^2b-26ab^2 x^5-x^3 (x^2+y^2)^2-4x^2y^2 25(x+y)^2+10(x+y)+1 a*(-a^2)^3*(a^3)^2

-4a^3+16a^2b-26ab^2
=-2a(2a^2-8ab+13b^2)
x^5-x^3
=x^3(x^2-1)
=x^3(x-1)(x+1)
(x^2+y^2)^2-4x^2y^2
=(x^2+y^2+2xy)(x^2+y^2-2xy)
=(x+y)^2(x-y)^2
25(x+y)^2+10(x+y)+1
=[5(x+y)+1]^2
=(5x+5y+1)^2
a*(-a^2)^3*(a^3)^2
= a*(-a^6)*(a^6)
=-a^13

-16a^2b^2c÷(1/4a^2b)=

6x-1 = 2x = 1 / 2Y & sup2; - Y-2 = 0 (Y-2) (y + 1) = 0y = 2, y = - 1 (5x & sup2; y + 5xy-7x) - (4x & sup2; y + 5xy-7x) = 5x & sup2; y + 5xy-7x-4x & sup2; y-5xy + 7x = x & sup2; y = (1 / 2) & sup2; × 2 or (1 / 2) & sup2; × (- 1) = 1 / 2 or - 1 / 4

It is known that AB is not equal to 1, a, B satisfy a square + 4A + 2 = 0, 2b square + 4B + 1 = 0, then a cube + B cube is 1 / 2=

a^2+4a+2=0
2b^2+4b+1=0 （1）
b≠0
(1) B ^ 2
2+4*1/b+1/b^2=0
(1/b)^2+4/1/b+2=0
a. B can be regarded as two roots of the equation x ^ 2 + 4x + 2 = 0
a+1/b=-4,a*1/b=2
a^3+1/b^3=(a+1/b)(a^2-a*1/b+1/b^2)
=(a+1/b)[(a+1/b)^2-3a*1/b]
=-4*[(-4)^2-3*2]
=-4*[16-6]
=-4*10
=-40

A + B + C = 6, A-B + C = 0, 4A + 2B + C = 12, how much a, B, C is equal to, write the process

Because a + B + C = 6 (1) A-B + C = 0 (2) 4A + 2B + C = 12 (3)
(1) + (2) gives 2 (a + C) = 6
So a + C = 3 (4)
Substituting (4) into (1) yields B = 3
Substituting B = 3 into (3) yields 4A + C = 3 (5)
(5) (4) a = 1
Substituting a = 1D into (3) yields C = 2
So a = 1, B = 3, C = 2

(3b) ^ 2 - (AB) ^ 2 (- 4A ^ 2) ^ 3 (y ^ 2Z ^ 3) ^ 3 (XY ^ 4) ^ m - (P ^ 2q) ^ n (- 3x ^ 3) ^ 2 - [(2x) ^ 2] ^ 3 has a process

(3b）^2 -（ab）^2 =9b^2 -a^2b^2=(9-a^2)b^2=(3+a)(3-a)b^2(-4a^2)^3+（y^2z^3）^3 =（y^2z^3）^3-(4a^2)^3=（y^2z^3 -4a^2)(y^4z^6 +4a^2y^2z^3+16a^4)（xy^4）^m -（p^2q）^n =x^my^4^m -p^2nq^n（-3x^3）^2-[（2...