Given that a and B are two of the equations x2 + (M + 2) x + 1 = 0, then the value of (A2 + Ma + 1) (B2 + MB + 1) is______ ．

Given that a and B are two of the equations x2 + (M + 2) x + 1 = 0, then the value of (A2 + Ma + 1) (B2 + MB + 1) is______ ．

∵ a, B are two of the equations x2 + (M + 2) x + 1 = 0, ∵ a + B = - (M + 2), ab = 1, A2 + (M + 2) a + 1 = 0, B2 + (M + 2) B + 1 = 0, ∵ A2 + 1 = - (M + 2) a, B2 + 1 = - (M + 2) B, ∵ A2 + Ma + 1) (B2 + MB + 1) = [- (M + 2) a + Ma] [- (M + 2) B + MB] = (- 2A) · (- 2b) = 4AB =

a. B is the two roots of the equation x ^ 2 + (m-5) x + 7, then (a ^ 2 + Ma + 7) (b ^ 2 + MB + 7)=

a+b=5-m;
ab=7;
(a^2+ma+7)(b^2+mb+7)
=(0+5a)(0+5b)
=25ab
=25×7
=135;
If you don't understand this question, you can ask,

Factorization (M + n) ② - 4 (M + n-1)
② Square

＝(m+n)②-4(m+n）+4
＝(m+n-2)②

How does factorization m (M + n) (n-m) - n (M + n) (m-n) decompose

First (m-n) is transformed into - (n-m), and then the common factor is extracted
m(m+n)(n-m)-n(m+n)(m-n)
=m(m+n)(n-m)+n(m+n)(n-m)
=(m+n)(n-m)(m+n)
=(m+n)^2(n-m)