Maths(F.2)

1(a). Prove that (m+1)^2 -3m-1=m^2-m is an identity

(m+1)^2 -3m-1=m^2-m
LHS:(m+1)^2 -3m-1
=m^2+2m+1-3m-1
=m^2-m
=RHS

1(b). By substituting m=x-1 into the identity in (a), find the unknown constants A and B in x^2 - 3x + 2 identity to (Ax-B)(x-2).

m^2-m
=(x-1)^2-(x-1)
=x^2-2x+1-x+1
=x^2-3x+2
=(x-2)(x-1)

comparing the coefficient,
A=1 B=1

1(a). Prove that (m+1)^2 -3m-1=m^2-m is an identity.
L.H.S. = (m+1)^2 -3m-1
=m^2+2m+1-3m-1
=m^2-1
=R.H.S.

1(b). By substituting m=x-1 into the identity in (a), find the unknown constants A and B in x^2 - 3x + 2 identity to (Ax-B)(x-2).

sub m=x-1 into (m+1)^2 -3m-1=m^2-m
[(x-1)+1]^2 -3(x-1)-1=(x-1)^2-(x-1)
x^2 -3x+2 = x^2-2x+1-x+1
(Ax-B)(x-2)=x^2-3x+2
Ax^2-Bx-2Ax+2B=(x-1)(x-2)
Ax^2-Bx-2Ax+2B=x^2-3x+2
A=1 and -B-2A=-3
B+2=3
B=1

A=1, B=1

2007-03-02 22:27:43 補充：
1(b) :since this is an identity, the coefficients on each side is equal.