form 2 maths

If(x+A)(Bx-1)≡2(x^2-1)+Cx,find the values of A,B,C

(x+A)(Bx-1)≡2(x^2-1)+Cx
Bx^2+ABx-x-A≡2x^2+Cx-2
Compare the codfficients of both sides, we have
B=2
AB-1=C
-A=-2

therefore A=2, B=2 and C=AB-1=2*2-1=3

If x^2+y^2=17 and xy=7.5,find the values of
a) (x+y)^2
b) (x-y)^2
c) x^4-y^4

a)(x+y)^2=x^2+2xy+y^2
=(x^2+y^2)+2xy
=17+2*7.5
=17+15
=32

b)(x-y)^2=x^2-2xy+y^2
=(x^2+y^2)-2xy
=17-2*7.5
=17-15
=2

c)x^4-y^4 =(x^2+y^2)(x^2-y^2)
=(x^2+y^2)(x+y)(x-y)
=17(32^1/2)(2^1/2) <--by (a)x+y=32^1/2 and (b) x-y=2^1/2
=17(64^1/2)
=17*8
=136

2.a) Expand (x-1)^2
b) Hence,expand (x-1+y)^2
C)If (x-1+y)^2≡Ax^2+Bxy+Cy^2+Dx+Ey+F,find A,B,C,D,E and F

a)(x-1)^2=x^2-2x+1

b)(x-1+y)^2=(x-(1-y))^2
=x^2-2(1-y)x+(1-y)^2
=x^2-2x+2xy+1-2y+y^2
=x^2+2xy+y^2-2x-2y+1

c)(x-1+y)^2≡Ax^2+Bxy+Cy^2+Dx+Ey+F
x^2+2xy+y^2-2x-2y+1≡Ax^2+Bxy+Cy^2+Dx+Ey+F (by the result of (b))
Compare the coefficients, we have
A=1
B=2
C=1
D=-2
E=-2
F=1

Lefthand=(x+A)(Bx-1)
=Bx^2 -x +ABx -A
=2(Bx^2/2 -A/2) +ABx
Righthand=2(x^2 -1) +Cx
So B=1/2,A=2,C=AB=1

x^2+y^2=17 and xy=7.5,???

Expand (x-1)^2

(x-1)^2
=x^2 -2x +1
(a-b)^2= a^2 -2ab +b^2!!!


Hence,expand (x-1+y)^2

(x-1+y)^2
=[(x-1)+y]^2
=(x-1)^2 +2y(x-1) +y^2
=(x^2 -2x +1) +2xy-2y +y^2
=x^2 +2xy +y^2 -2x -2y +1

If (x-1+y)^2≡Ax^2+Bxy+Cy^2+Dx+Ey+F,find A,B,C,D,E and F

x^2 +2xy +y^2 -2x -2y +1≡Ax^2+Bxy+Cy^2+Dx+Ey+F
A=1,B=2,C=1,D= -2,E= -2,F=1