(x+a) (x+b)
x2+ab x+ab
x2+ (a+b)x+ ab
x2+(ab+b)x+ab
ax2+bx+ab
If one factor of 8 x3+27 y3+18 xy-1 is (2 x+3 y - 1) then the other is
4 x2+9 y2+1
4 x2+9 y2-1
4 x2+9 y2+1 -6 xy+3 y+2 x
4 x2+9 y2+1 +6 xy-3 y-2 x
Factors of x2+7 x+12 are
(x+5) (x+2)
(x+3) (x+4)
(x+6) (x+2)
(x+3)2
Factors of 9 x2-16 y2 are
(3 x+4 y)2
(3 x- 4 y)2
(3 x+4 y)(3 x- 4 y)
(3 x-8 y)2
Factorize 1- 8 x+16 x2
(1-4 x)2
(1+4 x)2
(2-4 x)2
(1-8 x)2
Which is the constant term of 5 x2+ 10 x+1
1
10
5
10x
(a+b)2- (a-b)2=
2 a2
2 b2
4 ab
8 ab
(x - 4) (x + 3) =
x2 + x - 12
x2 - 7 x - 12
x2 + 7 x + 12
x2 - x - 12
Splitting of x- term in 3 x2- 11 x + 6 is
-11, -1
-9,-2
-5,-6
-7,-4
x3+27 =
(x-3) (x2-3 x+9)
(x+3) (x2+3 x+9)
(x+3) (x2-3 x+9)
(x-3) (x2+3 x+9)
