Question

Factorise:
1) 12-(x+x²) (8-x-x²)
2) 9(x-y)²-(x+2y)²

Answer 1

do you think factorise completely ?

Answer 2

but first of all this is a math question please re-post it on the math subject - ty.

Answer 3

Moved to mathematics

Answer 4

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Vocaloid
Moved to mathematics
\(\color{#0cbb34}{\text{End of Quote}}\)
Alr

Answer 5

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jhonyy9
do you think factorise completely ?
\(\color{#0cbb34}{\text{End of Quote}}\)
Wym?

Answer 6

1.
\[12-(x+x^2)[8-(x+x^2)]\]
\[=12-8(x+x^2)+(x+x^2)^2\]
\[put~x+x^2=y\]
\[y^2-8y+12\]
\[=y^2-2y-6y+12\]
\[=y(y-2)-6(y-2)\]
\[=(y-2)(y-6)\]
\[=(x^2+x-2)(x^2+x-6)\]
\[=[x^2+2x-x-2][x^2+3x-2x+6]\]
\[=[x(x+2)-1(x+2)][x(x+3)-2(x+3)]\]
\[=(x+2)(x-1)(x+3)(x-2)\]
\[=(x-2)(x-1)(x+2)(x+3)\]
2.
\[9(x-y)^2-(x+2y)^2\]
\[=[3(x-y)]^2-(x+2y)^2\]
\[=(3x-3y)^2-(x+2y)^2\]
\[=[3x-3y+x+2y][3x-3y-x-2y]\]
\[=(4x-y)(2x-5y)\]

Answer 7

\(\color{#0cbb34}{\text{Originally Posted by}}\) @surjithayer
1.
\[12-(x+x^2)[8-(x+x^2)]\]
\[=12-8(x+x^2)+(x+x^2)^2\]
\[put~x+x^2=y\]
\[y^2-8y+12\]
\[=y^2-2y-6y+12\]
\[=y(y-2)-6(y-2)\]
\[=(y-2)(y-6)\]
\[=(x^2+x-2)(x^2+x-6)\]
\[=[x^2+2x-x-2][x^2+3x-2x+6]\]
\[=[x(x+2)-1(x+2)][x(x+3)-2(x+3)]\]
\[=(x+2)(x-1)(x+3)(x-2)\]
\[=(x-2)(x-1)(x+2)(x+3)\]
2.
\[9(x-y)^2-(x+2y)^2\]
\[=[3(x-y)]^2-(x+2y)^2\]
\[=(3x-3y)^2-(x+2y)^2\]
\[=[3x-3y+x+2y][3x-3y-x-2y]\]
\[=(4x-y)(2x-5y)\]
\(\color{#0cbb34}{\text{End of Quote}}\)
Thnx

Answer 8

yw