It has a repeating part made by commutative property.
[tex]\;[/tex]
We have [tex]\rm({x}^{2}+2x-3)({x}^{2}+2x-8)+4[/tex].
[tex]\;[/tex]
How did I do this? The commutative property as said. We can prioritize different factors. It's to observe two similar terms.
[tex]\boxed{\rm(x-1)(x+3)={x}^{2}+2x-3}[/tex]
[tex]\boxed{\rm(x-2)(x+4)={x}^{2}+2x-8}[/tex]
[tex]\;[/tex]
Observing a pattern is very important in algebra. We find a repeating part in both factors. Any variable you choose does not matter. I would assume [tex]\rm X={x}^{2}+2x[/tex] here.
(It's not your fault, so don't feel bad. Solving is a term for equations. Factorization is expressing polynomials in the form of products. I hope this helps.)
Answers & Comments
Verified answer
It has a repeating part made by commutative property.
[tex]\;[/tex]
We have [tex]\rm({x}^{2}+2x-3)({x}^{2}+2x-8)+4[/tex].
[tex]\;[/tex]
How did I do this? The commutative property as said. We can prioritize different factors. It's to observe two similar terms.
[tex]\boxed{\rm(x-1)(x+3)={x}^{2}+2x-3}[/tex]
[tex]\boxed{\rm(x-2)(x+4)={x}^{2}+2x-8}[/tex]
[tex]\;[/tex]
Observing a pattern is very important in algebra. We find a repeating part in both factors. Any variable you choose does not matter. I would assume [tex]\rm X={x}^{2}+2x[/tex] here.
[tex]\rm\Longrightarrow(x-1)(x-2)(x+3)(x+4)+4=\{(x-1)(x+3)\}\{(x-2)(x+4)\}+4[/tex]
[tex]\rm\Longrightarrow\{(x-1)(x+3)\}\{(x-2)(x+4)\}+4=({x}^{2}+2x-3)({x}^{2}+2x-8)+4[/tex]
[tex]\rm\Longrightarrow({x}^{2}+2x-3)({x}^{2}+2x-8)+4=(X-3)(X-8)+4[/tex]
[tex]\rm\Longrightarrow(X-3)(X-8)+4={X}^{2}-11X+28[/tex]
[tex]\rm\Longrightarrow{X}^{2}-11X+28=(X-4)(X-7)[/tex]
[tex]\rm\Longrightarrow(X-4)(X-7)=({x}^{2}+2x-4)({x}^{2}+2x-7)[/tex]
Hence the factorization ends here.
(It's not your fault, so don't feel bad. Solving is a term for equations. Factorization is expressing polynomials in the form of products. I hope this helps.)