Respuesta :
I suppose that the expression you wanted to write is [tex] \frac{3+n-2n^2}{1+n} [/tex]. If so, then we have:
[tex]\frac{3+n-2n^2}{1+n}= \frac{3+3n-2n-2n^2}{1+n} \\ \\ = \frac{3(1+n)-2n(1+n)}{1+n} = \frac{(3-2n)(1+n)}{1+n} \\ \\ =3-2n[/tex]
[tex]\frac{3+n-2n^2}{1+n}= \frac{3+3n-2n-2n^2}{1+n} \\ \\ = \frac{3(1+n)-2n(1+n)}{1+n} = \frac{(3-2n)(1+n)}{1+n} \\ \\ =3-2n[/tex]
Answer:
Equivalent Polynomial of given polynomial is 3 - 2n.
Step-by-step explanation:
We are given following polynomial:
[tex]\frac{3+n-2n^2}{1+n}[/tex]
We have top find equivalent polynomial such that n ≠ -1
Consider,
[tex]\frac{3+n-2n^2}{1+n}[/tex]
[tex]=\frac{3+3n-2n-2n^2}{1+n}[/tex]
[tex]=\frac{3(1+n)-2n(1+n)}{1+n}[/tex]
[tex]=\frac{(1+n)(3-2n)}{1+n}[/tex]
[tex]=3-2n[/tex]
Therefore, Equivalent Polynomial of given polynomial is 3 - 2n.