What is expand (n^4-2n^3+0n^2+2n-1)/(n^2-2n+1) ?

The solution to expand (n^4-2n^3+0n^2+2n-1)/(n^2-2n+1) is n^2-1

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expand (n^4-2n^3+0n^2+2n-1)/(n^2-2n+1)

expand

\frac{n ^{4} - 2 n ^{3} + 0 n ^{2} + 2 n - 1}{n ^{2} - 2 n + 1}

n^{2} - 1

Solution steps

\frac{n ^{4} - 2 n ^{3} + 0 \cdot n ^{2} + 2 n - 1}{n ^{2} - 2 n + 1}

n^{4} - 2 n^{3} + 0 \cdot n^{2} + 2 n - 1 = n^{4} - 2 n^{3} + 2 n - 1

= \frac{n ^{4} - 2 n ^{3} + 2 n - 1}{n ^{2} - 2 n + 1}

Factor

n^{4} - 2 n^{3} + 2 n - 1 : (n + 1) (n - 1)^{3}

= \frac{( n + 1 ) ( n - 1 ) ^{3}}{n ^{2} - 2 n + 1}

Factor

n^{2} - 2 n + 1 : (n - 1)^{2}

= \frac{( n + 1 ) ( n - 1 ) ^{3}}{( n - 1 ) ^{2}}

Cancel

\frac{( n + 1 ) ( n - 1 ) ^{3}}{( n - 1 ) ^{2}} : (n + 1) (n - 1)

= (n + 1) (n - 1)

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = n, b = 1

= n^{2} - 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= n^{2} - 1

\frac{2 x ^{4} - 8 x ^{2} - 10 + 3 x ^{3} - 9 x}{x ( x ^{2} + 1 ) ( x ^{2} - 5 )}

e^{3 x} (1 + 3 x)

(y = 3 \frac{,}{6} x - 5) (y = 25 x + 3)

\frac{d}{d t} ((x) (2 x^{2} + 1))

\frac{2 ( 216 - 216 x + 72 x ^{2} - 8 x ^{3} )}{27}

What is expand (n^4-2n^3+0n^2+2n-1)/(n^2-2n+1) ?
The solution to expand (n^4-2n^3+0n^2+2n-1)/(n^2-2n+1) is n^2-1