What is factor-2k^5+k^{10}+1 ?

The solution to factor-2k^5+k^{10}+1 is (k-1)^2(k^4+k^3+k^2+k+1)^2

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factor-2k^5+k^{10}+1

factor

- 2 k^{5} + k^{10} + 1

(k - 1)^{2} (k^{4} + k^{3} + k^{2} + k + 1)^{2}

Solution steps

- 2 k^{5} + k^{10} + 1

Write in the standard form

= k^{10} - 2 k^{5} + 1

Use the rational root theorem

= (k - 1) \frac{k ^{10} - 2 k ^{5} + 1}{k - 1}

\frac{k ^{10} - 2 k ^{5} + 1}{k - 1} = k^{9} + k^{8} + k^{7} + k^{6} + k^{5} - k^{4} - k^{3} - k^{2} - k - 1

= (k - 1) (k^{9} + k^{8} + k^{7} + k^{6} + k^{5} - k^{4} - k^{3} - k^{2} - k - 1)

Factor

k^{9} + k^{8} + k^{7} + k^{6} + k^{5} - k^{4} - k^{3} - k^{2} - k - 1 : (k^{4} + k^{3} + k^{2} + k + 1) (k - 1) (k^{4} + k^{3} + k^{2} + k + 1)

= (k - 1) (k^{4} + k^{3} + k^{2} + k + 1) (k - 1) (k^{4} + k^{3} + k^{2} + k + 1)

Simplify

= (k - 1)^{2} (k^{4} + k^{3} + k^{2} + k + 1)^{2}

\frac{( cos ^{2} ( a ) )}{( sin ( a ) ) + sin ( a )}

0 + 5 x - (\frac{1}{2}) (9.81) x^{2}

(3 x^{5})^{3} (2 x^{7})^{2}

x^{2} - y^{2} + 2 x + 1

cos^{2} (a) - sin^{2} (b)

What is factor-2k^5+k^{10}+1 ?
The solution to factor-2k^5+k^{10}+1 is (k-1)^2(k^4+k^3+k^2+k+1)^2