What is factor 1+128x^{14} ?

The solution to factor 1+128x^{14} is (2x^2+1)(64x^{12}-32x^{10}+16x^8-8x^6+4x^4-2x^2+1)

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factor 1+128x^{14}

factor

1 + 128 x^{14}

(2 x^{2} + 1) (64 x^{12} - 32 x^{10} + 16 x^{8} - 8 x^{6} + 4 x^{4} - 2 x^{2} + 1)

Solution steps

1 + 128 x^{14}

Write in the standard form

= 128 x^{14} + 1

Let

u = x^{2}

= 128 u^{7} + 1

Factor

128 u^{7} + 1 : (2 u + 1) (2^{6} u^{6} - 2^{5} u^{5} + 2^{4} u^{4} - 2^{3} u^{3} + 2^{2} u^{2} - 2 u + 1)

= (2 u + 1) (2^{6} u^{6} - 2^{5} u^{5} + 2^{4} u^{4} - 2^{3} u^{3} + 2^{2} u^{2} - 2 u + 1)

Substitute back

u = x^{2}

= (2 x^{2} + 1) (2^{6} x^{12} - 2^{5} x^{10} + 2^{4} x^{8} - 2^{3} x^{6} + 2^{2} x^{4} - 2 x^{2} + 1)

Simplify

(2 x^{2} + 1) (2^{6} x^{12} - 2^{5} x^{10} + 2^{4} x^{8} - 2^{3} x^{6} + 2^{2} x^{4} - 2 x^{2} + 1) : (2 x^{2} + 1) (64 x^{12} - 32 x^{10} + 16 x^{8} - 8 x^{6} + 4 x^{4} - 2 x^{2} + 1)

= (2 x^{2} + 1) (64 x^{12} - 32 x^{10} + 16 x^{8} - 8 x^{6} + 4 x^{4} - 2 x^{2} + 1)

(5 x + 2 y) + (3 x - 5 y) - (6 x + 4 y)

(\frac{( 3 a ^{2} )}{( a ^{- 2} \cdot b ^{3} \cdot c ^{2} )})^{- 2}

a \cdot a^{\frac{1}{3}}

(x - 5) (250000 x + 9)

r^{2} x cos (f) \cdot (2 - r^{2})

What is factor 1+128x^{14} ?
The solution to factor 1+128x^{14} is (2x^2+1)(64x^{12}-32x^{10}+16x^8-8x^6+4x^4-2x^2+1)