What is factor 16(x^2+1)^2-4(2x)^2 ?

The solution to factor 16(x^2+1)^2-4(2x)^2 is 16(x^2+x+1)(x^2-x+1)

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factor 16(x^2+1)^2-4(2x)^2

factor

16 (x^{2} + 1)^{2} - 4 (2 x)^{2}

16 (x^{2} + x + 1) (x^{2} - x + 1)

Solution steps

16 (x^{2} + 1)^{2} - 4 (2 x)^{2}

Factor out common term

4 : 4 (4 (x^{2} + 1)^{2} - (2 x)^{2})

= 4 (4 (x^{2} + 1)^{2} - (2 x)^{2})

Factor

4 (x^{2} + 1)^{2} - (2 x)^{2} : 4 (x^{2} + x + 1) (x^{2} - x + 1)

= 4 \cdot 4 (x^{2} + x + 1) (x^{2} - x + 1)

Multiply the numbers:

4 \cdot 4 = 16

= 16 (x^{2} + x + 1) (x^{2} - x + 1)

\frac{( x - 2 )}{3} < 1

(\frac{3}{4} x^{2}) (\frac{4}{5} x^{5})

4 - 25 a^{2}

lo g_{4} (\frac{1}{64})

\frac{245}{5}

What is factor 16(x^2+1)^2-4(2x)^2 ?
The solution to factor 16(x^2+1)^2-4(2x)^2 is 16(x^2+x+1)(x^2-x+1)