Question

factorize x^4-1/64 pls solve fast

Answer 1

Answer:X^4–1/64

=(X^2–1/8)(X^2+1/8)=(X-1/2rOOT(2))(X+1/2rOOT(2))(X^2+1/8)

If you can't understand here's the pic

Answer 2

Factors are

$\bf \red{{x}^{4} - \frac{1}{64} = \left( {x}^{2} + \frac{1}{8} \right)\left(x + \frac{1}{2 \sqrt{2} } \right) \left( x- \frac{1}{2 \sqrt{2} } \right)}$

Given:

• ${x}^{4} - \frac{1}{64}$

To find:

• Factors of given polynomial.

Solution:

Identity to be used:

$\boxed{{a}^{2} - {b}^{2} = (a + b)(a - b)}$

Step 1:

Rewrite the polynomial.

$( { {x}^{2} })^{2} - \left( { \frac{1}{8} } \right)^{2}$

here

$a = {x}^{2}$

and

$b = \frac{1}{8}$

So,

$( { {x}^{2} })^{2} - \left( { \frac{1}{8} } \right)^{2} = ( {x}^{2} + \frac{1}{8} )( {x}^{2} - \frac{1}{8} )$

Step 2:

Again rewrite the factors having -ve sign and apply the same identity again.

$( {x}^{2} - \frac{1}{8} ) = {x}^{2} - \left({ \frac{1}{2 \sqrt{2} } } \right)^{2}$

or

${x}^{2} - \left({ \frac{1}{2 \sqrt{2} } } \right)^{2} =\left(x + \frac{1}{2 \sqrt{2} } \right) \left( x- \frac{1}{2 \sqrt{2} } \right)$

Thus,

The factors are

$\bf {x}^{4} - \frac{1}{64} = \left( {x}^{2} + \frac{1}{8} \right)\left(x + \frac{1}{2 \sqrt{2} } \right) \left( x- \frac{1}{2 \sqrt{2} } \right)$

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