How do you factor $256 z^{2} - 4 - 192 z^{2} + 3$ $256z^2-4-192z^2+3$ ?

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Answer 1 · 2018-03-25T10:19:11

Simplify & use identities

= $256 z^{2} - 192 z^{2} - 1$ $256z^2-192z^2-1$

= $64 z^{2} - 1$ $64z^2-1$

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

= ${(8 z)}^{2} - {(1)}^{2}$ $(8z)^2-(1)^2$

$(8 z + 1) (8 z - 1 = 0$ $(8z+1)(8z-1=0$

$z = \frac{1}{8}, - \frac{1}{8}$ $z=1/8 , -1/8$

How do you factor 256z2−4−192z2+3256z^2-4-192z^2+3?