How do you multiply $(7 z + 8 x) (7 z - 8 x)$ $(7z + 8x ) ( 7z - 8x )$ ?

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How do you multiply $(7 z + 8 x) (7 z - 8 x)$ $(7z + 8x ) ( 7z - 8x )$ ?

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Answer 1 · 2018-03-03T19:29:49

$(7 z + 8 x) (7 z - 8 x)$ $(7z+8x)(7z-8x)$
$7 z \cdot 7 z - 7 z \cdot 8 x + 8 x \cdot 7 z - 8 x \cdot 8 x$ $7z*7z-7z*8x+8x*7z-8x*8x$
$49 z^{2} - 56 z x + 56 x z - 64 x^{2}$ $49z^2-56zx+56xz-64x^2$
$49 z^{2} - 64 x^{2}$ $49z^2-64x^2$

Answer 2 · 2018-03-03T19:34:58

$49 z^{2} - 64 x^{2}$ $49z^2-64x^2$

Explanation:

The trick to solving binomial expansions is to start with the leftmost term and multiply it with the other two numbers in the next bracket, and move to the second number in the first bracket and then multiply that number with the two numbers in the other bracket, separately.

Hence:

$(7 z \times 7 z) = 49 z^{2}$ $(7z times 7z) = 49z^2$

2. $(7 z \times - 8 x) = - 56 x z$ $(7z times -8x)= -56xz$

3.Start using 8x: $(8 x \times 7 z) = 56 x z$ $(8x times 7z)= 56xz$

4. $(8 x \times - 8 x) = - 64 x^{2}$ $(8x times -8x)= -64x^2$

Then we simply add all the terms together:
$49 z^{2} + (- 56 x z) + 56 x z + (- 64 x^{2}) = 49 z^{2} - 64 x^{2}$ $49z^2+(-56xz)+56xz+(-64x^2) = 49z^2-64x^2$

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