How do you multiply ${(7 b^{2} - 3 c)}^{2}$ $(7b^2-3c)^2$ ?

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How do you multiply ${(7 b^{2} - 3 c)}^{2}$ $(7b^2-3c)^2$ ?

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Answer 1 · 2017-02-01T12:45:28

See the entire solution process below:

Explanation:

We can rewrite this expression as:

$(7 b^{2} - 3 c) (7 b^{2} - 3 c)$ $(7b^2 - 3c)(7b^2 - 3c)$

To multiply these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.

$(7 b^{2} - 3 c) (7 b^{2} - 3 c)$ $(color(red)(7b^2) - color(red)(3c))(color(blue)(7b^2) - color(blue)(3c))$ becomes:

$(7 b^{2} \times 7 b^{2}) - (7 b^{2} \times 3 c) - (3 c \times 7 b^{2}) + (3 c \times 3 c)$ $(color(red)(7b^2) xx color(blue)(7b^2)) - (color(red)(7b^2) xx color(blue)(3c)) - (color(red)(3c) xx color(blue)(7b^2)) + (color(red)(3c) xx color(blue)(3c))$

$49 b^{4} - 21 b^{2} c - 21 b^{2} c + 9 c^{2}$ $49b^4 - 21b^2c - 21b^2c + 9c^2$

We can now combine like terms:

$49 b^{4} + (- 21 - 21) b^{2} c + 9 c^{2}$ $49b^4 + (-21 - 21)b^2c + 9c^2$

$49 b^{4} - 42 b^{2} c + 9 c^{2}$ $49b^4 - 42b^2c + 9c^2$

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How do you multiply ${(7 b^{2} - 3 c)}^{2}$ $(7b^2-3c)^2$ ?

1 Answer

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How do you multiply ${(7 b^{2} - 3 c)}^{2}$ $(7b^2-3c)^2$ ?