How do you simplify (3+ 2c ) ( 7- 4c )(3+2c)(7−4c)?

2 Answers
smendyka
Jul 29, 2017

See a solution process below:

Explanation:

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

(color(red)(3) + color(red)(2c))(color(blue)(7) - color(blue)(4c))(3+2c)(7−4c) becomes:

(color(red)(3) xx color(blue)(7)) - (color(red)(3) xx color(blue)(4c)) + (color(red)(2c) xx color(blue)(7)) - (color(red)(2c) xx color(blue)(4c))(3×7)−(3×4c)+(2c×7)−(2c×4c)

21 - 12c + 14c - 8c^221−12c+14c−8c2

We can now combine like terms:

21 + (-12 + 14)c - 8c^221+(−12+14)c−8c2

21 + 2c - 8c^221+2c−8c2

Or in standard form:

-8c^2 + 2c + 21−8c2+2c+21

Jim G.
Jul 29, 2017

21+2c-8c^221+2c−8c2

Explanation:

"simplify means to multiply the factors together"simplify means to multiply the factors together

"each term in the second factor is multiplied by each term"each term in the second factor is multiplied by each term
"in the first factor"in the first factor

rArr(color(red)(3+2c))(7-4c)⇒(3+2c)(7−4c)

=color(red)(3)(7-4c)color(red)(+2c)(7-4c)=3(7−4c)+2c(7−4c)

"distributing the brackets gives"distributing the brackets gives

=(color(red)(3)xx7)+(color(red)(3)xx-4c)+(color(red)(2c)xx7)+(color(red)(2c)xx-4c)=(3×7)+(3×−4c)+(2c×7)+(2c×−4c)

=21+(-12c)+14c+(-8c^2)=21+(−12c)+14c+(−8c2)

=21+2c-8c^2=21+2c−8c2

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