How do you find the product of $(3 r + 2) (9 r^{2} + 6 r + 4)$ $(3r+2)(9r^2+6r+4)$ ?

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How do you find the product of $(3 r + 2) (9 r^{2} + 6 r + 4)$ $(3r+2)(9r^2+6r+4)$ ?

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Answer 1 · 2017-08-16T10:49:58

$27 r^{3} + 36 r^{2} + 24 r + 8$ $27r^3+36r^2+24r+8$

Explanation:

$each term in the second factor must be multiplied by$ $"each term in the second factor must be multiplied by "$
$each term in the first factor$ $"each term in the first factor"$

$(3 r + 2) (9 r^{2} + 6 r + 4)$ $(color(red)(3r+2))(9r^2+6r+4)$

$= 3 r (9 r^{2} + 6 r + 4) + 2 (9 r^{2} + 6 r + 4)$ $=color(red)(3r)(9r^2+6r+4)color(red)(+2)(9r^2+6r+4)$

$= (3 r \times 9 r^{2}) + (3 r \times 6 r) + (3 r \times 4)$ $=(color(red)(3r)xx9r^2)+(color(red)(3r)xx6r)+(color(red)(3r)xx4)$
$\times + (2 \times 9 r^{2}) + (2 \times 6 r) + (2 \times 4)$ $color(white)(xx)+(color(red)(2)xx9r^2)+(color(red)(2)xx6r)+(color(red)(2)xx4)$

$= 27 r^{3} + 18 r^{2} + 12 r + 18 r^{2} + 12 r + 8$ $=27r^3+18r^2+12r+18r^2+12r+8$

$= 27 x^{3} + 36 r^{2} + 24 r + 8$ $=27x^3+36r^2+24r+8$

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How do you find the product of $(3 r + 2) (9 r^{2} + 6 r + 4)$ $(3r+2)(9r^2+6r+4)$ ?

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How do you find the product of (3r+2)(9r^2+6r+4)(3r+2)(9r2+6r+4)(3r+2)(9r^2+6r+4)?

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How do you find the product of $(3 r + 2) (9 r^{2} + 6 r + 4)$ $(3r+2)(9r^2+6r+4)$ ?