What is (8n^3-16/5 n^2-24/7 n)(3/8 n^2) ?

The solution to (8n^3-16/5 n^2-24/7 n)(3/8 n^2) is (3n^3(35n^2-14n-15))/(35)

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simplify (8n^3-16/5 n^2-24/7 n)(3/8 n^2)

simplify

(8 n^{3} - \frac{16}{5} n^{2} - \frac{24}{7} n) (\frac{3}{8} n^{2})

\frac{3 n ^{3} ( 35 n ^{2} - 14 n - 15 )}{35}

Solution steps

(8 n^{3} - \frac{16}{5} n^{2} - \frac{24}{7} n) (\frac{3}{8} n^{2})

Remove parentheses:

(a) = a

= (8 n^{3} - \frac{16}{5} n^{2} - \frac{24}{7} n) \frac{3}{8} n^{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{3 ( 8 n ^{3} - \frac{16}{5} n ^{2} - \frac{24}{7} n ) n ^{2}}{8}

Multiply

\frac{16}{5} n^{2} : \frac{16 n ^{2}}{5}

= \frac{3 n ^{2} ( 8 n ^{3} - \frac{24}{7} n - \frac{16 n ^{2}}{5} )}{8}

Multiply

\frac{24}{7} n : \frac{24 n}{7}

= \frac{3 n ^{2} ( 8 n ^{3} - \frac{24 n}{7} - \frac{16 n ^{2}}{5} )}{8}

Join

8 n^{3} - \frac{16 n ^{2}}{5} - \frac{24 n}{7} : \frac{280 n ^{3} - 112 n ^{2} - 120 n}{35}

= \frac{3 n ^{2} \frac{280 n ^{3} - 112 n ^{2} - 120 n}{35}}{8}

Multiply

3 \cdot \frac{280 n ^{3} - 112 n ^{2} - 120 n}{35} n^{2} : \frac{3 n ^{2} ( 280 n ^{3} - 112 n ^{2} - 120 n )}{35}

= \frac{\frac{3 n ^{2} ( 280 n ^{3} - 112 n ^{2} - 120 n )}{35}}{8}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{( 280 n ^{3} - 112 n ^{2} - 120 n ) \cdot 3 n ^{2}}{35 \cdot 8}

Multiply the numbers:

35 \cdot 8 = 280

= \frac{3 n ^{2} ( 280 n ^{3} - 112 n ^{2} - 120 n )}{280}

Factor

(280 n^{3} - 112 n^{2} - 120 n) \cdot 3 n^{2} : 24 n^{3} (35 n^{2} - 14 n - 15)

= \frac{24 n ^{3} ( 35 n ^{2} - 14 n - 15 )}{280}

Cancel the common factor:

8

= \frac{3 n ^{3} ( 35 n ^{2} - 14 n - 15 )}{35}

- 3 (x - 2) \leq \frac{1}{3}

2

\frac{\frac{3}{2} + \frac{1}{4}}{\frac{5}{6} - \frac{1}{3}}

12 y^{2} + 13 y + 3

- 2^{3} (m + 32) < 16 m - 112

What is (8n^3-16/5 n^2-24/7 n)(3/8 n^2) ?
The solution to (8n^3-16/5 n^2-24/7 n)(3/8 n^2) is (3n^3(35n^2-14n-15))/(35)