How do you simplify $6 p q (3 p^{2} + 4 q)$ $6pq(3p^2+4q)$ ?

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Answer 1 · 2017-04-22T20:48:23

$18 p^{3} q + 24 p q^{2}$ $18p^3q+24pq^2$

Distribute $6 p q$ $6pq$ to $3 p^{2}$ $3p^2$ and $4 q$ $4q$

$6 p q \cdot 3 p^{2} + 6 p q \cdot 4 q$ $color(red)(6pq)*3p^2+color(red)(6pq)*4q$

note: when multiplying two of the same variables we ADD the exponents such that $x^{a} \cdot x^{b} = x^{a + b}$ $x^a*x^b= x^(a+b)$

Thus,
$6 p q \cdot 3 p^{2} + 6 p q \cdot 4 q \to 18 p^{3} q + 24 p q^{2}$ $color(red)(6pq)*3p^2+color(red)(6pq)*4q -> 18p^3q+24pq^2$

How do you simplify 6pq(3p2+4q)6pq(3p^2+4q)?