Question

(3p³– 9p²q - 6pq²) ÷ (–3p), Divide the given polynomial by the given monomia

Answer 1

$\frac{{ \:3 {p}^{3} - 9 {p}^{2}q - 6p {q}^{2} } }{ - 3p} \\ \frac{(3p)( {p}^{2} - 3pq - 2 {q}^{2} }{ - 3p} \\ = > - {p}^{2} + 3pq + 2 {q}^{2}$

Answer 2

Answer:

(2q²+3pq-p²)

Step-by-step explanation:

We have to get the value of (3p³-9p²q-6pq²)÷(-3p)

To get the value of the above expression we have to first factorize (3p³-9p²q-6pq²) in order to cancel the (-3p) term.

Now, taking 3p common from the expression we get the following

(3p³-9p²q-6pq²)

=3p(p²-3pq-2q²)

Now, take an (-1) common and get

= -3p(2q²+3pq-p²)

Hence, we can get the value of the division (3p³-9p²q-6pq²)÷(-3p)

= -3p(2q²+3pq-p²)÷(-3p)

= (2q²+3pq-p²). (Answer)