Question

$2p ^{2}q ^{2} + 4p ^{2} + 3q ^{2}p - 4q ^{2} + 6p - 8$
please answer fast ​

Answer 1

Answer:

A(B+C)=AB+AC

\displaystyle (2p+3q)(4p^2-6pq+9q^2)(2p+3q)(4p2−6pq+9q2)

First, you have to solve with parenthesis.

\displaystyle 2p*4p^2+2p(-6pq)+2p*9q^2+3q*4p^2+3q(-6pq)+3q*9q^22p∗4p2+2p(−6pq)+2p∗9q2+3q∗4p2+3q(−6pq)+3q∗9q2

Change negative sign to positive sign.

\Rightarrow \displaystyle +(-b)=-b⇒+(−b)=−b

Rewrite the problem.

\displaystyle 2\cdot \:4p^2p-2\cdot \:6ppq+2\cdot \:9pq^2+3\cdot \:4p^2q-3\cdot \:6pqq+3\cdot \:9q^2q2⋅4p2p−2⋅6ppq+2⋅9pq2+3⋅4p2q−3⋅6pqq+3⋅9q2q

Solve.

\displaystyle 2*\:4p^2p-2* \:6ppq+2* \:9pq^2+3* \:4p^2q-3*\:6pqq+3*\:9q^2q2∗4p2p−2∗6ppq+2∗9pq2+3∗4p2q−3∗6pqq+3∗9q2q

\displaystyle 2*4p^2p2∗4p2p

Multiply.

\Rightarrow\displaystyle 2\times4p^2p=8p^3⇒2×4p2p=8p3

\displaystyle 2\times6ppq=12p^2q2×6ppq=12p2q

\displaystyle 2\times9=182×9=18

\displaystyle 4\times3=124×3=12

\displaystyle 6\times3=186×3=18

\displaystyle 9\times3=279×3=27

\Rightarrow \displaystyle 8p^3-12p^2q+18pq^2+12p^2q-18pq^2+27q^3⇒8p3−12p2q+18pq2+12p2q−18pq2+27q3

Combined like terms. (Group like terms.)

\Rightarrow\displaystyle 8p^3-12p^2q+12p^2q+18pq^2-18pq^2+27q^3⇒8p3−12p2q+12p2q+18pq2−18pq2+27q3

Then, you add or subtract elements to the numbers from left to right.

\displaystyle 12-12=012−12=0

\displaystyle -12p^2q+12p^2q=0−12p2q+12p2q=0

Rewrite the problem again.

\displaystyle 8p^3+18pq^2-18pq^2+27q^38p3+18pq2−18pq2+27q3

Subtract.

\displaystyle 18-18=018−18=0

\Rightarrow \Large\boxed{8p^3+27q^3}⇒8p3+27q3

Ans=

\Rightarrow \Large\boxed{8p^3+27q^3}⇒8p3+27q3