Question

Expand the given:

$(2p + 3q)(4 {p}^{2} - 6pq + 9 {q}^{2} )$
​

Answer 1

Step-by-step explanation:

(2p+3q)(4p²-6pq+9q²)

⇒(2p+3q){(2p)²-2(2p)(3q)+(3q)²}

⇒(2p+3q)(2p-3q)²

⇒(2p+3q)(2p-3q)(2p-3q)

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Answer 2

ANSWERS:

$\longrightarrow \Large\boxed{8p^3+27q^3}$

EXPLANATIONS:

To solve this problem, first you have to use distributive property by using with expanding expression.

SOLUTIONS:

$\mathsf{Distributive\quad Property}$

$\mathsf{A(B+C)=AB+AC}$

$\displaystyle (2p+3q)(4p^2-6pq+9q^2)$

First, you have to solve with parenthesis.

$\displaystyle 2p*4p^2+2p(-6pq)+2p*9q^2+3q*4p^2+3q(-6pq)+3q*9q^2$

Change negative sign to positive sign.

$\Rightarrow \displaystyle +(-b)=-b$

Rewrite the problem.

$\displaystyle 2\cdot \:4p^2p-2\cdot \:6ppq+2\cdot \:9pq^2+3\cdot \:4p^2q-3\cdot \:6pqq+3\cdot \:9q^2q$

Solve.

$\displaystyle 2*\:4p^2p-2* \:6ppq+2* \:9pq^2+3* \:4p^2q-3*\:6pqq+3*\:9q^2q$

$\displaystyle 2*4p^2p$

Multiply.

$\Rightarrow\displaystyle 2\times4p^2p=8p^3$

$\displaystyle 2\times6ppq=12p^2q$

$\displaystyle 2\times9=18$

$\displaystyle 4\times3=12$

$\displaystyle 6\times3=18$

$\displaystyle 9\times3=27$

$\Rightarrow \displaystyle 8p^3-12p^2q+18pq^2+12p^2q-18pq^2+27q^3$

Combined like terms. (Group like terms.)

$\Rightarrow\displaystyle 8p^3-12p^2q+12p^2q+18pq^2-18pq^2+27q^3$

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

$\displaystyle 12-12=0$

$\displaystyle -12p^2q+12p^2q=0$

Rewrite the problem again.

$\displaystyle 8p^3+18pq^2-18pq^2+27q^3$

Subtract.

$\displaystyle 18-18=0$

$\Rightarrow \Large\boxed{8p^3+27q^3}$

So, the correct answer is 8p³+27q³.