Q. 184 - Intermediate Algebra

Question

In the following exercises, multiply. 

$$\begin{gathered} \left(a\right) -8y\left(y^2+2y-15\right) \\ \left(b\right) -4y^2{z}^3\left(3y^2+12yz-z^2\right) \end{gathered}$$

Step-by-Step Solution

Answer

The solution is 

$$\begin{gathered} \left(a\right) -8y\left(y^2+2y-15\right)=-8y^3-16y^2+120y \\ \left(b\right) -4y^2{z}^3\left(3y^2+12yz-z^2\right)=-12y^4{z}^3-48y^3{z}^4+4y^2{z}^5 \end{gathered}$$

Step 1. Apply the concept of Distributive Property

If $A,B and C$ are any real numbers or algebraic expressions, then

$A\left(B+C\right)=\left(B+C\right)A=AB+AC$

Part (a) Step 1. Given information

The given expression is 

$-8y\left(y^2+2y-15\right)$

Part (a) Step 2. Simplify

Using Distributive property,

$$\begin{gathered} -8y\left(y^2+2y-15\right) \\ =-8y·y^2-8y·2y+8y·15 \\ =-8y^3-16y^2+120y \end{gathered}$$

Part (b) Step 1. Given information

The given expression is 

$-4y^2{z}^3\left(3y^2+12yz-z^2\right)$

Part (b) Step 2. Simplify

Using Distributive property,

$$\begin{gathered} -4y^2{z}^3\left(3y^2+12yz-z^2\right) \\ =-4y^2{z}^3·3y^2-4y^2{z}^3·12yz+4y^2{z}^3·z^2 \\ =-12y^4{z}^3-48y^3{z}^4+4y^2{z}^5 \end{gathered}$$