Q.210 - Intermediate Algebra

Question

In the following exercises, multiply using a) the Distributive Property; b) the Vertical Method

$\left(y^2-3y+8\right)\left(4y^2+y-7\right)$

Step-by-Step Solution

Answer

The product of the given polynomials is $4y^4-11y^3+22y^2+29y-56$

Step 1. Given Information

The given polynomials are $\left(y^2-3y+8\right)\left(4y^2+y-7\right)$

The product is to be determined using a) the Distributive Property; b) the Vertical Method

Part (a) Step 1. Using distributive property

According to distributive property:

$$\begin{gathered} \left(y^2-3y+8\right)\left(4y^2+y-7\right)=y^2\left(4y^2+y-7\right)-3y\left(4y^2+y-7\right)+8\left(4y^2+y-7\right) \\ \left(y^2-3y+8\right)\left(4y^2+y-7\right)=4y^4+y^3-7y^2-12y^3-3y^2+21y+32y^2+8y-56 \\ \left(y^2-3y+8\right)\left(4y^2+y-7\right)=4y^4-11y^3+22y^2+29y-56 \end{gathered}$$

Part (a) Step 2. Conclusion

The product of the polynomials is $4y^4-11y^3+22y^2+29y-56$

Part (b) Step 1. Using Vertical Method

Multiplying using vertical method:

$\begin{array}{rrrr}& 4y^2+& y-& 7\\ \times & y^2-& 3y+& 8\\ & 32y^2+& 8y-& 56\\ -12y^3-& 3y^2+& 21y & \\ 4y^4+ y^3-& 7y^2 & & \\ 4y^4-11y^3+& 22y^2+& 29y-& 56\end{array}$

Part (b) Step 2. Conclusion

The product of the polynomials is $4y^4-11y^3+22y^2+29y-56$