Questions: Find (3y-4)(2y^2+y-1). (A) 6y^3-5y^2-7y-4 (B) 6y^3-7y^2-7y+4 (C) 6y^3-5y^2-7y+4 (D) 6y^3-5y^2+7y+4

Transcript text: 14. $\qquad$ Find $(3 y-4)\left(2 y^{2}+y-1\right)$. (A) $6 y^{3}-5 y^{2}-7 y-4$ (B) $6 y^{3}-7 y^{2}-7 y+4$ (C) $6 y^{3}-5 y^{2}-7 y+4$ (D) $6 y^{3}-5 y^{2}+7 y+4$

Solution

Solution Steps

To solve the problem of finding the product $(3y-4)(2y^2+y-1)$, we will use the distributive property (also known as the FOIL method for binomials) to expand the expression. This involves multiplying each term in the first polynomial by each term in the second polynomial and then combining like terms.

Step 1: Expand the Expression

We start with the expression $(3y - 4)(2y^2 + y - 1)$. To expand this, we apply the distributive property:

\[ (3y - 4)(2y^2 + y - 1) = 3y(2y^2) + 3y(y) + 3y(-1) - 4(2y^2) - 4(y) - 4(-1) \]

Step 2: Calculate Each Term

Calculating each term gives us:

\[ 3y(2y^2) = 6y^3 \] \[ 3y(y) = 3y^2 \] \[ 3y(-1) = -3y \] \[ -4(2y^2) = -8y^2 \] \[ -4(y) = -4y \] \[ -4(-1) = 4 \]

Step 3: Combine Like Terms

Now, we combine all the terms:

\[ 6y^3 + (3y^2 - 8y^2) + (-3y - 4y) + 4 = 6y^3 - 5y^2 - 7y + 4 \]