Question

if $b = -3q - 8$ and $c = 2q^2 - q - 2$, find an expression that equals $3b - 3c$ in standard form.

Explanation:

Step1: Substitute B and C into the expression

We know \( B = -3q - 8 \) and \( C = 2q^{2}-q - 2 \), so substitute them into \( 3B - 3C \):
\( 3(-3q - 8)-3(2q^{2}-q - 2) \)

Step2: Distribute the 3s

Using the distributive property \( a(b + c)=ab+ac \), we get:
\( 3\times(-3q)+3\times(-8)-3\times(2q^{2})+3\times(q)+3\times(2) \)
\( = -9q - 24 - 6q^{2}+ 3q + 6 \)

Step3: Combine like terms

Combine the \( q \)-terms and the constant terms:
For \( q \)-terms: \( -9q+3q=-6q \)
For constant terms: \( -24 + 6=-18 \)
So the expression becomes \( -6q^{2}-6q - 18 \) (we also order the terms in descending powers of \( q \) for standard form)

Answer:

\( -6q^{2}-6q - 18 \)