Question

If C, equals, 1, plus, 5, n, minus, 6, n, squaredC=1+5n−6n
2
and B, equals, n, plus, 4, n, squared, commaB=n+4n
2
, find an expression that equals 2, C, minus, 3, B2C−3B in standard form.

Answer 1

To find the expression that equals \(2C - 3B\) and put it in standard form, we first need to express \(C\) and \(B\) based on the provided definitions.

### Step 1: Write down the expressions for \(C\) and \(B\).
1. For \(C\):
\[
C = 1 + 5n - 6n^2
\]
2. For \(B\):
\[
B = n + 4n^2
\]

### Step 2: Calculate \(3B\).
Now, let's compute \(3B\):
\[
3B = 3(n + 4n^2) = 3n + 12n^2
\]

### Step 3: Calculate \(2C\).
Next, we compute \(2C\):
\[
2C = 2(1 + 5n - 6n^2) = 2 + 10n - 12n^2
\]

### Step 4: Compute \(2C - 3B\).
Now we subtract \(3B\) from \(2C\):
\[
2C - 3B = (2 + 10n - 12n^2) - (3n + 12n^2)
\]

### Step 5: Simplify the expression.
Combine like terms:
\[
= 2 + (10n - 3n) + (-12n^2 - 12n^2)
\]
\[
= 2 + 7n - 24n^2
\]

### Step 6: Write the expression in standard form.
The final expression in standard form, where we typically write terms in descending powers of \(n\), is:
\[
-24n^2 + 7n + 2
\]

Thus, the expression that equals \(2C - 3B\) in standard form is:
\[
\boxed{-24n^2 + 7n + 2}
\]