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.

1 answer

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: \[ \boxed{-24n^2 + 7n + 2} \]