Question

If D, equals, 1, plus, 4, p, minus, 6, p, squaredD=1+4p−6p
2
and C, equals, 1, minus, p, commaC=1−p, find an expression that equals 2, D, minus, 3, C2D−3C in standard form.

Answer 1

To find the expression \(2D - 3C^2\), we need to first determine \(D\) and \(C\), then calculate \(C^2\), and finally substitute these values into the expression.

1. Given:
\[
D = 1 + 4p - 6p^2
\]

2. Given:
\[
C = 1 - p
\]

3. First, calculate \(C^2\):
\[
C^2 = (1 - p)^2
\]
Expanding this using the formula \((a - b)^2 = a^2 - 2ab + b^2\):
\[
C^2 = 1^2 - 2(1)(p) + p^2 = 1 - 2p + p^2
\]

4. Next, calculate \(3C^2\):
\[
3C^2 = 3(1 - 2p + p^2) = 3 - 6p + 3p^2
\]

5. Now calculate \(2D\):
\[
2D = 2(1 + 4p - 6p^2) = 2 + 8p - 12p^2
\]

6. Now substitute \(2D\) and \(3C^2\) into the expression \(2D - 3C^2\):
\[
2D - 3C^2 = (2 + 8p - 12p^2) - (3 - 6p + 3p^2)
\]
Simplifying this:
\[
= 2 + 8p - 12p^2 - 3 + 6p - 3p^2
\]
\[
= (2 - 3) + (8p + 6p) + (-12p^2 - 3p^2)
\]
\[
= -1 + 14p - 15p^2
\]

7. Finally, write the expression in standard form:
\[
-15p^2 + 14p - 1
\]

Thus, the expression that equals \(2D - 3C^2\) in standard form is:
\[
\boxed{-15p^2 + 14p - 1}
\]