Question
-x^2-1/2+x in standard form
Answers
Answered by
GPT-4o mini
To write the expression \(-x^2 - \frac{1}{2} + x\) in standard form, we want to rearrange it in the format \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants.
Let's rearrange the expression:
1. Identify the terms: \(-x^2\), \(x\), and \(-\frac{1}{2}\).
2. Rewrite the expression in standard form:
\[
-x^2 + x - \frac{1}{2}
\]
This is already in standard form, but we typically like the leading coefficient to be positive if possible. So we can factor out \(-1\):
\[
-1(x^2 - x + \frac{1}{2})
\]
Thus, the standard form of the expression in terms of the original structure is:
\[
-x^2 + x - \frac{1}{2}
\]
If you want to emphasize the factorization with a negative leading coefficient, you can write it as \(-1(x^2 - x + \frac{1}{2})\).
Let's rearrange the expression:
1. Identify the terms: \(-x^2\), \(x\), and \(-\frac{1}{2}\).
2. Rewrite the expression in standard form:
\[
-x^2 + x - \frac{1}{2}
\]
This is already in standard form, but we typically like the leading coefficient to be positive if possible. So we can factor out \(-1\):
\[
-1(x^2 - x + \frac{1}{2})
\]
Thus, the standard form of the expression in terms of the original structure is:
\[
-x^2 + x - \frac{1}{2}
\]
If you want to emphasize the factorization with a negative leading coefficient, you can write it as \(-1(x^2 - x + \frac{1}{2})\).