Q. Rewrite the following polynomial in standard form: (x – 9 + frac{x^{2}}{2}).

Q: What does "standard form" of a polynomial mean?

Write terms in descending powers of x, combine like terms, and include coefficients (even fractions). Example: \frac{x^2}{2}+x-9..

Q: How do you rewrite x - 9 + \frac{x^2}{2} in standard form?.

Arrange by degree: \frac{x^2}{2}+x-9.

Q: What is the degree and leading coefficient of \frac{x^2}{2}+x-9 ?.

Degree is 2 (highest exponent). Leading coefficient is \tfrac{1}{2}, the coefficient of x^2..

Q: How can I convert \frac{x^2}{2}+x-9 to integer coefficients?.

Multiply by the least common denominator (2): 2\left(\frac{x^2}{2}+x-9\right)=x^2+2x-18.

Q: How do I factor \frac{x^2}{2}+x-9 ?

How do I factor \frac{x^2}{2}+x-9 ?

Q: How can I check my standard-form rewrite is correct?

Substitute a few x-values into both expressions or expand any factored form; results must match. Example: x=2 gives \frac{4}{2}+2-9=-3 for both.

Q: What if the original expression meant 2x^2 + x - 9 instead?

Then standard form is 2x^2+x-9; degree 2, leading coefficient 2. It does not factor nicely over integers.

Answer

Reorder terms by descending powers:
\[
\frac{1}{2}x^2 + x – 9
\]

Detailed Explanation

Set \( x = 0 \):

\[ -3(0) + 6y = 21, \quad 6y = 21, \quad y = \dfrac{7}{2} \]

Intercepts: \( (-7, 0) \) and \( \left(0, \dfrac{7}{2}\right) \)

Given expression:

\[ x – 9 + \frac{x^2}{2} \]

Identify each term and its degree.

Term of degree 2: \( \dfrac{x^2}{2} \)
Term of degree 1: \( x \)
Constant term (degree 0): \( -9 \)

Standard form (polynomial written with powers in descending order):
Place the degree-2 term first, then degree-1, then constant. This gives:

\[ \frac{x^2}{2} + x – 9 \]

Optional: rewrite the coefficient \( \dfrac{1}{2} \) explicitly next to \( x^2 \):

\[ \frac{1}{2}x^2 + x – 9 \]

Optional: write as a single fraction (common denominator 2) to check consistency.
Convert \( x \) to \( \dfrac{2x}{2} \) and \( -9 \) to \( \dfrac{-18}{2} \), then combine:

\[ \frac{x^2 + 2x – 18}{2} \]

Factor numerator if desired: \[ \frac{(x+6)(x-3)}{2} \]

Final answer (standard polynomial form):

\[ \frac{1}{2}x^2 + x – 9 \]

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Algebra FAQs

What does "standard form" of a polynomial mean?

Write terms in descending powers of \(x\), combine like terms, and include coefficients (even fractions). Example: \(\frac{x^2}{2}+x-9\)..

How do you rewrite \(x - 9 + \frac{x^2}{2}\) in standard form?.

Arrange by degree: \(\frac{x^2}{2}+x-9\).

How do I handle fractional coefficients when ordering terms?

Ordering ignores coefficient type: always sort by exponent size. Fractions remain as coefficients, e.g., \(\frac{x^2}{2}\) comes before \(x\).

What is the degree and leading coefficient of \( \frac{x^2}{2}+x-9 \) ?.

Degree is 2 (highest exponent). Leading coefficient is \(\tfrac{1}{2}\), the coefficient of \(x^2\)..

How can I convert \(\frac{x^2}{2}+x-9\) to integer coefficients?.

Multiply by the least common denominator (2): \(2\left(\frac{x^2}{2}+x-9\right)=x^2+2x-18.\)

How do I factor \( \frac{x^2}{2}+x-9 \)?

How can I check my standard-form rewrite is correct?

Substitute a few \(x\)-values into both expressions or expand any factored form; results must match. Example: \(x=2\) gives \(\frac{4}{2}+2-9=-3\) for both.

What if the original expression meant \(2x^2 + x - 9\) instead?

Then standard form is \(2x^2+x-9\); degree 2, leading coefficient 2. It does not factor nicely over integers.

Order polynomials to standard form
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