Q. ( -frac{2}{3} x_{1}^{1} – frac{2}{sqrt{3}} x_{1}^{2} ).

Step 1 — Write the original expression

\( -\frac{2}{3}x_{1}^{1} – \frac{2}{\sqrt{3}}x_{1}^{2} \)

Step 2 — Simplify the exponents (any variable to the first power is itself)

\( x_{1}^{1} = x_{1} \), so the expression becomes

\( -\frac{2}{3}x_{1} – \frac{2}{\sqrt{3}}x_{1}^{2} \)

Step 3 — Identify a common factor. Both terms contain the factor \( -2 \) and at least one factor \( x_{1} \). Factor out \( -2x_{1} \).

Divide each term by \( -2x_{1} \):

First term: \( \dfrac{-\tfrac{2}{3}x_{1}}{-2x_{1}} = \dfrac{1}{3} \).

Second term: \( \dfrac{-\tfrac{2}{\sqrt{3}}x_{1}^{2}}{-2x_{1}} = \dfrac{x_{1}}{\sqrt{3}} \).

Thus

\( -\frac{2}{3}x_{1} – \frac{2}{\sqrt{3}}x_{1}^{2} = -2x_{1}\!\left(\frac{1}{3} + \frac{x_{1}}{\sqrt{3}}\right) \)

Step 4 — (Optional) Present an equivalent form with a rationalized or simpler inner coefficient. Note that \( \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \). Using that, put the factor outside as \( -\frac{2x_{1}}{3} \):

\( -2x_{1}\!\left(\frac{1}{3} + \frac{x_{1}}{\sqrt{3}}\right)
= -\frac{2x_{1}}{3}\!\left(1 + \sqrt{3}\,x_{1}\right) \)

Final simplified (factorized) forms:

• \( -\dfrac{2}{3}x_{1} – \dfrac{2}{\sqrt{3}}x_{1}^{2} = -2x_{1}\!\left(\dfrac{1}{3} + \dfrac{x_{1}}{\sqrt{3}}\right) \)
• Equivalent alternative: \( -\dfrac{2}{3}x_{1} – \dfrac{2}{\sqrt{3}}x_{1}^{2} = -\dfrac{2x_{1}}{3}\!\left(1 + \sqrt{3}\,x_{1}\right) \)