问题

Factor the following quadratic: $10 x^2+\frac{260 x}{\sqrt{3}}-1050$. Please put your final answer in a $\\boxed{}$.

答案

模型 模型答案 正确答案
金声玉亮2.0 To factor the quadratic expression \(10 x^2 + \frac{260 x}{\sqrt{3}} - 1050\), we will follow these steps:

1. **Factor out the common factor**: Notice that each term in the quadratic expression has a common factor of 10. We can factor out 10 from each term:
\[
10 x^2 + \frac{260 x}{\sqrt{3}} - 1050 = 10 \left( x^2 + \frac{26 x}{\sqrt{3}} - 105 \right)
\]

2. **Simplify the expression inside the parentheses**: We need to simplify the quadratic expression inside the parentheses:
\[
x^2 + \frac{26 x}{\sqrt{3}} - 105
\]

3. **Rationalize the coefficient of \(x\)**: To make the coefficient of \(x\) a rational number, we multiply the entire quadratic expression by \(\sqrt{3}\):
\[
\sqrt{3} \left( x^2 + \frac{26 x}{\sqrt{3}} - 105 \right) = \sqrt{3} x^2 + 26 x - 105 \sqrt{3}
\]

4. **Factor the simplified quadratic expression**: Now, we need to factor the quadratic expression \(\sqrt{3} x^2 + 26 x - 105 \sqrt{3}\). We look for two numbers that multiply to \(-105 \sqrt{3}\) and add to 26. These numbers are 35 and -3:
\[
\sqrt{3} x^2 + 26 x - 105 \sqrt{3} = (\sqrt{3} x + 35)(x - 3)
\]

5. **Combine the factors**: Recall that we initially factored out 10. Therefore, we need to include this factor in our final answer:
\[
10 \left( \sqrt{3} x + 35 \right) \left( x - 3 \right)
\]

Thus, the factored form of the quadratic expression is:
\[
\boxed{10 (\sqrt{3} x + 35)(x - 3)}
\]
10 \left(-x-\frac{35}{\sqrt{3}}\right) \left(3 \sqrt{3}-x\right)
deepseek-r1

To factor the quadratic \(10 x^2 + \frac{260 x}{\sqrt{3}} - 1050\), we start by rationalizing the denominator of the middle term:

\[
\frac{260 x}{\sqrt{3}} = \frac{260 x \sqrt{3}}{3}
\]

Rewriting the quadratic with this term:

\[
10 x^2 + \frac{260 \sqrt{3}}{3} x - 1050
\]

Next, we factor out the common factor of 10:

\[
10 \left( x^2 + \frac{26 \sqrt{3}}{3} x - 105 \right)
\]

We use the substitution \(y = x \sqrt{3}\), which transforms the quadratic inside the brackets. Substituting \(x = \frac{y}{\sqrt{3}}\) into the quadratic expression:

\[
\left( \frac{y}{\sqrt{3}} \right)^2 + \frac{26 \sqrt{3}}{3} \left( \frac{y}{\sqrt{3}} \right) - 105
\]

Simplifying each term:

\[
\frac{y^2}{3} + \frac{26 y}{3} - 105
\]

Factoring out \(\frac{1}{3}\):

\[
\frac{1}{3} \left( y^2 + 26 y - 315 \right)
\]

We factor the quadratic \(y^2 + 26 y - 315\) by finding two numbers that multiply to \(-315\) and add to \(26\). These numbers are \(35\) and \(-9\):

\[
(y + 35)(y - 9)
\]

Substituting back \(y = x \sqrt{3}\):

\[
\frac{1}{3} (x \sqrt{3} + 35)(x \sqrt{3} - 9)
\]

Multiplying by the factored-out 10:

\[
10 \cdot \frac{1}{3} (x \sqrt{3} + 35)(x \sqrt{3} - 9)
\]

Simplifying the expression:

\[
\frac{10}{3} (x \sqrt{3} + 35)(x \sqrt{3} - 9)
\]

Rationalizing the terms inside the factors:

\[
x \sqrt{3} + 35 = \sqrt{3} \left( x + \frac{35 \sqrt{3}}{3} \right)
\]
\[
x \sqrt{3} - 9 = \sqrt{3} \left( x - 3 \sqrt{3} \right)
\]

Thus, the factored form is:

\[
10 \left( x - 3 \sqrt{3} \right) \left( x + \frac{35 \sqrt{3}}{3} \right)
\]

The final answer is:

\[
\boxed{10 \left( x - 3\sqrt{3} \right) \left( x + \frac{35\sqrt{3}}{3} \right)}
\]
10 \left(-x-\frac{35}{\sqrt{3}}\right) \left(3 \sqrt{3}-x\right)
gpto1 To factor the quadratic expression \(10 x^2 + \dfrac{260 x}{\sqrt{3}} - 1050\), we'll aim to break it down into its component factors. Here's a step-by-step approach:

**Step 1: Simplify the expression by eliminating radicals in denominators.**

First, to eliminate denominators, multiply both sides of the equation by \(\sqrt{3}\):
\[
\sqrt{3} \times \left(10 x^2 + \dfrac{260 x}{\sqrt{3}} - 1050\right) = 0.
\]

Simplify:
\[
10 \sqrt{3} x^2 + 260 x - 1050 \sqrt{3} = 0.
\]

**Step 2: Factor the simplified quadratic expression.**

We can rewrite the quadratic as:
\[
10 \sqrt{3} x^2 + 260 x - 1050 \sqrt{3} = 10 \left( \sqrt{3} x^2 + 26 x - 105 \sqrt{3} \right).
\]

Next, factor the quadratic inside the parentheses. To do this, consider factoring it as \((\sqrt{3} x + a)(x + b)\). Expanding the product, we get:
\[
\sqrt{3} x^2 + (\sqrt{3} b + a)x + a b = \sqrt{3} x^2 + 26 x - 105 \sqrt{3}.
\]

Set up the system of equations:
\[
\begin{cases}
\sqrt{3} b + a = 26,\\
a b = -105 \sqrt{3}.
\end{cases}
\]

Let \(a = 26 - \sqrt{3} b\). Substitute back into \(a b\):
\[
(26 - \sqrt{3} b) b = -105 \sqrt{3} \implies 26 b - \sqrt{3} b^2 = -105 \sqrt{3}.
\]

Multiply both sides by \(-1\) to make calculations easier:
\[
\sqrt{3} b^2 - 26 b + 105 \sqrt{3} = 0.
\]

Let \(b = k \sqrt{3}\). Substituting and simplifying, we get the quadratic equation:
\[
3 k^2 - 26 k - 105 = 0.
\]

Solving this quadratic equation, we find \(k = \dfrac{35}{3}\) or \(k = -3\).

**Step 3: Determine the factors.**

Corresponding to \(k = \dfrac{35}{3}\):
\[
b = \dfrac{35}{3} \sqrt{3}, \quad a = 26 - 3 k = -9.
\]

Corresponding to \(k = -3\):
\[
b = -3 \sqrt{3}, \quad a = 26 - 3 (-3) = 35.
\]

Using the solution that results in integer coefficients in \(a\) and \(b\), we select \(k = -3\).

Therefore, the factors are:
\[
\sqrt{3} x^2 + 26 x - 105 \sqrt{3} = (\sqrt{3} x + 35)(x - 3 \sqrt{3}).
\]

**Step 4: Write the final factored form.**

Including the factor of \(10\) we factored out earlier, the original quadratic expression factors completely as:
\[
10 x^2 + \dfrac{260 x}{\sqrt{3}} - 1050 = 10 (\sqrt{3} x + 35)(x - 3 \sqrt{3}).
\]

Therefore, the factored form is:

\(\boxed{10\,(\sqrt{3}x+35)\left(x-3\sqrt{3}\right)}\)
10 \left(-x-\frac{35}{\sqrt{3}}\right) \left(3 \sqrt{3}-x\right)
gpt-4o-2024-08-06 To factor the quadratic \(10x^2 + \frac{260x}{\sqrt{3}} - 1050\), we start by simplifying the expression. First, let's clear the fraction by multiplying the entire quadratic by \(\sqrt{3}\):

\[
\sqrt{3} \cdot \left(10x^2 + \frac{260x}{\sqrt{3}} - 1050\right) = 10\sqrt{3}x^2 + 260x - 1050\sqrt{3}
\]

Now, we have the quadratic:

\[
10\sqrt{3}x^2 + 260x - 1050\sqrt{3}
\]

Next, we will factor this quadratic expression. We need to find two numbers that multiply to the product of the coefficient of \(x^2\) term and the constant term, and add up to the coefficient of the \(x\) term.

The product is:

\[
10\sqrt{3} \times (-1050\sqrt{3}) = -10500 \times 3 = -31500
\]

We need two numbers that multiply to \(-31500\) and add to \(260\).

After testing several pairs, we find that \(450\) and \(-70\) work because:

\[
450 \times (-70) = -31500 \quad \text{and} \quad 450 + (-70) = 380
\]

However, we need to find numbers that add to \(260\), not \(380\). Let's re-evaluate our approach. We need to find numbers that multiply to \(-31500\) and add to \(260\).

Upon further inspection, the correct numbers are \(525\) and \(-60\) because:

\[
525 \times (-60) = -31500 \quad \text{and} \quad 525 + (-60) = 465
\]

This is incorrect. Let's try another approach by using the quadratic formula to find the roots and then factor.

The quadratic formula is:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \(a = 10\sqrt{3}\), \(b = 260\), and \(c = -1050\sqrt{3}\).

Calculate the discriminant:

\[
b^2 - 4ac = 260^2 - 4 \times 10\sqrt{3} \times (-1050\sqrt{3})
\]

\[
= 67600 + 4 \times 10 \times 1050 \times 3
\]

\[
= 67600 + 126000
\]

\[
= 193600
\]

The discriminant is a perfect square, \(\sqrt{193600} = 440\).

Now, find the roots:

\[
x = \frac{-260 \pm 440}{20\sqrt{3}}
\]

Calculate the roots:

1. \(x = \frac{-260 + 440}{20\sqrt{3}} = \frac{180}{20\sqrt{3}} = \frac{9}{\sqrt{3}} = 3\sqrt{3}\)

2. \(x = \frac{-260 - 440}{20\sqrt{3}} = \frac{-700}{20\sqrt{3}} = \frac{-35}{\sqrt{3}} = -35\sqrt{3}\)

Thus, the quadratic factors as:

\[
10\sqrt{3}(x - 3\sqrt{3})(x + 35\sqrt{3})
\]

Finally, the factorization of the original quadratic is:

\[
\boxed{(x - 3\sqrt{3})(x + 35\sqrt{3})}
\]
10 \left(-x-\frac{35}{\sqrt{3}}\right) \left(3 \sqrt{3}-x\right)
DeepSeek-Coder-V2-Lite-Instruct To factor the quadratic expression \(10 x^2 + \frac{260 x}{\sqrt{3}} - 1050\), we will follow these steps:

1. **Factor out the greatest common factor (GCF) from the terms**:
The GCF of the coefficients \(10\), \(\frac{260}{\sqrt{3}}\), and \(-1050\) is \(10\).

\[
10 x^2 + \frac{260 x}{\sqrt{3}} - 1050 = 10 \left( x^2 + \frac{26 x}{\sqrt{3}} - 105 \right)
\]

2. **Simplify the expression inside the parentheses**:
To simplify the expression inside the parentheses, we need to rationalize the denominator. However, it is often easier to factor the quadratic expression directly without rationalizing the denominator.

3. **Factor the quadratic expression \(x^2 + \frac{26 x}{\sqrt{3}} - 105\)**:
We need to find two numbers that multiply to \(-105\) (the constant term) and add up to \(\frac{26}{\sqrt{3}}\) (the coefficient of the linear term).

Let's denote the quadratic as \(x^2 + \frac{26 x}{\sqrt{3}} - 105\).

To factor this, we look for two numbers that multiply to \(-105\) and add up to \(\frac{26}{\sqrt{3}}\).

The numbers that satisfy these conditions are \(\frac{15}{\sqrt{3}}\) and \(-\frac{7}{\sqrt{3}}\) because:
\[
\left( \frac{15}{\sqrt{3}} \right) \left( -\frac{7}{\sqrt{3}} \right) = -105
\]
and
\[
\frac{15}{\sqrt{3}} + \left( -\frac{7}{\sqrt{3}} \right) = \frac{15 - 7}{\sqrt{3}} = \frac{8}{\sqrt{3}} = \frac{26}{\sqrt{3}}
\]

4. **Rewrite the quadratic expression using these numbers**:
\[
x^2 + \frac{26 x}{\sqrt{3}} - 105 = \left( x + \frac{15}{\sqrt{3}} \right) \left( x - \frac{7}{\sqrt{3}} \right)
\]

5. **Combine the factored terms with the GCF**:
\[
10 \left( x + \frac{15}{\sqrt{3}} \right) \left( x - \frac{7}{\sqrt{3}} \right)
\]

6. **Simplify the terms inside the parentheses**:
To simplify \(\frac{15}{\sqrt{3}}\) and \(\frac{7}{\sqrt{3}}\), we rationalize the denominators:
\[
\frac{15}{\sqrt{3}} = \frac{15 \sqrt{3}}{3} = 5 \sqrt{3}
\]
\[
\frac{7}{\sqrt{3}} = \frac{7 \sqrt{3}}{3} = \frac{7 \sqrt{3}}{3}
\]

Therefore, the factored form becomes:
\[
10 \left( x + 5 \sqrt{3} \right) \left( x - \frac{7 \sqrt{3}}{3} \right)
\]

7. **Final factored form**:
\[
10 \left( x + 5 \sqrt{3} \right) \left( x - \frac{7 \sqrt{3}}{3} \right)
\]

However, this form is not in the simplest factored form. Let's re-evaluate the factorization:

\[
10 \left( x + \frac{15}{\sqrt{3}} \right) \left( x - \frac{7}{\sqrt{3}} \right)
\]

Simplifying further:

\[
10 \left( x + 5 \sqrt{3} \right) \left( x - \frac{7 \sqrt{3}}{3} \right)
\]

This is the correct factored form. Therefore, the final answer is:

\[
\boxed{10 \left( x + 5 \sqrt{3} \right) \left( x - \frac{7 \sqrt{3}}{3} \right)}
\]
10 \left(-x-\frac{35}{\sqrt{3}}\right) \left(3 \sqrt{3}-x\right)
Meta-Llama-3.1-405B-Instruct-Turbo ## Step 1: To factor the given quadratic, we first need to simplify it by getting rid of the fraction and making the coefficient of $x^2$ equal to 1.
We can achieve this by dividing the entire quadratic by 10 and then multiplying the $x$ term by $\sqrt{3}$ to rationalize the denominator.

## Step 2: After simplifying, the quadratic becomes $x^2+\frac{26}{\sqrt{3}}x-105$.
Now, we need to find two numbers whose product is -105 and whose sum is $\frac{26}{\sqrt{3}}$.

## Step 3: To make things easier, let's multiply the entire quadratic by $\sqrt{3}$ to get rid of the fraction.
This gives us $\sqrt{3}x^2+26x-105\sqrt{3}$.

## Step 4: Now, we need to find two numbers whose product is $-105\sqrt{3}$ and whose sum is 26.
Let's try to factor the quadratic by finding these two numbers.

## Step 5: The factors of $-105\sqrt{3}$ are $-15\sqrt{3}$ and 7, or $-21\sqrt{3}$ and 5, or $-35\sqrt{3}$ and 3, or $-105\sqrt{3}$ and 1.
We need to find the pair whose sum is 26.

## Step 6: After trying different pairs, we find that $-15\sqrt{3}$ and 7 do not add up to 26, $-21\sqrt{3}$ and 5 do not add up to 26, and $-35\sqrt{3}$ and 3 do not add up to 26.
However, $-15\sqrt{3}$ and 7 are not the correct pair, $-21\sqrt{3}$ and 5 are not the correct pair, and $-35\sqrt{3}$ and 3 are not the correct pair. But $-15\sqrt{3}$ is not one of the factors of the quadratic.

## Step 7: Upon re-examining the factors, we find that $-15\sqrt{3}$ is not one of the factors, but 15 is a factor of 105, and $\sqrt{3}$ is already present in the quadratic.
Let's try to factor the quadratic again, and we find that $-15\sqrt{3}$ is actually $-15$ times $\sqrt{3}$.

## Step 8: Now, we can rewrite the quadratic as $(\sqrt{3}x-15)(x+7)$.
This is the factored form of the given quadratic.

## Step 9: However, we still need to multiply the entire quadratic by 10 to get the original quadratic.
So, the final answer will be $10(\sqrt{3}x-15)(x+7)$.

## Step 10: We can now simplify the expression to get the final answer.
$10(\sqrt{3}x-15)(x+7)$ is already simplified.

The final answer is: $\boxed{10(\sqrt{3}x-15)(x+7)}$
10 \left(-x-\frac{35}{\sqrt{3}}\right) \left(3 \sqrt{3}-x\right)
Meta-Llama-3.1-70B-Instruct-Turbo ## Step 1: Identify the given quadratic equation
The given quadratic equation is $10 x^2+\frac{260 x}{\sqrt{3}}-1050$.

## Step 2: Determine the coefficients of the quadratic equation
The coefficients of the quadratic equation are $a = 10$, $b = \frac{260}{\sqrt{3}}$, and $c = -1050$.

## Step 3: Use the quadratic formula to find the roots of the equation
The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Plugging in the values of $a$, $b$, and $c$, we get $x = \frac{-\frac{260}{\sqrt{3}} \pm \sqrt{\left(\frac{260}{\sqrt{3}}\right)^2 - 4(10)(-1050)}}{2(10)}$.

## Step 4: Simplify the expression under the square root
Simplifying the expression under the square root, we get $\left(\frac{260}{\sqrt{3}}\right)^2 - 4(10)(-1050) = \frac{67600}{3} + 42000 = \frac{67600 + 126000}{3} = \frac{193600}{3}$.

## Step 5: Simplify the quadratic formula expression
Simplifying the quadratic formula expression, we get $x = \frac{-\frac{260}{\sqrt{3}} \pm \sqrt{\frac{193600}{3}}}{20} = \frac{-\frac{260}{\sqrt{3}} \pm \frac{440}{\sqrt{3}}}{20}$.

## Step 6: Simplify the expression further
Simplifying the expression further, we get $x = \frac{-260 \pm 440}{20\sqrt{3}} = \frac{-260 + 440}{20\sqrt{3}}$ or $x = \frac{-260 - 440}{20\sqrt{3}}$.

## Step 7: Simplify the two possible values of x
Simplifying the two possible values of x, we get $x = \frac{180}{20\sqrt{3}} = \frac{9}{\sqrt{3}} = 3\sqrt{3}$ or $x = \frac{-700}{20\sqrt{3}} = \frac{-35}{\sqrt{3}} = -\frac{35\sqrt{3}}{3}$.

## Step 8: Write the factored form of the quadratic equation
The factored form of the quadratic equation is $10(x - 3\sqrt{3})(x + \frac{35\sqrt{3}}{3})$.

## Step 9: Simplify the factored form
Simplifying the factored form, we get $10(x - 3\sqrt{3})(x + \frac{35\sqrt{3}}{3}) = 10(x - 3\sqrt{3})(\frac{3x + 35\sqrt{3}}{3}) = \frac{10}{3}(3x - 9\sqrt{3})(x + \frac{35\sqrt{3}}{3}) = \frac{10}{3}(3x - 9\sqrt{3})(3x + 35\sqrt{3}) = \frac{10}{3}(9x^2 + 26x\sqrt{3} - 315)$.

## Step 10: Write the final factored form
The final factored form is $\boxed{10(x - 3\sqrt{3})(x + \frac{35\sqrt{3}}{3})}$.

The final answer is: $\boxed{10(x - 3\sqrt{3})(x + \frac{35\sqrt{3}}{3})}$
10 \left(-x-\frac{35}{\sqrt{3}}\right) \left(3 \sqrt{3}-x\right)
Qwen2-72B-Instruct To factor the given quadratic expression, we first simplify it by factoring out the greatest common factor (GCF). The GCF of \(10x^2\), \(\frac{260x}{\sqrt{3}}\), and \(-1050\) is \(10\). So, we can write:

\[10x^2 + \frac{260x}{\sqrt{3}} - 1050 = 10(x^2 + \frac{26x}{\sqrt{3}} - 105)\]

Next, to simplify further, let's rationalize the denominator of the second term:

\[\frac{26x}{\sqrt{3}} = \frac{26x\sqrt{3}}{\sqrt{3}\sqrt{3}} = \frac{26x\sqrt{3}}{3}\]

So, our expression becomes:

\[10(x^2 + \frac{26x\sqrt{3}}{3} - 105)\]

Now, we need to factor the quadratic inside the parentheses. To do this, we look for two numbers that multiply to give us the product of the coefficient of \(x^2\) (which is \(1\)) and the constant term (\(-105\)), and add up to the coefficient of \(x\) (which is \(\frac{26\sqrt{3}}{3}\)).

The product of the coefficient of \(x^2\) and the constant term is \((-105) \times 1 = -105\).

We need to find two numbers that multiply to \(-105\) and add up to \(\frac{26\sqrt{3}}{3}\). These numbers are \(\frac{35\sqrt{3}}{3}\) and \(-\frac{30\sqrt{3}}{3}\) because:

\[\frac{35\sqrt{3}}{3} \times -\frac{30\sqrt{3}}{3} = -105\]

and

\[\frac{35\sqrt{3}}{3} + -\frac{30\sqrt{3}}{3} = \frac{5\sqrt{3}}{3} \times 6 = \frac{30\sqrt{3}}{3} = \frac{26\sqrt{3}}{3}\]

Thus, we can rewrite the quadratic as:

\[x^2 + \frac{35\sqrt{3}}{3}x - \frac{30\sqrt{3}}{3}x - 105\]

This can be factored by grouping:

\[x(x + \frac{35\sqrt{3}}{3}) - 35(x + \frac{35\sqrt{3}}{3})\]

\[= (x - 35)(x + \frac{35\sqrt{3}}{3})\]

Therefore, the fully factored form of the original expression is:

\[10(x - 35)(x + \frac{35\sqrt{3}}{3})\]

Putting the final answer in a \(\boxed{}\):

\[\boxed{10(x - 35)(x + \frac{35\sqrt{3}}{3})}\]
10 \left(-x-\frac{35}{\sqrt{3}}\right) \left(3 \sqrt{3}-x\right)
baichuan4 To factor the quadratic equation $10x^2 + \frac{260x}{\sqrt{3}} - 1050$, we first need to make sure that the coefficients are integers. We can do this by multiplying the entire equation by $\sqrt{3}$ to remove the radical from the coefficient of the $x$ term.

Multiplying through by $\sqrt{3}$ gives us:
$$10x^2\sqrt{3} + \frac{260x}{\sqrt{3}}\sqrt{3} - 1050\sqrt{3}$$

Simplifying, we get:
$$10\sqrt{3}x^2 + 260x - 1050\sqrt{3}$$

Now, we can try to factor this quadratic using the standard form $(ax + b)(cx + d)$. To do this, we need to find two numbers whose product is $ac$ and whose sum is $b$. In this case, $a = 10\sqrt{3}$, $b = 260$, and $c = 1$. So we need to find two numbers whose product is $10\sqrt{3}$ and whose sum is $260$.

We can factor $10\sqrt{3}$ as $(5\sqrt{3})(2)$. Now, we need to find two numbers that multiply to $10\sqrt{3}$ and add to $260$. By trial and error or by using the quadratic formula, we can find that these numbers are $5\sqrt{3}$ and $52$.

Therefore, we can write the quadratic as:
$$(5\sqrt{3}x + 52)(2x - 21)$$

This is because $(5\sqrt{3})(2) = 10\sqrt{3}$ and $(5\sqrt{3})(-21) + (52)(2) = -105\sqrt{3} + 104 = 260 - 1050/\sqrt{3}$.

So, the factored form of the quadratic is:
$$\boxed{(5\sqrt{3}x + 52)(2x - 21)}$$
10 \left(-x-\frac{35}{\sqrt{3}}\right) \left(3 \sqrt{3}-x\right)
doubao \[
\begin{align*}
&10x^2+\frac{260x}{\sqrt{3}}-1050\\
=&10(x^2+\frac{26x}{\sqrt{3}}-105)\\
=&10(x^2+\frac{26\sqrt{3}x}{3}-105)\\
=&10(x^2+\frac{26\sqrt{3}x}{3}+\frac{169}{3}-\frac{169}{3}-105)\\
=&10[(x+\frac{13\sqrt{3}}{3})^2-(\frac{\sqrt{409}}{3})^2]\\
=&10(x+\frac{13\sqrt{3}}{3}+\frac{\sqrt{409}}{3})(x+\frac{13\sqrt{3}}{3}-\frac{\sqrt{409}}{3})
\end{align*}
\]
\[
\boxed{10(x+\frac{13\sqrt{3}}{3}+\frac{\sqrt{409}}{3})(x+\frac{13\sqrt{3}}{3}-\frac{\sqrt{409}}{3})}
\]
10 \left(-x-\frac{35}{\sqrt{3}}\right) \left(3 \sqrt{3}-x\right)
gpto1 To factor the quadratic expression \(10 x^2 + \dfrac{260 x}{\sqrt{3}} - 1050\), we'll aim to break it down into its component factors. Here's a step-by-step approach:

**Step 1: Simplify the expression by eliminating radicals in denominators.**

First, to eliminate denominators, multiply both sides of the equation by \(\sqrt{3}\):
\[
\sqrt{3} \times \left(10 x^2 + \dfrac{260 x}{\sqrt{3}} - 1050\right) = 0.
\]

Simplify:
\[
10 \sqrt{3} x^2 + 260 x - 1050 \sqrt{3} = 0.
\]

**Step 2: Factor the simplified quadratic expression.**

We can rewrite the quadratic as:
\[
10 \sqrt{3} x^2 + 260 x - 1050 \sqrt{3} = 10 \left( \sqrt{3} x^2 + 26 x - 105 \sqrt{3} \right).
\]

Next, factor the quadratic inside the parentheses. To do this, consider factoring it as \((\sqrt{3} x + a)(x + b)\). Expanding the product, we get:
\[
\sqrt{3} x^2 + (\sqrt{3} b + a)x + a b = \sqrt{3} x^2 + 26 x - 105 \sqrt{3}.
\]

Set up the system of equations:
\[
\begin{cases}
\sqrt{3} b + a = 26,\\
a b = -105 \sqrt{3}.
\end{cases}
\]

Let \(a = 26 - \sqrt{3} b\). Substitute back into \(a b\):
\[
(26 - \sqrt{3} b) b = -105 \sqrt{3} \implies 26 b - \sqrt{3} b^2 = -105 \sqrt{3}.
\]

Multiply both sides by \(-1\) to make calculations easier:
\[
\sqrt{3} b^2 - 26 b + 105 \sqrt{3} = 0.
\]

Let \(b = k \sqrt{3}\). Substituting and simplifying, we get the quadratic equation:
\[
3 k^2 - 26 k - 105 = 0.
\]

Solving this quadratic equation, we find \(k = \dfrac{35}{3}\) or \(k = -3\).

**Step 3: Determine the factors.**

Corresponding to \(k = \dfrac{35}{3}\):
\[
b = \dfrac{35}{3} \sqrt{3}, \quad a = 26 - 3 k = -9.
\]

Corresponding to \(k = -3\):
\[
b = -3 \sqrt{3}, \quad a = 26 - 3 (-3) = 35.
\]

Using the solution that results in integer coefficients in \(a\) and \(b\), we select \(k = -3\).

Therefore, the factors are:
\[
\sqrt{3} x^2 + 26 x - 105 \sqrt{3} = (\sqrt{3} x + 35)(x - 3 \sqrt{3}).
\]

**Step 4: Write the final factored form.**

Including the factor of \(10\) we factored out earlier, the original quadratic expression factors completely as:
\[
10 x^2 + \dfrac{260 x}{\sqrt{3}} - 1050 = 10 (\sqrt{3} x + 35)(x - 3 \sqrt{3}).
\]

Therefore, the factored form is:

\(\boxed{10\,(\sqrt{3}x+35)\left(x-3\sqrt{3}\right)}\)
10 \left(-x-\frac{35}{\sqrt{3}}\right) \left(3 \sqrt{3}-x\right)
kimi To factor the quadratic \(10x^2 + \frac{260x}{\sqrt{3}} - 1050\), we will follow these steps:

1. **Simplify the coefficients**: First, let's simplify the coefficient of \(x\) by rationalizing the denominator.
\[
\frac{260}{\sqrt{3}} = \frac{260 \sqrt{3}}{3}
\]
So the quadratic becomes:
\[
10x^2 + \frac{260\sqrt{3}}{3}x - 1050
\]

2. **Find the product and sum**: To factor the quadratic, we need to find two numbers that multiply to the product of the coefficient of \(x^2\) and the constant term, and add up to the coefficient of \(x\). The product is:
\[
10 \times (-1050) = -10500
\]
We need two numbers that multiply to \(-10500\) and add up to \(\frac{260\sqrt{3}}{3}\).

3. **Rationalize the quadratic**: To make it easier to find these numbers, we can multiply the entire quadratic by 3 to clear the fraction:
\[
3 \left(10x^2 + \frac{260\sqrt{3}}{3}x - 1050\right) = 30x^2 + 260\sqrt{3}x - 3150
\]
Now, we need to find two numbers that multiply to \(30 \times (-3150) = -94500\) and add up to \(260\sqrt{3}\).

4. **Find the numbers**: We can use the quadratic formula to find the roots of the quadratic equation \(30x^2 + 260\sqrt{3}x - 3150 = 0\):
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = 30\), \(b = 260\sqrt{3}\), and \(c = -3150\). Plugging in these values, we get:
\[
x = \frac{-260\sqrt{3} \pm \sqrt{(260\sqrt{3})^2 - 4 \cdot 30 \cdot (-3150)}}{2 \cdot 30}
\]
Simplifying inside the square root:
\[
x = \frac{-260\sqrt{3} \pm \sqrt{202800 + 378000}}{60} = \frac{-260\sqrt{3} \pm \sqrt{580800}}{60} = \frac{-260\sqrt{3} \pm 762}{60}
\]
This gives us two solutions:
\[
x = \frac{-260\sqrt{3} + 762}{60} = \frac{-130\sqrt{3} + 381}{30}
\]
and
\[
x = \frac{-260\sqrt{3} - 762}{60} = \frac{-130\sqrt{3} - 381}{30}
\]

5. **Write the factored form**: The quadratic can be factored as:
\[
30(x - \frac{-130\sqrt{3} + 381}{30})(x - \frac{-130\sqrt{3} - 381}{30})
\]
Simplifying the factors:
\[
30 \left( x + \frac{130\sqrt{3} - 381}{30} \right) \left( x + \frac{130\sqrt{3} + 381}{30} \right)
\]
This can be written as:
\[
\left( \sqrt{3}x + 13 \right) \left( 10x - 30\sqrt{3} \right)
\]

6. **Final answer**: The factored form of the quadratic is:
\[
\boxed{(\sqrt{3}x + 13)(10x - 30\sqrt{3})}
\]
10 \left(-x-\frac{35}{\sqrt{3}}\right) \left(3 \sqrt{3}-x\right)
minimax To factor the quadratic $10x^2 + \frac{260x}{\sqrt{3}} - 1050$, we first need to make the coefficients more manageable by eliminating the fraction. We can do this by multiplying the entire equation by $\sqrt{3}$ to get rid of the denominator:

$\sqrt{3} \cdot 10x^2 + \sqrt{3} \cdot \frac{260x}{\sqrt{3}} - \sqrt{3} \cdot 1050 = 0$

This simplifies to:

$10\sqrt{3}x^2 + 260x - 1050\sqrt{3} = 0$

Now, we can factor out the greatest common factor (GCF) from the quadratic, which is 2 in this case:

$2(5\sqrt{3}x^2 + 130x - 525\sqrt{3}) = 0$

Next, we look for two numbers that multiply to $5\sqrt{3} \times -525\sqrt{3} = -2625 \times 3$ and add up to
10 \left(-x-\frac{35}{\sqrt{3}}\right) \left(3 \sqrt{3}-x\right)
wenxinyiyan To factor the given quadratic, we first need to find two numbers that multiply to the constant term (-1050) and add to the coefficient of the linear term ($\frac{260}{\sqrt{3}}$).

Let's first simplify the coefficient of the linear term:
$\frac{260}{\sqrt{3}} = \frac{260\sqrt{3}}{3}$

Now, we need to find two numbers that multiply to -1050 and add to $\frac{260\sqrt{3}}{3}$.

After some trial and error, we find that the two numbers are $70\sqrt{3}$ and $-\frac{30}{\sqrt{3}}$.

Now, we can factor the quadratic as follows:
$10x^2 + \frac{260x}{\sqrt{3}} - 1050 = 10(x^2 + \frac{26x}{\sqrt{3}} - 105)$
$= 10(x + 7\sqrt{3})(x - \frac{30}{\sqrt{3}})$

However, we can further simplify the second factor:
$x - \frac{30}{\sqrt{3}} = x - 10\sqrt{3}$

So, the final factored form of the quadratic is:
$\boxed{10(x + 7\sqrt{3})(x - 10\sqrt{3})}$
10 \left(-x-\frac{35}{\sqrt{3}}\right) \left(3 \sqrt{3}-x\right)
zhipu First, we factor out the common factor of 10 from each term:
\[
10 x^2+\frac{260 x}{\sqrt{3}}-1050 = 10(x^2+\frac{26 x}{\sqrt{3}}-105).
\]

Next, we focus on factoring the quadratic inside the parentheses. We look for two numbers that multiply to $-105$ and add to $\frac{26}{\sqrt{3}}$.

To rationalize the coefficients, multiply the entire quadratic by $\sqrt{3}$:
\[
\sqrt{3} \cdot (x^2+\frac{26 x}{\sqrt{3}}-105) = \sqrt{3}x^2 + 26x - 105\sqrt{3}.
\]

Now, factor the quadratic $\sqrt{3}x^2 + 26x - 105\sqrt{3}$. We need two numbers that multiply to $-105\sqrt{3} \cdot \sqrt{3} = -315$ and add to 26. These numbers are 35 and -9:
\[
\sqrt{3}x^2 + 26x - 105\sqrt{3} = (\sqrt{3}x + 35)(x - 3\sqrt{3}).
\]

Divide back by $\sqrt{3}$ to return to the original coefficients:
\[
x^2+\frac{26 x}{\sqrt{3}}-105 = \left(x + \frac{35}{\sqrt{3}}\right)\left(x - 3\sqrt{3}\right).
\]

Now, include the factored out 10:
\[
10(x^2+\frac{26 x}{\sqrt{3}}-105) = 10\left(x + \frac{35}{\sqrt{3}}\right)\left(x - 3\sqrt{3}\right).
\]

To simplify further, rationalize the denominator in the first factor:
\[
x + \frac{35}{\sqrt{3}} = x + \frac{35\sqrt{3}}{3}.
\]

Thus, the final factorization is:
\[
\boxed{10\left(x + \frac{35\sqrt{3}}{3}\right)(x - 3\sqrt{3})}.
\]
10 \left(-x-\frac{35}{\sqrt{3}}\right) \left(3 \sqrt{3}-x\right)
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