| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To factor the quadratic expression \(2 x^2 + \frac{104 x}{3} - \frac{320}{3}\), we will follow these steps: 1. **Multiply the entire expression by 3 to clear the fractions:** \[ 3 \left( 2 x^2 + \frac{104 x}{3} - \frac{320}{3} \right) = 6 x^2 + 104 x - 320 \] 2. **Factor the resulting polynomial \(6 x^2 + 104 x - 320\):** - First, find the greatest common factor (GCF) of the coefficients: \[ \text{GCF of } 6, 104, \text{ and } -320 \text{ is } 2 \] - Factor out the GCF: \[ 6 x^2 + 104 x - 320 = 2 (3 x^2 + 52 x - 160) \] 3. **Factor the quadratic \(3 x^2 + 52 x - 160\):** - We need to find two numbers that multiply to \(3 \cdot (-160) = -480\) and add to \(52\). - These numbers are \(60\) and \(-8\) because: \[ 60 \cdot (-8) = -480 \quad \text{and} \quad 60 + (-8) = 52 \] - Rewrite the middle term using these numbers: \[ 3 x^2 + 52 x - 160 = 3 x^2 + 60 x - 8 x - 160 \] - Group the terms and factor by grouping: \[ (3 x^2 + 60 x) + (-8 x - 160) = 3 x (x + 20) - 8 (x + 20) \] - Factor out the common binomial factor \((x + 20)\): \[ 3 x (x + 20) - 8 (x + 20) = (3 x - 8) (x + 20) \] 4. **Substitute back into the factored form with the GCF:** \[ 6 x^2 + 104 x - 320 = 2 (3 x - 8) (x + 20) \] 5. **Divide the entire expression by 3 to revert to the original form:** \[ 2 x^2 + \frac{104 x}{3} - \frac{320}{3} = \frac{2 (3 x - 8) (x + 20)}{3} \] Thus, the factored form of the original quadratic expression is: \[ \boxed{\frac{2 (3 x - 8) (x + 20)}{3}} \] |
-2 (-x-20) \left(x-\frac{8}{3}\right) |
| deepseek-r1 | To factor the quadratic expression \(2x^2 + \frac{104x}{3} - \frac{320}{3}\), we start by eliminating the fractions by multiplying the entire expression by 3: \[ 3 \left(2x^2 + \frac{104x}{3} - \frac{320}{3}\right) = 6x^2 + 104x - 320 \] Next, we factor the quadratic \(6x^2 + 104x - 320\). We look for two numbers that multiply to \(6 \times (-320) = -1920\) and add up to 104. These numbers are 120 and -16. We split the middle term and factor by grouping: \[ 6x^2 + 120x - 16x - 320 \] Grouping the terms: \[ (6x^2 + 120x) + (-16x - 320) \] Factoring out the greatest common factors from each group: \[ 6x(x + 20) - 16(x + 20) \] Factoring out the common factor \((x + 20)\): \[ (6x - 16)(x + 20) \] Since we initially multiplied by 3, we now divide by 3 to return to the original expression: \[ \frac{(6x - 16)(x + 20)}{3} \] We factor out 2 from \(6x - 16\): \[ \frac{2(3x - 8)(x + 20)}{3} \] Thus, the factored form of the quadratic is: \[ \boxed{\dfrac{2}{3}(3x - 8)(x + 20)} \] |
-2 (-x-20) \left(x-\frac{8}{3}\right) |
| gpto1 | To factor the quadratic \(2x^2 + \frac{104x}{3} - \frac{320}{3}\): **Step 1: Eliminate Fractions** Multiply the entire expression by 3 to eliminate denominators: \[ 3 \times \left(2x^2 + \frac{104x}{3} - \frac{320}{3}\right) = 6x^2 + 104x - 320 \] **Step 2: Factor Out Common Factors** Notice that all coefficients are divisible by 2: \[ 2(3x^2 + 52x - 160) \] **Step 3: Factor the Quadratic** We need to factor \(3x^2 + 52x - 160\). To do this, look for two numbers that multiply to \(a \times c = 3 \times (-160) = -480\) and add up to \(b = 52\). The numbers are \(-8\) and \(60\): \[ (-8) \times 60 = -480 \quad \text{and} \quad (-8) + 60 = 52 \] **Step 4: Rewrite and Factor by Grouping** Rewrite \(52x\) as \(-8x + 60x\): \[ 3x^2 -8x +60x -160 \] Group terms and factor each group: \[ (3x^2 -8x) + (60x -160) = x(3x -8) +20(3x -8) \] Extract the common factor \((3x -8)\): \[ (3x -8)(x + 20) \] **Step 5: Combine Factors** Include the common factor pulled out earlier: \[ 2(3x -8)(x + 20) \] **Final Answer:** \[ \boxed{2(3x -8)(x + 20)} \] |
-2 (-x-20) \left(x-\frac{8}{3}\right) |
| gpt-4o-2024-08-06 | To factor the quadratic \(2x^2 + \frac{104x}{3} - \frac{320}{3}\), we start by eliminating the fractions. We can do this by multiplying the entire expression by 3, the common denominator: \[ 3 \left(2x^2 + \frac{104x}{3} - \frac{320}{3}\right) = 6x^2 + 104x - 320 \] Now, we need to factor the quadratic \(6x^2 + 104x - 320\). We will use the method of factoring by grouping, which involves finding two numbers that multiply to the product of the leading coefficient and the constant term, and add to the middle coefficient. The product of the leading coefficient and the constant term is: \[ 6 \times (-320) = -1920 \] We need two numbers that multiply to \(-1920\) and add to \(104\). After testing several pairs, we find that \(120\) and \(-16\) work because: \[ 120 \times (-16) = -1920 \quad \text{and} \quad 120 + (-16) = 104 \] We can now rewrite the middle term \(104x\) using these two numbers: \[ 6x^2 + 120x - 16x - 320 \] Next, we factor by grouping: \[ (6x^2 + 120x) + (-16x - 320) \] Factor out the greatest common factor from each group: \[ 6x(x + 20) - 16(x + 20) \] Notice that \((x + 20)\) is a common factor: \[ (6x - 16)(x + 20) \] We can simplify \(6x - 16\) by factoring out the greatest common factor, which is 2: \[ 2(3x - 8)(x + 20) \] Thus, the factored form of the original quadratic is: \[ \boxed{2(3x - 8)(x + 20)} \] |
-2 (-x-20) \left(x-\frac{8}{3}\right) |
| DeepSeek-Coder-V2-Lite-Instruct | To factor the quadratic expression \(2 x^2 + \frac{104 x}{3} - \frac{320}{3}\), we will follow these steps: 1. **Combine the terms over a common denominator**: \[ 2 x^2 + \frac{104 x}{3} - \frac{320}{3} \] The common denominator is 3. So, we rewrite each term with this common denominator: \[ \frac{6 x^2}{3} + \frac{104 x}{3} - \frac{320}{3} \] Combine these fractions: \[ \frac{6 x^2 + 104 x - 320}{3} \] 2. **Factor out the greatest common divisor (GCD) from the numerator**: The GCD of the terms \(6 x^2\), \(104 x\), and \(-320\) is 2. Factor out 2: \[ \frac{2 (3 x^2 + 52 x - 160)}{3} \] 3. **Factor the quadratic expression inside the parentheses**: We need to factor \(3 x^2 + 52 x - 160\). We look for two numbers that multiply to \(3 \times (-160) = -480\) and add to \(52\). The numbers that satisfy these conditions are \(60\) and \(-8\): \[ 3 x^2 + 60 x - 8 x - 160 \] Group the terms: \[ (3 x^2 + 60 x) + (-8 x - 160) \] Factor by grouping: \[ 3 x (x + 20) - 8 (x + 20) \] Factor out the common binomial factor \((x + 20)\): \[ (3 x - 8)(x + 20) \] 4. **Combine the factored form with the outer factor**: \[ \frac{2 (3 x - 8)(x + 20)}{3} \] Thus, the factored form of the quadratic expression is: \[ \boxed{\frac{2 (3 x - 8)(x + 20)}{3}} \] |
-2 (-x-20) \left(x-\frac{8}{3}\right) |
| Meta-Llama-3.1-405B-Instruct-Turbo | ## Step 1: To factor the given quadratic, we first need to identify the coefficients of the quadratic equation. The given quadratic equation is $2x^2 + \frac{104x}{3} - \frac{320}{3}$. The coefficients are $a = 2$, $b = \frac{104}{3}$, and $c = -\frac{320}{3}$. ## Step 2: Next, we need to find two numbers whose product is $ac$ and whose sum is $b$. The product $ac$ is $2 \times -\frac{320}{3} = -\frac{640}{3}$. We are looking for two numbers that multiply to $-\frac{640}{3}$ and add up to $\frac{104}{3}$. ## Step 3: Finding the two numbers that satisfy the conditions mentioned in step 2. To simplify the process, we can multiply all coefficients by 3 to get rid of the fractions, turning the equation into $6x^2 + 104x - 320$. Now, we look for factors of $6 \times -320 = -1920$ that add up to $104$. These numbers are $160$ and $-12$ because $160 - 12 = 148$ does not equal $104$, we made a mistake. ## Step 4: Correct the mistake made in step 3. The correct factors to look for should multiply to $-1920$ and add to $104$. By trial and error, we find that $120$ and $-16$ are the correct numbers because $120 \times -16 = -1920$ and $120 + (-16) = 104$. ## Step 5: Rewrite the middle term using the numbers found in step 4. Rewrite the quadratic equation with the correct factors: $6x^2 + 120x - 16x - 320$. ## Step 6: Factor by grouping. Group the terms: $(6x^2 + 120x) + (-16x - 320)$. Factor out the greatest common factor (GCF) from each group: $6x(x + 20) - 16(x + 20)$. ## Step 7: Factor out the common binomial factor. The common binomial factor is $(x + 20)$. Factoring it out gives: $(6x - 16)(x + 20)$. ## Step 8: Simplify the expression if necessary. The expression $(6x - 16)(x + 20)$ can be simplified by dividing through by the common factor we initially removed (which was 3 in this case) to keep the equation equivalent to the original. However, since we adjusted our method to work with the multiplied coefficients, our factorization is correct for the adjusted equation. To match the original, ensure the factorization correctly reflects the original coefficients, thus we simplify by realizing we do not need to adjust back since our factorization correctly represents the original quadratic in its multiplied form. The final answer is: $\boxed{(6x - 16)(x + 20)/3}$ |
-2 (-x-20) \left(x-\frac{8}{3}\right) |
| Meta-Llama-3.1-70B-Instruct-Turbo | ## Step 1: Identify the coefficients of the quadratic equation The given quadratic equation is $2x^2 + \frac{104x}{3} - \frac{320}{3}$. The coefficients are $a = 2$, $b = \frac{104}{3}$, and $c = -\frac{320}{3}$. ## Step 2: Use the factoring method to factor the quadratic equation To factor the quadratic equation, we need to find two numbers whose product is $ac$ and whose sum is $b$. In this case, $ac = 2 \times -\frac{320}{3} = -\frac{640}{3}$ and $b = \frac{104}{3}$. ## Step 3: Find the two numbers whose product is $ac$ and whose sum is $b$ We need to find two numbers whose product is $-\frac{640}{3}$ and whose sum is $\frac{104}{3}$. These numbers are $\frac{80}{3}$ and $-\frac{8}{3}$, since $\frac{80}{3} \times -\frac{8}{3} = -\frac{640}{3}$ and $\frac{80}{3} + (-\frac{8}{3}) = \frac{104}{3}$. ## Step 4: Write the factored form of the quadratic equation Using the numbers found in Step 3, we can write the factored form of the quadratic equation as $2x^2 + \frac{104x}{3} - \frac{320}{3} = 2(x + \frac{80}{3})(x - \frac{8}{3})$. ## Step 5: Simplify the factored form Simplifying the factored form, we get $2(x + \frac{80}{3})(x - \frac{8}{3}) = 2(x + \frac{80}{3})(x - \frac{8}{3})$. ## Step 6: Write the final answer in the required format The final answer is: $\boxed{2(x + \frac{80}{3})(x - \frac{8}{3})}$ |
-2 (-x-20) \left(x-\frac{8}{3}\right) |
| Qwen2-72B-Instruct | To factor the given quadratic expression, we first need to find two numbers that multiply to give the product of the coefficient of \(x^2\) and the constant term, and add up to the coefficient of \(x\). The given quadratic is: \[2x^2 + \frac{104}{3}x - \frac{320}{3}\] First, let's simplify it by multiplying through by 3 to eliminate the fractions: \[6x^2 + 104x - 320\] Now, we need to find two numbers that multiply to \(6 \times (-320) = -1920\) and add up to \(104\). These numbers are \(120\) and \(-16\), since: \[120 \times (-16) = -1920\] \[120 + (-16) = 104\] So, we can write the middle term as: \[104x = 120x - 16x\] Thus, the expression becomes: \[6x^2 + 120x - 16x - 320\] We can then group the terms and factor by grouping: \[(6x^2 + 120x) - (16x + 320)\] Factoring out common factors from each group gives: \[6x(x + 20) - 16(x + 20)\] Now, we can factor out the common binomial factor \((x + 20)\): \[(6x - 16)(x + 20)\] Simplifying further: \[2(3x - 8)(x + 20)\] Therefore, the factored form of the given quadratic expression is: \[\boxed{2(3x - 8)(x + 20)}\] |
-2 (-x-20) \left(x-\frac{8}{3}\right) |
| baichuan4 | To factor the quadratic equation $2x^2 + \frac{104x}{3} - \frac{320}{3}$, we first need to make sure that the equation is in standard form, which means that the coefficient of the $x^2$ term is equal to 1. We can do this by dividing the entire equation by the coefficient of the $x^2$ term, which is 2. So, let's divide the equation by 2: $$ \frac{2x^2}{2} + \frac{\frac{104x}{3}}{2} - \frac{\frac{320}{3}}{2} $$ This simplifies to: $$ x^2 + \frac{52x}{3} - \frac{160}{3} $$ Now, we need to find two numbers whose product is equal to the constant term (the term without an $x$), which is $-\frac{160}{3}$, and whose sum is equal to the coefficient of the $x$ term, which is $\frac{52}{3}$. To find these numbers, we can use the fact that the product of two numbers is negative if one is positive and the other is negative. And since the sum of these numbers is positive, the positive number must be larger in absolute value than the negative number. Let's try some numbers and see if we can find a pair that works. We know that the product should be $-\frac{160}{3}$, so we can try factors of $160$ and adjust them as necessary. Factors of $160$ include $1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 80, 160$. We need a pair of numbers whose product is $-\frac{160}{3}$, so we can start by looking at the pairs $(8, 20)$ and $(-8, -20)$. We adjust these pairs by dividing them by $3$ to get the correct product: $$ \frac{8}{3} \cdot \frac{20}{3} = \frac{160}{9} $$ This is too large, so we need to find a smaller pair. Let's try $(4, 40)$ and $(-4, -40)$. Adjusting for the product: $$ \frac{4}{3} \cdot \frac{40}{3} = \frac{160}{9} $$ Still too large. Let's try $(2, 80)$ and $(-2, -80)$. Adjusting for the product: $$ \frac{2}{3} \cdot \frac{80}{3} = \frac{160}{9} $$ Still too large. Finally, let's try $(1, 160)$ and $(-1, -160)$. Adjusting for the product: $$ \frac{1}{3} \cdot \frac{160}{3} = \frac{160}{9} $$ This is getting closer, but the product is still not $-\frac{160}{3}$. We need to find a pair where one number is negative and the other is positive, and their product is $-\frac{160}{3}$. Consider the pair $(8, -20)$. Adjusting for the product: $$ \frac{8}{3} \cdot \left(-\frac{20}{3}\right) = -\frac{160}{9} $$ This is not quite right, but it's close. We need to find a pair that, when divided by $3$, gives us the correct product. The pair $(10, -16)$ works: $$ \frac{10}{3} \cdot \left(-\frac{16}{3}\right) = -\frac{160}{9} $$ Now, let's check the sum: $$ \frac{10}{3} + \left(-\frac{16}{3}\right) = \frac{10 - 16}{3} = -\frac{6}{3} = -2 $$ This is not the correct sum. We need a sum of $\frac{52}{3}$. Let's try the pair $(16, -10)$: $$ \frac{16}{3} + \left(-\frac{10}{3}\right) = \frac{16 - 10}{3} = \frac{6}{3} = 2 $$ This is also not the correct sum. We need to find a pair that sums to $\frac{52}{3}$. The pair $(20, -8)$ works: $$ \frac{20}{3} + \left(-\frac{8}{3}\right) = \frac{20 - 8}{3} = \frac{12}{3} = 4 $$ This is still not the correct sum. We need to find a pair that sums to $\frac{52}{3}$. The pair $(24, -6)$ works: $$ \frac{24}{3} + \left(-\frac{6}{3}\right) = \frac{24 - 6}{3} = \frac{18}{3} = 6 $$ This is still not the correct sum. We need to find a pair that sums to $\frac{52}{3}$. The pair $(26, -4)$ works: $$ \frac{26}{3} + \left(-\frac{4}{3}\right) = \frac{26 - 4}{3} = \frac{22}{3} $$ This is still not the correct sum. We need to find a pair that sums to $\frac{52}{3}$. The pair $(28, -2)$ works: $$ \frac{28}{3} + \left(-\frac{2}{3}\right) = \frac{28 - 2}{3} = \frac{26}{3} $$ This is still not the correct sum. We need to find a pair that sums to $\frac{52}{3}$. The pair $(30, -10)$ works: $$ \frac{30}{3} + \left(-\frac{10}{3}\right) = \frac{30 - 10}{3} = \frac{20}{3} $$ This is still not the correct sum. We need to find a pair that sums to $\frac{52}{3}$. The pair $(32, -8)$ works: $$ \frac{32}{3}+ \left+ \left(-\frac{8}{3}\right) = \frac{32 - 8}{3} = \frac{24}{3} = 8 $$ This is still not the correct sum. We need to find a pair that sums to $\frac{52}{3}$. The pair $(34, -6)$ works: $$ \frac{34}{3} + \left(-\frac{6}{3}\right) = \frac{34 - 6}{3} = \frac{28}{3} $$ This is still not the correct sum. We need to find a pair that sums to $\frac{52}{3}$. The pair $(36, -4)$ works: $$ \frac{36}{3} + \left(-\frac{4}{3}\right) = \frac{36 - 4}{3} = \frac{32}{3} $$ This is still not the correct sum. We need to find a pair that sums to $\frac{52}{3}$. The pair $(38, -2)$ works: $$ \frac{38}{3} + \left(-\frac{2}{3}\right) = \frac{38 - 2}{3} = \frac{36}{3} = 12 $$ This is still not the correct sum. We need to find a pair that sums to $\frac{52}{3}$. The pair $(40, -10)$ works: $$ \frac{40}{3} + \left(-\frac{10}{3}\right) = \frac{40 - 10}{3} = \frac{30}{3} = 10 $$ |
-2 (-x-20) \left(x-\frac{8}{3}\right) |
| doubao | \[ \begin{align*} &2x^2+\frac{104x}{3}-\frac{320}{3}\\ =&2(x^2+\frac{52x}{3}-\frac{160}{3})\\ =&2(x^2+(\frac{55-3}{3})x-\frac{160}{3})\\ =&2(x^2+(\frac{55}{3}-\frac{3}{3})x-\frac{160}{3})\\ =&2(x^2+\frac{52}{3}x-\frac{160}{3})\\ =&2(x^2+(\frac{65-13}{3})x-\frac{160}{3})\\ =&2(x^2+(\frac{65}{3}-\frac{13}{3})x-\frac{160}{3})\\ =&2(x^2+\frac{65}{3}x-\frac{13}{3}x-\frac{160}{3})\\ =&2[x(x+\frac{65}{3})- \frac{13}{3}(x+\frac{160}{13})]\\ =&2(x+\frac{160}{13})(x-\frac{13}{3}) \end{align*} \] So the answer is $\boxed{2(x+\frac{160}{13})(x-\frac{13}{3})}$. |
-2 (-x-20) \left(x-\frac{8}{3}\right) |
| gpto1 | To factor the quadratic \(2x^2 + \frac{104x}{3} - \frac{320}{3}\): **Step 1: Eliminate Fractions** Multiply the entire expression by 3 to eliminate denominators: \[ 3 \times \left(2x^2 + \frac{104x}{3} - \frac{320}{3}\right) = 6x^2 + 104x - 320 \] **Step 2: Factor Out Common Factors** Notice that all coefficients are divisible by 2: \[ 2(3x^2 + 52x - 160) \] **Step 3: Factor the Quadratic** We need to factor \(3x^2 + 52x - 160\). To do this, look for two numbers that multiply to \(a \times c = 3 \times (-160) = -480\) and add up to \(b = 52\). The numbers are \(-8\) and \(60\): \[ (-8) \times 60 = -480 \quad \text{and} \quad (-8) + 60 = 52 \] **Step 4: Rewrite and Factor by Grouping** Rewrite \(52x\) as \(-8x + 60x\): \[ 3x^2 -8x +60x -160 \] Group terms and factor each group: \[ (3x^2 -8x) + (60x -160) = x(3x -8) +20(3x -8) \] Extract the common factor \((3x -8)\): \[ (3x -8)(x + 20) \] **Step 5: Combine Factors** Include the common factor pulled out earlier: \[ 2(3x -8)(x + 20) \] **Final Answer:** \[ \boxed{2(3x -8)(x + 20)} \] |
-2 (-x-20) \left(x-\frac{8}{3}\right) |
| kimi | To factor the quadratic \(2x^2 + \frac{104x}{3} - \frac{320}{3}\), we will follow these steps: 1. **Extract the greatest common factor (GCF):** The GCF of the coefficients \(2\), \(\frac{104}{3}\), and \(-\frac{320}{3}\) is \(2\). \[ 2x^2 + \frac{104x}{3} - \frac{320}{3} = 2 \left( x^2 + \frac{52x}{3} - \frac{160}{3} \right) \] 2. **Simplify the expression inside the parentheses:** Divide each term inside the parentheses by \(2\): \[ x^2 + \frac{52x}{3} - \frac{160}{3} = \frac{3x^2 + 52x - 160}{3} \] 3. **Factor the quadratic expression inside the parentheses:** The quadratic expression \(\frac{3x^2 + 52x - 160}{3}\) can be factored by finding two numbers that multiply to \(3 \times (-160) = -480\) and add up to \(52\). The two numbers that satisfy these conditions are \(60\) and \(-8\). Rewrite the middle term using these numbers: \[ 3x^2 + 60x - 8x - 160 \] Group the terms in pairs: \[ (3x^2 + 60x) - (8x + 160) \] Factor out the GCF from each pair: \[ 3x(x + 20) - 8(x + 20) \] Factor out the common binomial factor \((x + 20)\): \[ (3x - 8)(x + 20) \] 4. **Combine the factored form with the extracted GCF:** Multiply the factored form by the extracted GCF \(2\): \[ 2 \left( \frac{3x - 8}{3} \right) \left( x + 20 \right) \] 5. **Simplify the fraction:** Multiply the GCF \(2\) by the numerator of the fraction: \[ 2 \cdot \frac{3x - 8}{3} = \frac{2(3x - 8)}{3} = \frac{6x - 16}{3} \] So, the final factored form is: \[ \frac{6x - 16}{3} (x + 20) \] 6. **Write the final answer within a box:** \[ \boxed{\frac{6x - 16}{3} (x + 20)} \] |
-2 (-x-20) \left(x-\frac{8}{3}\right) |
| minimax | To factor the quadratic $2x^2 + \frac{104}{3}x - \frac{320}{3}$, we first need to find two numbers that multiply to give the product of the coefficient of $x^2$ (which is 2) and the constant term (which is $-\frac{320}{3}$), and at the same time add up to the coefficient of $x$ (which is $\frac{104}{3}$). Let's find the product of the coefficient of $x^2$ and the constant term: $2 \times -\frac{320}{3} = -\frac{640}{3}$ Now we need to find two numbers that multiply to $-\frac{640}{3}$ and add up to $\frac{104}{3}$. Since the numbers are fractions, let's first convert them to whole numbers by multiplying by 3 to eliminate the denominators: $2 \times -640 = -1280$ $\frac{104}{3} \times 3 = 104$ Now we need to find two numbers that multiply to $-12 |
-2 (-x-20) \left(x-\frac{8}{3}\right) |
| wenxinyiyan | To factor the given quadratic, we first need to find two numbers that multiply to the constant term $-\frac{320}{3}$ and add to the coefficient of the linear term $\frac{104}{3}$. Let's start by finding two numbers that multiply to $-\frac{320}{3}$. These numbers could be $-\frac{40}{3}$ and $8$, since $-\frac{40}{3} \times 8 = -\frac{320}{3}$. Next, we check if these two numbers add to $\frac{104}{3}$. We find that $-\frac{40}{3} + 8 = \frac{-40 + 24}{3} = \frac{-16}{3}$, which does not equal $\frac{104}{3}$. It seems we chose the wrong numbers. Let's try again. This time, we look for two numbers that multiply to $-\frac{320}{3}$ and add to $\frac{104}{3}$. After some trial and error, we find that $16$ and $-\frac{20}{3}$ work. They multiply to $-\frac{320}{3}$ and add to $\frac{104}{3}$: $16 \times -\frac{20}{3} = -\frac{320}{3}$ $16 + -\frac{20}{3} = \frac{48 - 20}{3} = \frac{28}{3}$ Now that we have found the correct numbers, we can factor the quadratic: $2 x^2+\frac{104 x}{3}-\frac{320}{3} = 2x^2 + 16x - \frac{20}{3}x - \frac{320}{3}$ $= 2x(x + 8) - \frac{20}{3}(x + 8)$ $= (2x - \frac{20}{3})(x + 8)$ $= 2(x - \frac{10}{3})(x + 8)$ $= \boxed{2(x - \frac{10}{3})(x + 8)}$ |
-2 (-x-20) \left(x-\frac{8}{3}\right) |
| zhipu | First, we multiply the entire quadratic by $3$ to eliminate the fractions, resulting in: $$ 3(2x^2 + \frac{104x}{3} - \frac{320}{3}) = 6x^2 + 104x - 320. $$ Next, we factor out the common factor of $4$ from the quadratic: $$ 6x^2 + 104x - 320 = 4(1.5x^2 + 26x - 80). $$ To simplify further, we can multiply the terms inside the parentheses by $2$ to clear the decimal: $$ 4(1.5x^2 + 26x - 80) = 4(3x^2 + 52x - 160). $$ Now, we factor the quadratic inside the parentheses: We look for two numbers that multiply to $3 \times -160 = -480$ and add up to $52$. These numbers are $60$ and $-8$: $$ 3x^2 + 52x - 160 = 3x^2 + 60x - 8x - 160. $$ Grouping and factoring gives: $$ = 3x(x + 20) - 8(x + 20) = (3x - 8)(x + 20). $$ Thus, the fully factored form of the original quadratic is: $$ 4(3x - 8)(x + 20). $$ Putting it in a box, we get: $$ \boxed{4(3x - 8)(x + 20)}. $$ |
-2 (-x-20) \left(x-\frac{8}{3}\right) |
| deepseek-v3 | To factor the quadratic expression \( 2x^2 + \frac{104x}{3} - \frac{320}{3} \), follow these steps: 1. **Eliminate the Fractions:** Multiply every term by 3 to eliminate the denominators: \[ 3 \times 2x^2 + 3 \times \frac{104x}{3} - 3 \times \frac{320}{3} = 6x^2 + 104x - 320 \] 2. **Factor Out the Greatest Common Factor (GCF):** The GCF of the coefficients 6, 104, and -320 is 2: \[ 2(3x^2 + 52x - 160) \] 3. **Factor the Quadratic Expression Inside the Parentheses:** We need to factor \( 3x^2 + 52x - 160 \). Look for two numbers that multiply to \( 3 \times (-160) = -480 \) and add to 52. These numbers are 60 and -8. Rewrite the middle term using these numbers: \[ 3x^2 + 60x - 8x - 160 \] Factor by grouping: \[ (3x^2 + 60x) + (-8x - 160) = 3x(x + 20) - 8(x + 20) \] Factor out the common binomial factor: \[ (3x - 8)(x + 20) \] 4. **Combine the Factors:** Include the GCF that was factored out initially: \[ 2(3x - 8)(x + 20) \] So, the factored form of the quadratic expression is: \[ \boxed{2(3x - 8)(x + 20)} \] |
-2 (-x-20) \left(x-\frac{8}{3}\right) |