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Remove the largest possible common factor. Check your answer by multiplicat \[ 14 x^{4}+21 x^{3}+28 x \] Factor out the greatest common factor \( 14 x^{4}+21 x^{3}+28 x=\square \)

Ask by Barnett Vaughn. in the United States
Mar 11,2025

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Tutor-Verified Answer

Answer

The factored form is \(7x(2x^3 + 3x^2 + 4)\).

Solution

We begin with the polynomial: \[ 14 x^{4}+21 x^{3}+28 x \] **Step 1: Identify the greatest common factor (GCF).** - For the coefficients \(14\), \(21\), and \(28\), the GCF is \(7\). - For the variable part, the smallest power of \(x\) present is \(x\). Thus, the GCF is \(7x\). **Step 2: Factor out \(7x\).** Divide each term by \(7x\): - \(\displaystyle \frac{14x^4}{7x} = 2x^3\) - \(\displaystyle \frac{21x^3}{7x} = 3x^2\) - \(\displaystyle \frac{28x}{7x} = 4\) So, we have: \[ 14 x^{4}+21 x^{3}+28 x = 7x(2x^3+3x^2+4) \] **Step 3: Check the factorization by multiplication.** Multiply \(7x\) by each term in the parentheses: \[ 7x \cdot 2x^3 = 14x^4, \] \[ 7x \cdot 3x^2 = 21x^3, \] \[ 7x \cdot 4 = 28x. \] Adding these gives the original polynomial: \[ 14x^4 + 21x^3 + 28x. \] Thus, the factorization is correct. The final factorization is: \[ 14 x^{4}+21 x^{3}+28 x = 7x(2x^3+3x^2+4) \]

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To factor out the greatest common factor (GCF) from the expression \( 14 x^{4}+21 x^{3}+28 x \), we first identify the GCF of the coefficients (14, 21, and 28), which is 7. Additionally, the lowest power of \( x \) present in all terms is \( x \), so we can also factor out \( x \). Thus, the GCF is \( 7x \). Now, we can rewrite the expression: \[ 14 x^{4}+21 x^{3}+28 x = 7x(2x^{3}+3x^{2}+4) \] To check our work, we can multiply \( 7x \) back by \( (2x^{3}+3x^{2}+4) \): \[ 7x(2x^{3}+3x^{2}+4) = 14x^{4} + 21x^{3} + 28x \] This verifies that the factoring was done correctly! So, the final answer is: \[ 14 x^{4}+21 x^{3}+28 x=7x(2x^{3}+3x^{2}+4) \]

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