3. \( x^{2}+x-12 \) \( \rightarrow(x+3)(x+4) \) 4. \( 2 x^{2}-14 x+24 \) \( 5-2 x^{2}-4 x-30 \) 4 \( 2 x^{2}+4 x-16 \)

It looks like you're working on factoring quadratic polynomials. Let's go through each of the problems you've listed, verify any given factorizations, and provide the correct factorizations where necessary. --- ### **Problem 3** **Given Polynomial:** \[ x^{2} + x - 12 \] **Attempted Factorization:** \[ (x + 3)(x + 4) \] **Verification:** Let's expand \((x + 3)(x + 4)\): \[ (x + 3)(x + 4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12 \] This does **not** match the original polynomial \(x^{2} + x - 12\). **Correct Factorization:** We need two numbers that multiply to \(-12\) (the constant term) and add up to \(1\) (the coefficient of \(x\)). - **Numbers:** \(4\) and \(-3\) \(4 \times (-3) = -12\) \(4 + (-3) = 1\) **Factorization:** \[ x^{2} + x - 12 = (x + 4)(x - 3) \] --- ### **Problem 4** **Given Polynomial:** \[ 2x^{2} - 14x + 24 \] **Factorization Steps:** 1. **Factor out the Greatest Common Factor (GCF):** The GCF of \(2x^{2}\), \(-14x\), and \(24\) is \(2\). \[ 2x^{2} - 14x + 24 = 2(x^{2} - 7x + 12) \] 2. **Factor the Quadratic Inside the Parentheses:** We need two numbers that multiply to \(12\) and add up to \(-7\). - **Numbers:** \(-3\) and \(-4\) \(-3 \times -4 = 12\) \(-3 + (-4) = -7\) 3. **Complete Factorization:** \[ 2(x^{2} - 7x + 12) = 2(x - 3)(x - 4) \] --- ### **Problem 5** **Given Polynomial:** \[ -2x^{2} - 4x - 30 \] **Factorization Steps:** 1. **Factor out the GCF:** The GCF of \(-2x^{2}\), \(-4x\), and \(-30\) is \(-2\). \[ -2x^{2} - 4x - 30 = -2(x^{2} + 2x + 15) \] 2. **Attempt to Factor the Quadratic Inside the Parentheses:** We need two numbers that multiply to \(15\) and add up to \(2\). - **Possible Pairs:** \(5\) and \(3\), but \(5 + 3 = 8 \neq 2\). - Since there are no real numbers that satisfy both conditions, the quadratic \(x^{2} + 2x + 15\) **cannot** be factored further over the set of real numbers. 3. **Final Factorization:** \[ -2x^{2} - 4x - 30 = -2(x^{2} + 2x + 15) \] *Note:* If you're dealing with complex numbers, you can further factor \(x^{2} + 2x + 15\), but for real numbers, this is the simplest form. --- ### **Problem 4 (Another Instance or Possibly Problem 6)** **Given Polynomial:** \[ 2x^{2} + 4x - 16 \] **Factorization Steps:** 1. **Factor out the GCF:** The GCF of \(2x^{2}\), \(4x\), and \(-16\) is \(2\). \[ 2x^{2} + 4x - 16 = 2(x^{2} + 2x - 8) \] 2. **Factor the Quadratic Inside the Parentheses:** We need two numbers that multiply to \(-8\) and add up to \(2\). - **Numbers:** \(4\) and \(-2\) \(4 \times (-2) = -8\) \(4 + (-2) = 2\) 3. **Complete Factorization:** \[ 2(x^{2} + 2x - 8) = 2(x + 4)(x - 2) \] --- ### **Summary of Correct Factorizations** 1. **Problem 3:** \[ x^{2} + x - 12 = (x + 4)(x - 3) \] 2. **Problem 4:** \[ 2x^{2} - 14x + 24 = 2(x - 3)(x - 4) \] 3. **Problem 5:** \[ -2x^{2} - 4x - 30 = -2(x^{2} + 2x + 15) \] *(Cannot be factored further over real numbers)* 4. **Problem 4 (Second Instance):** \[ 2x^{2} + 4x - 16 = 2(x + 4)(x - 2) \] --- If you have any further questions or need additional explanations, feel free to ask!