Arrange the steps of the following factor by grouping problem: \( 18 x^{4}-9 x^{3}-8 x^{2}+4 x \) 1. \( x\left(18 x^{3}-9 x^{2}-8 x+4\right) \) 2. \( x\left(9 x^{2}(2 x-1)-4(2 x-1)\right) \) 3. \( x(2 x-1)\left((3 x)^{2}-2^{2}\right) \) 4. \( x(2 x-1)(3 x-2)(3 x+2) \) 5. \( x\left((2 x-1)\left(9 x^{2}-4\right)\right) \) Your Score: 2 of 5

To factor the expression \( 18 x^{4}-9 x^{3}-8 x^{2}+4 x \) by grouping, we can follow these steps: 1. **Factor out the greatest common factor (GCF)** from the entire expression. The GCF here is \( x \): \[ x(18 x^{3}-9 x^{2}-8 x+4) \] 2. **Group the terms** inside the parentheses: \[ x\left(18 x^{3}-9 x^{2}-8 x+4\right) \] 3. **Factor by grouping** the terms \( 18 x^{3}-9 x^{2} \) and \( -8 x+4 \): - From the first group \( 18 x^{3}-9 x^{2} \), we can factor out \( 9 x^{2} \): \[ 9 x^{2}(2 x-1) \] - From the second group \( -8 x+4 \), we can factor out \( -4 \): \[ -4(2 x-1) \] 4. **Combine the factored groups**: \[ x\left(9 x^{2}(2 x-1)-4(2 x-1)\right) \] 5. **Factor out the common binomial factor \( (2 x-1) \)**: \[ x(2 x-1)\left(9 x^{2}-4\right) \] 6. **Recognize that \( 9 x^{2}-4 \) is a difference of squares**: \[ 9 x^{2}-4 = (3 x)^{2}-2^{2} = (3 x-2)(3 x+2) \] 7. **Combine everything**: \[ x(2 x-1)(3 x-2)(3 x+2) \] Now, let's arrange the steps in the correct order: 1. \( x\left(18 x^{3}-9 x^{2}-8 x+4\right) \) (Step 1) 2. \( x\left(9 x^{2}(2 x-1)-4(2 x-1)\right) \) (Step 4) 3. \( x(2 x-1)\left((3 x)^{2}-2^{2}\right) \) (Step 5) 4. \( x(2 x-1)(3 x-2)(3 x+2) \) (Step 6) 5. \( x\left((2 x-1)\left(9 x^{2}-4\right)\right) \) (Step 3) So the correct order of steps is: 1. Step 1 2. Step 2 3. Step 5 4. Step 4 5. Step 3 This gives us the final factorization: \[ x(2 x-1)(3 x-2)(3 x+2) \]