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We are tasked with factoring the cubic polynomial \( 2x^3 + 5x^2 - 8x - 20 \), given that one factor is \( 2x + 5 \).

To completely factor the polynomial, we can perform polynomial division, dividing the given polynomial by \( 2x + 5 \).

### Step 1: Polynomial Division

We need to divide \( 2x^3 + 5x^2 - 8x - 20 \) by \( 2x + 5 \).

Let's proceed with the division:

1. Divide the first term:  
   \[
   \frac{2x^3}{2x} = x^2
   \]
   Multiply \( x^2 \) by \( 2x + 5 \):  
   \[
   x^2(2x + 5) = 2x^3 + 5x^2
   \]
   Subtract:  
   \[
   (2x^3 + 5x^2 - 8x - 20) - (2x^3 + 5x^2) = -8x - 20
   \]

2. Now, divide \( -8x \) by \( 2x \):  
   \[
   \frac{-8x}{2x} = -4
   \]
   Multiply \( -4 \) by \( 2x + 5 \):  
   \[
   -4(2x + 5) = -8x - 20
   \]
   Subtract:  
   \[
   (-8x - 20) - (-8x - 20) = 0
   \]

So, the quotient is \( x^2 - 4 \), and the remainder is 0, meaning the division is exact.

### Step 2: Factor the Quotient

Now we need to factor \( x^2 - 4 \), which is a difference of squares:
\[
x^2 - 4 = (x - 2)(x + 2)
\]

### Step 3: Final Factorization

Thus, the complete factorization of \( 2x^3 + 5x^2 - 8x - 20 \) is:
\[
2x^3 + 5x^2 - 8x - 20 = (2x + 5)(x - 2)(x + 2)
\]

Would you like further explanation on any part of this process?

Here are five related questions:
1. How do you apply synthetic division to factor polynomials?
2. Can a cubic polynomial have more than one factorization?
3. What are the steps to recognize a difference of squares?
4. How do you check if a factor is correct in polynomial factorization?
5. How do you solve polynomials once they are factored completely?

**Tip:** Always verify your factorization by multiplying the factors back together to ensure they give the original polynomial.

Factor completely given one factor: 2x^3 + 5x^2 - 8x - 20 ; 2x+5

Learn how to factor the cubic polynomial 2x^3 + 5x^2 - 8x - 20, given one factor 2x+5. This problem involves polynomial division and recognizing a difference of squares. Detailed step-by-step solution suitable for students in Grades 10-12.

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