Question

Your friend Chadwick is trying to factor the polynomial 6x3+8x2+15x+20 . "I noticed that I could group the first and the third terms; this would help me begin to factor the polynomial,” says Chadwick. You notice that another equally valid grouping would be to group the first and second terms, with a common factor of 2x2 , and the third and fourth terms, with a common factor of 5. Given that both of these first steps are valid, which of the following options is not equivalent to Chadwick’s polynomial? Option 1: 2x2(3x+4)+5(3x+4) Option 2: 3x(2x2+5)+4(2x2+5) Option 3: 5(2x2+3x)+4x(2x+5)

Answer 1

Let's analyze Chadwick's polynomial \( 6x^3 + 8x^2 + 15x + 20 \) and verify the factorizations for the given options.

### Step 1: Verify Chadwick's Grouping

Chadwick groups the polynomial as follows:
- Group 1: \( 6x^3 + 15x \)
- Group 2: \( 8x^2 + 20 \)

This grouping gives:
\[
3x(2x^2 + 5) + 4(2x^2 + 5) = (3x + 4)(2x^2 + 5)
\]

### Step 2: Analyze Each Option

Now, we will analyze the proposed options to see which one is not equivalent to \( 6x^3 + 8x^2 + 15x + 20 \).

**Option 1: \( 2x^2(3x + 4) + 5(3x + 4) \)**
- Factor out \( 3x + 4 \):
\[
(3x + 4)(2x^2 + 5)
\]
This expression is equivalent to Chadwick's polynomial.

**Option 2: \( 3x(2x^2 + 5) + 4(2x^2 + 5) \)**
- Factor out \( 2x^2 + 5 \):
\[
(2x^2 + 5)(3x + 4)
\]
This expression is also equivalent to Chadwick's polynomial.

**Option 3: \( 5(2x^2 + 3x) + 4x(2x + 5) \)**
- Expand this expression:
\[
5(2x^2 + 3x) + 4x(2x + 5) = 10x^2 + 15x + 8x^2 + 20 = 18x^2 + 15x + 20
\]
This simplification shows that the polynomial becomes \( 18x^2 + 15x + 20 \), which is not equal to \( 6x^3 + 8x^2 + 15x + 20 \).

### Conclusion
The option that is **not equivalent** to Chadwick's polynomial \( 6x^3 + 8x^2 + 15x + 20 \) is:

**Option 3: \( 5(2x^2 + 3x) + 4x(2x + 5) \)**