Question

Kevin is solving 6x^2−7x+2=0
by grouping. He performed the first three steps as shown below. However, he struggles to complete Step 4 because he could not come up with a shared binomial in Step 3. What change should Kevin make to arrive at a shared binomial in Step 3?

Step 1: 6x^2+(−3x)+(−4x)+2=0


Step 2: (6x^2+(−3x))+(−4x+2)=0


Step 3: 3x(2x−1)+2(−2x+1)=0

Answer 1

Kevin's steps for solving the equation \(6x^2−7x+2=0\) show that he is trying to factor by grouping, but he needs to ensure that the terms are arranged in a way that allows for a shared binomial factor.

In Kevin's Step 1, he rewrote \(-7x\) as \(-3x - 4x\). However, he can make a more effective choice in splitting the middle term, which will help him achieve the shared binomial.

**To arrive at a shared binomial in Step 3**, Kevin should split \(-7x\) in a way that allows for matching coefficients. Specifically, if we factor the quadratic directly:

1. **Find two numbers that multiply to the product of the coefficient of \(x^2\) (which is 6) and the constant term (which is 2)**:
- The product is \(6 \cdot 2 = 12\).
2. **Find two numbers that sum to \(-7\) (the coefficient of \(x\)) and multiply to \(12\)**:
- These numbers are \(-3\) and \(-4\), which Kevin correctly used, but let's consider how we might group them more effectively.

Instead of \(-3x\) and \(-4x\):

- Rearrange \(6x^2 - 4x - 3x + 2 = 0\) to help see if there's a common factor:

1. Write the equation as:
\[
6x^2 - 4x - 3x + 2 = 0
\]

2. Then group them as:
\[
(6x^2 - 4x) + (-3x + 2) = 0
\]

3. Now factor each part:
\[
2x(3x - 2) - 1(3x - 2) = 0
\]

Now, in Step 3, you will see the shared binomial:

- \( (2x - 1)(3x - 2) = 0 \)

Thus, to arrive at a shared binomial in Step 3, Kevin could revise his grouping or arrangement of terms in Step 1 to \(6x^2 - 4x - 3x + 2 = 0\), which would lead to the correct shared binomial during the factoring process.

**So, the change Kevin should make is to split \(-7x\) into \(-4x\) and \(-3x\) instead.** This will allow him to create the shared binomial in Step 3 effectively.