Kevin is solving 6x2−7x+2=0

Question

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: 6x2+(−3x)+(−4x)+2=0


Step 2: (6x2+(−3x))+(−4x+2)=0


Step 3: 3x(2x−1)+2(−2x+1)=0


Change the grouping of pairs in Step 2.
Change the grouping of pairs in Step 2.

Use −3x
as the GCF of the first grouping in Step 3.
Use negative 3 x as the GCF of the first grouping in Step 3.

Use different factors for −7x
in Step 1.
Use different factors for negative 7 x in Step 1.

Use −2
as the GCF of the second grouping in Step 3.

Answer 1

To help Kevin arrive at a shared binomial in Step 3, he should use different factors for \(-7x\) in Step 1.

Currently, he has expressed \(-7x\) as \(-3x - 4x\). While this decomposition is valid, it does not lend itself to a shared binomial in the factored form. If Kevin tried using different factors, for example, \(-2x - 5x\), his grouping might yield:

Step 1: \(6x^2 + (-2x) + (-5x) + 2 = 0\)

Rearranging gives:

Step 2: \((6x^2 - 2x) + (-5x + 2) = 0\)

Then, in Step 3, the factorizations would look like:

\[(2x)(3x - 1) - 1(5x - 2) = 0\]

He can reach a shared binomial this way, instead of using the original grouping of \(-3x\) and \(-4x\), which does not provide the desired shared binomial.

Thus, the correct change for Kevin is:

Use different factors for \(-7x\) in Step 1.

Answer 2

incorrect!

Answer 3

I apologize for the misunderstanding. Let's clarify the problem and find the correct adjustment to help Kevin arrive at a shared binomial in Step 3.

Given that Kevin has:

1. Step 1: \(6x^2 + (-3x) + (-4x) + 2 = 0\)
2. Step 2: \((6x^2 + (-3x)) + (-4x + 2) = 0\)
3. Step 3: \(3x(2x - 1) + 2(-2x + 1) = 0\)

To reach a shared binomial, he needs to adjust his factorization in Step 3. The correct adjustment from the provided options is:

Use \(-2\) as the GCF of the second grouping in Step 3.

If Kevin changes his second grouping to use \(-2\) as the GCF instead of the current factor of 2, it should look like this:

In Step 2, the second grouping would be:

\(-4x + 2\) can be factored as \(-2(2x - 1)\).

Then in Step 3, after adjusting that grouping, he would have:

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

This results in a shared binomial of \((2x - 1)\).

Thank you for your patience, and I hope this helps clarify the correct approach!