First Step in Completing the Problem

[thunc14]

What would be the most efficient step to suggest to a student attempting to complete this problem?

Factor:

x^2+(2a+5)x+(a^2+5a+6)

a) Distribute (2a+5)
b) Factor in the form (x+c)(x+d)
c) Simplify and group terms with a
d) This is not factorable
e) Answer is not there

So through double checking with wolfram alpha, I know this isn't factorable in the intermediate algebra sense. However, if the student is ATTEMPTING the problem, shouldn't they try simplifying, like factoring that third term, or distributing the (2a+5)x? I copied and pasted the exact wording in its entirety and I'm not given wheether or not I got this question right or wrong. Seems like a poor question imo but I wanted a second look

 

[Otis]

… shouldn't they try simplifying, like factoring that third term …

That could be a start, thunc, but I'd say the answer is 'e'.

My suggestion: Try to find two numbers whose product is a^2+5a+6 and whose sum is 2a+5.

… Seems like a poor question …

I don't like the fuzzy wording.

?

 

[Dr.Peterson]

What would be the most efficient step to suggest to a student attempting to complete this problem?

Factor:

x^2+(2a+5)x+(a^2+5a+6)

a) Distribute (2a+5)
b) Factor in the form (x+c)(x+d)
c) Simplify and group terms with a
d) This is not factorable
e) Answer is not there

So through double checking with wolfram alpha, I know this isn't factorable in the intermediate algebra sense. However, if the student is ATTEMPTING the problem, shouldn't they try simplifying, like factoring that third term, or distributing the (2a+5)x? I copied and pasted the exact wording in its entirety and I'm not given whether or not I got this question right or wrong. Seems like a poor question imo but I wanted a second look

I'd do exactly what you suggested, factoring (a^2+5a+6) as (a+2)(a+3), then observe that (2a+5) = (a+2)+(a+3) as Otis said, so that the whole thing can be factored as (x+a+2)(x+a+3).

I would not distribute, which only makes things harder; the whole goal is to factor it as (x+c)(x+d), so that is not a "step"; it's possible that grouping terms with a could work, but that would take a lot of work (and (a) would be part of it); it is factorable. So you could answer (b), but suggesting that to a student probably wouldn't get him started. Or maybe it would, since that's why you'd factor the last term. I guess I have to vote for that.

I myself would avoid writing a problem like this because a student might just come up with a more efficient method than I had in mind ...

 

[JeffM]

I'm sorry. but I cannot see the issue in part because I have no clue what "factorable in the intermediate algebra sense" even means. Do students in that course factor differently from anyone else? Do students at that level not know the quadratic formula?

[MATH]x^2 +(2a + 5)x + a^2 + 5a + 6 = 0 \implies\\ x = \dfrac{-(2a + 5) \pm \sqrt{(2a + 5)^2 - 4(1)(a^2 + 5a + 6}}{2} \implies\\ x = \dfrac{-(2a + 5) \pm \sqrt{4a^2 + 20a + 25 - 4a^2 - 20a - 24}}{2} =\\ x = \dfrac{- 2a - 5 \pm \sqrt{1}}{2} = - a - \dfrac{5 \pm 1}{2}.[/MATH]Basic stuff.

[MATH]\therefore x^2 + (2a + 5)x + a^2 + 5a + 6 = \\ \left \{x - \left ( - a - \dfrac{5 - 1}{2} \right ) \right \}\left \{x - \left (- a - \dfrac{5 + 1}{2} \right ) \right \} =\\ (x + a + 2)(x + a + 3).[/MATH]Wow. I seem to be more competent than a machine. We had definitely better check my work.

[MATH](x + a + 2)(x + a + 3) =\\ x^2 + x(a + 3 + a + 2) + (a + 2)(a + 3) =\\ x^2 + (2a + 5)x + a^2 + 5a + 6. \ \checkmark[/MATH]None of this is beyond a competent second year student.

I GUESS the answer is e. The whole problem with the exercise is in the inane wording. If you are going to factor this quadratic, then you obviously are going to end up with an answer in the form (x + c)(x + d) so that conceivably might be the expected answer. But specifying what the answer would look like is not really much of a method except as a primitive heuristic. If you don't "see" a way to factor a quadratic, the alternatives are to complete the square or use the quadratic formula. And they are not listed so probably e is what is wanted, but I am not certain because "method" is not defined. In any case, any method that gives the correct answer is valid.

Another example of why multiple choice questions are absurd. In real life, your boss does not come and say, "Is one of these four the correct answer, and if so which one is it?" Nor do bosses usually say"I don't want to know the answer; I just want to know what method you would use to get an answer."