Update all PDFs

# Equivalent Expressions

Alignments to Content Standards: A-SSE.A.2

Find a value for $a$, a value for $k$, and and a value for $n$ so that $$(3x + 2) (2x - 5) = ax^2 + kx + n.$$

## IM Commentary

This is a standard problem phrased in a non-standard way. Rather than asking students to perform an operation, expanding, it expects them to choose the operation for themselves in response to a question about structure. Students must understand the need to transform the factored form of the quadratic expression (a product of sums) into a sum of products in order to easily see $a$, the coefficient of the $x^2$ term; $k$, the leading coefficient of the $x$ term; and $n$, the constant term.

The problem aligns with A-SSE.2 because it requires students to see the factored form as a product of sums, to which the distributive law can be applied.

## Solution

Using the distributive property of multiplication over addition, we have, for all real numbers $x$, $$(3x+2)(2x-5) = (3x+2)(2x) + (3x+2)(-5) = 6x^2+4x-15x-10 = 6x^2-11x-10.$$ So, $a = 6$, $k = -11$, and $n = -10.$

#### Jason says:

3 months

Just thought it would be useful to note that the context for this problem is expressions, not equations. That is, we are thinking of "$x$" here as a symbol to be pushed around, not as an unknown number to be rooted out.

In keeping with this, it seems to me that the key idea of the solution is:

  If $a = 6$, $k = -11$, and $n = -10$, then the expressions on either side of the equals sign are equivalent [[as expressions]].


More concretely:

 If $a = 6$, $k = -11$, and $n = -10$, then $(3x+2)(2x−5) = ax^2+kx+n$ {\it for all values of} $x$.


And one can show that no other values of $a$, $k$, $n$ have this property (although doing so goes well beyond the task as stated). But if all we had wanted was for the equation to hold for {\it some} value of $x$, then we'd be able to choose lots of other values for $a$, $k$, and $n$. For example, if $a = k = n = 0$, then the equation [[thinking of it now as an equation]] holds for the particular values $x = -\frac{2}{3}$ and $x = \frac{5}{2}$.

Anyway I hope this note was useful or interesting or helps to illuminate the task.

#### Howard Phillips says:

over 2 years

I feel that this task misses the point of the SMP What else can you do with a pair of brackets (UK speak) but expand them. "Look for and use structure" is surely to be applied in the context of a more general or more involved task, as PART of the process of getting somewhere with it.

#### Cam says:

over 2 years

Well, you could just leave them alone! I think one of the points of seeing structure is the different ways of presenting a given expression, and the value in switching from one to the other. I definitely agree that many students will have a gut response to a pair of brackets which compels them to distribute, despite the fact that this is often to their detriment!

That said, I agree that this is only a marginal justification for the use of the tag. To be honest, I think it was one of our first examples of using the tagging system -- we have much stronger illustrations of MP7 elsewhere in the database. So thanks very much for your insightful comments!