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# 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

No matter what the value of $x$, the distributive property of multiplication over addition tells us that $$(3x+2)(2x-5) = (3x+2)(2x) + (3x+2)(-5) = 6x^2+4x-15x-10 = 6x^2-11x-10.$$ So, if $a = 6$, $k = -11$, and $n = -10$, then the expression on the left has the same value as the expression on the right for all values of $x$; that is, the two expressions are equivalent.

#### Bill says:

5 months

Thanks Jason, I might incorporate some of this into the commentary (once our task editing interface becomes unbroken).

#### Jason says:

11 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.