# Equivalent Expressions

## Task

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.

## Equivalent Expressions

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