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# The Missing Coefficient

## Task

Consider the polynomial function $$ P(x) = x^4 - 3x^3 + ax^2 - 6x + 14, $$ where $a$ is an unknown real number. If $(x - 2)$ is a factor of this polynomial, what is the value of $a$?

## IM Commentary

The purpose of this task is to emphasize the use of the Remainder Theorem (a discussion of which should obviously be considered as a prerequisite for the task) as a method for determining structure in polynomial in equations, and in this particular instance, as a replacement for division of polynomials.

Indeed, one possible solution path is to use polynomial division to divide $P(x)$ by $(x - 2)$ and determine the remainder in terms of $a$, and then solve for $a$ by setting the remainder equal to zero. However, the division operation becomes unwieldy with the unknown parameter $a$ in play. A more straightforward approach is to use the Remainder Theorem (A-APR.2), which states that if $(x - 2)$ is to be a factor of $P(x)$, then $P(2)$ must equal zero.

## Solution

By the Remainder Theorem, if $(x - 2)$ is a factor of $P(x)$, then $P(2)$ must equal zero. Therefore, we must have $$ P(2) = 16 - 3 \cdot 8 + a \cdot 4 - 6 \cdot 2 + 14 = 0. $$ Simplifying, we find that $4a - 6 = 0$, and thus $a = 3/2$.

## The Missing Coefficient

Consider the polynomial function $$ P(x) = x^4 - 3x^3 + ax^2 - 6x + 14, $$ where $a$ is an unknown real number. If $(x - 2)$ is a factor of this polynomial, what is the value of $a$?

## Comments

Log in to comment## Harlem Mathematics says:

almost 7 yearsI'm having a hard time understanding how this task is different than anything in a high school algebra course circa the past hundred years? Where's the innovation?

## Cam says:

almost 7 yearsHi Harlem Mathematics,

This is a good question, and common enough that we've included it in our FAQ. I doubt I can respond any better than the FAQ does, but I'll just emphasize that innovation is only one role of the task bank. Probably the top priority is providing clear illustrations of the standards, and this task does so quite directly for A-APR.2. As the task bank grows in size, you'll see more comprehensive coverage of this and related standards, which will almost certainly include some rather innovative approaches.