Zeroes and factorization of a quadratic polynomial I

Task

Suppose $f$ is a quadratic function given by the equation $f(x) = ax^2 + bx + c$ where $a,b,c$ are real numbers and we will assume that $a$ is non-zero.

1. If $0$ is a root of $f$ explain why $c = 0$ or, in other words, $ax^2 + bx +c$ is evenly divisible by $x$.
2. If $1$ is a root of $f$ explain why $ax^2 + bx + c$ is evenly divisible by $x-1$.
3. Suppose $r$ is a real number. If $r$ is a root of $f$ explain why $ax^2 + bx + c$ is evenly divisible by $x-r$.

IM Commentary

For a polynomial function $p$, a real number $r$ is a root of $p$ if and only if $p(x)$ is evenly divisible by $x - r$. This fact leads to one of the important properties of polynomial functions: a polynomial of degree $d$ can have at most $d$ roots. This is the first of a sequence of problems aiming at showing this fact. The teacher should pay close attention to the logic used in the solution to part (c) where the divisibility of $ax^2 + bx + c$ by $x -r$ is obtained not by performing long division but by using the result of long division of these polynomials; namely, that said division will result in an expression of the following form: $$ ax^2 + bx + c = (x-r)\ell(x) + k $$ where $\ell$ is a linear polynomial and $k$ is a number.

This task could be used either for assessment or for instructional purposes. If it is used for assessment, parts (a) and (b) are more suitable than part (c). Each of the questions in this task could be formulated as an if and only if statement but the other implication, namely that $f(x)$ is divisible by $x-r$ if and only if $r$ is a root of $f$. The direction not presented in this task is more straightforward and so has been left out.

Solution

1. If $0$ is a root of the function $f$ this means that $f(0) = 0$. Since $f$ is given by the formula $f(x) = ax^2 + bx + c$ this means that $$ a\cdot 0^2 + b\cdot 0 + c = 0. $$ From here we see that $c = 0$. Since $c = 0$, we have $$ f(x) = ax^2 + bx = x(ax + b) $$ so $f(x)$ is evenly divisible by $x$.

2. If $1$ is a root of the function $f$ this means that $f(1) = 0$. Since $f$ is given by the formula $f(x) = ax^2 + bx + c$ this means that $$ a\cdot 1^2 + b\cdot 1 + c = 0. $$ So we have $a + b + c = 0$. If we perform long division of $ax^2 + bx + c$ by $(x-1)$ we find that $x-1$ goes in $ax + a + b$ times with a remainder of $a+b+c$. But we just saw that, when $1$ is a root of $f$, $a+b+c$ is zero so this means that $(x-1)$ divides $ax^2+bx+c$ evenly.

3. If $r$ is a root of $f$ this means that $f(r) = 0$. We could perform long division as in part (b) but it is complicated because we have an extra variable $r$. The result of performing long division will be a linear polynomial function $l$ and a constant $k$ so that $$ ax^2 + bx + c = (x-r)l(x) + k. $$ To evaluate $k$ we could perform the long division as in part (b) or we can plug $r$ into the above equation to find

   \begin{eqnarray} ar^2 + br + c &=& (r-r)l(r) + k \\ &=& k. \end{eqnarray}

   So $k = ar^2 + br + c = f(r)$. But we know that $f(r) = 0$ because $r$ is a root of $f$. Hence $ax^2+bx+c = (x-r)l(x)$ and so $x-r$ divides $ax^2+bx+c$ evenly as desired.

   Notice in part (c) that it was not necessary to calculate $l(x)$ explicitly, just to know that such a linear polynomial exists and can be found by performing long division of polynomials. This is important because the division process is tricky to perform when there are so many variables and it becomes even more difficult when the polynomials involved are of larger degree.