In algebra classes, polynomials are usually kept as simple as possible. In particular, we usually deal with polynomials whose coefficients are integers. Now, suppose that we attempt to perform synthetic division with a proposed root \(c\) that is also an integer. Each step of synthetic division involves either multiplication by \(c\) or addition of that product with a coefficient. Consequently, each step of the process will be an integer. The last step of the process involves adding the constant term coefficient \(a_0\) with the most recent product \(c \cdot b_0\) where \(b_0\) will be the final coefficient of the quotient polynomial.
If the number \(c\) is going to be a root, then the remainder \(r\) must equal zero, meaning that \(a_0 + c \cdot b_0 = 0\). Rewriting this, we find \(a_0 = - c \cdot b_0\). In other words, \(c\) must be one of the integer factors of the constant term in the polynomial of interest. This fact has a name called the Integer Root Test.
The practical consequence of this test is that by finding all possible integer factors of the constant term, we greatly reduce the number of \(x\)-values that we might test to be roots of a polynomial with integer coefficients. There will likely be other roots, but they will either be rational numbers or irrational numbers.
\(f(x) = x^3-5x-12\) has integer coefficients. The constant term is \(a_0 = -12\). The integer factors of \(a_0\) are ±1, ±2, ±3, ±4, ±6, and ±12. So when looking for integer roots, there are at most 12 numbers to try.
\(g(x) = x^3-5x^2+3x+4\) has integer coefficients. The constant term is \(a_0 = 4\). The integer factors of \(a_0\) are ±1, ±2, and ±4. So when looking for integer roots, there are at most 6 numbers to try.
\(h(x) = x^4+x^3-6x^2+1\) has a graph that looks like it has roots at \(x=-3\) and at \(x=2\). (Use your calculator and take a look!) Explain why it is impossible that these are actual roots.
Once we find one factor that is a root, any additional roots must be factors of what we obtain when dividing the original constant term by the root we found. Consequently, I always personally start with possible roots of a small size, starting at ±1, since the larger factors are less likely to be roots.
This is my own phrase, to “peel off” factors. A more common phrase is that we “deflate” the polynomial. In either case, the idea is that we find a single root, probably by testing the possible integer roots, and use that root \(x=c\) to factor a single factor \(x-c\) using synthetic division. This gives us a quotient polynomial, \[f(x) = (x-c) \cdot q(x). \] We then turn our attention to \(q(x)\) as a polynomial and try to find a single root of that polynomial and factor out another term. Each time we find a root, we “peel off” another factor and deflate the resulting quotient polynomial by a single degree.
Example: Factor the polynomial \(f(x) = x^4+3 x^3-3 x^2-7 x+6\).
The polynomial \(f\) has integer coefficients and a constant term \(a_0=6\). So the only possible integer roots are ±1, ±2, ±3, and ±6. We start trying the possible roots one at a time using synthetic division.
Test \(c=-1\). The resulting synthetic division table is given below.
| 1 | 3 | -3 | -7 | 6 |
| 0 | -1 | -2 | 5 | 2 |
| 1 | 2 | -5 | -2 | 8 |
Test \(c=1\). The resulting synthetic division table is given below.
| 1 | 3 | -3 | -7 | 6 |
| 0 | 1 | 4 | 1 | -6 |
| 1 | 4 | 1 | -6 | 0 |
The quotient polynomial is now \(q(x) = x^3+4x^2+x-6\), which has a constant term \(-6\). This quotient polynomial has the same possible integer roots ±1, ±2, ±3, and ±6. So we will repeat the process but with the simpler polynomial.
We have already eliminated \(c=-1\). It didn't work for \(f(x)\) and it can not work for \(q(x)\). Although we already tried \(c=1\), that root might still be a root for \(q(x)\). If it is, it is called a repeated root. Let's give it a try.
Try \(c=1\) (again), but this time on the quotient polynomial \(q(x) = x^3+4x^2+x-6\).| 1 | 4 | 1 | -6 |
| 0 | 1 | 5 | 6 |
| 1 | 5 | 6 | 0 |
We could repeat the process on the new quotient polynomial \(x^2+5x+6\), but this is a quadratic polynomial that we know how to factor directly. So the final factorization of our function is \[f(x) = (x-1) \cdot (x-1) \cdot (x+2) \cdot (x+3).\]
In calculus problems involving polynomials, it is often very difficult to create a problem where all of the solutions to a relevant equation are integers. Furthermore, when they are integers, many students rely too much on their calculator to guess the answers instead of practicing algebraic methods that truly demonstrate the results.
It is much easier to create a problem where some of the solutions are integers and others are either rational numbers or solutions to a quadratic equation. So the strategy for the student is to first identify integer roots and use polynomial deflation to peel off the easy factors. The remaining quotient should then be in a form where you can either factor by eye or use the quadratic formula to find remaining roots.
Example: Find all solutions to \(x^4-2 x^3-14 x^2+37 x-12 = 0\).
The polynomial in this problem is not a quadratic, so the quadratic formula can not be used. However, the coefficients are all integers and the constant term is \(-12\). The Integer Root Test therefore applies, and the only possible integer roots are ±1, ±2, ±3, ±4, ±6, and ±12.
If you do not have a graphing calculator, then you're stuck trying synthetic division on each of these possible roots until you happen upon one that has a zero remainder. However, if you do have access to a graphing calculator, then you can take a peek at the graph and see if the graph crosses at any of those points. Or you could look at the table using the formula. Either way, it is much faster to type in the formula for the polynomial and let the calculator compute some values than to do the synthetic divisions.
Try it! Look at the graph and identify which possible integer roots correspond to where the graph crosses the axis.
I took a peek and saw that \(c=-4\) looks like a root. But now I need to use algebra to verify this result using synthetic division. The advantage is that I didn't waste my time on possible roots that ultimately wouldn't work.
Try \(c=-4\) as a root of \(f(x) = x^4-2 x^3-14 x^2+37 x-12\):| 1 | -2 | -14 | 37 | -12 |
| 0 | -4 | 24 | -40 | 12 |
| 1 | -6 | 10 | -3 | 0 |
The quotient in this factorization is \(q(x) = x^3-6x^2+10x-3\), which has a constant term -3. So the only remaining possible integer roots are ±1 and ±3. From our peek at the graph, we saw that there might be a root near \(x=3\), so we try that one next using the deflated polynomial.
Try \(c=3\) as a root of \(q(x) = x^3-6x^2+10x-3\):
| 1 | -6 | 10 | -3 |
| 0 | 3 | -9 | 3 |
| 1 | -3 | 1 | 0 |
The resulting quotient in this deflation is \(x^2-3x+1\) which does not factor nicely. So we go back to the question of solving the equation. Through our factoring, we have shown that the original equation \[x^4-2 x^3-14 x^2+37 x-12 = 0\] is equivalent to the factored equation \[(x+4) \cdot (x-3) \cdot (x^2-3x+1) = 0.\] Even though we have not completely factored the polynomial, we can still apply the zero product principle, which says that a product can equal zero only if one of the factors equals zero. We have three possibilities: \[ \begin{aligned} x+4 &= 0 \qquad \Leftrightarrow \qquad x=-4 \\ x-3 &= 0 \qquad \Leftrightarrow \qquad x=3 \\ x^2-3x+1 &= 0 \qquad \Leftrightarrow \qquad x = \frac{3 \pm \sqrt{9-4}}{2} = \frac{3 \pm \sqrt{5}}{2} \end{aligned} \] Because we have deflated the problem to a quadratic, we could finish solving the equation using the quadratic formula.
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