**Polynomial functions** are functions of a single independent variable, in which that variable can appear more than once, raised to any integer power. For example, the function

$$f(x) = 8x^4 - 4x^3 + 3x^2 - 2x + 22$$

is a polynomial. Polynomial functions are sums of terms consisting of a numerical **coefficient** multiplied by a unique power of the independent variable.

We generally write these **terms** in decreasing order of the power of the variable, from left to right*. Here is a summary of the structure and nomenclature of a polynomial function:

$f(x) = 3x - 2$ | Linear polynomial (linear function) |

$f(x) = x^2 - 4x + 1$ | Quadratic polynomial |

$f(x) = -3x^3 + x - 6$ | Cubic polynomial with no quadratic term |

$f(x) = (x - 3)^2(2x - 1)$ | Cubic polynomial (convince yourself that the largest power will be three when expanded) |

$f(x) = 2x^4 - 5x^3 + 2x^2 - x + 17$ | Quartic polynomial |

$f(x) = 18x^5 - 7$ | Quintic polynomial with only the 5^{th} degree and constant terms. |

Polynomial functions (we usually just say "polynomials") are used to model a wide variety of real phenomena. In physics and chemistry particularly, special sets of named polynomial functions like Legendre, Laguerre and Hermite polynomials (thank goodness for the French!) are the solutions to some very important problems.

It is important that you become adept at sketching the graphs of polynomial functions and finding their zeros (roots), and that you become familiar with the shapes and other characteristics of their graphs.

The appearance of the graph of a polynomial is largely determined by the **leading term** – it's exponent and its coefficient. Because the leading term has the largest power, its size outgrows that of all other terms as the value of the independent variable grows. For example, in $f(x) = 8x^4 - 4x^3 + 3x^2 - 2x + 22,$ as **x** grows, the term $8x^4$ dominates all other terms.

You can check this out yourself by making a quick spreadsheet. Label one column x and fill it with integer values from 1-10, then calculate the value of each term (4 more columns) as x grows. Sum them and add the constant term (22) to find the value of the polynomial. The leading term will grow most rapidly. Here's what I mean:

Each algebraic feature of a polynomial equation has a consequence for the graph of the function. Here is a table of those algebraic features, such as single and double roots, and how they are reflected in the graph of **f(x)**.

Term | Definition |
---|---|

Single root |
A solution of f(x) = 0 where the graph crosses the x-axis. For example, the quadratic function f(x) = (x+2)(x-4) has single roots at x = -2 and x = 4. |

Double root |
A solution of f(x) = 0 where the graph just touches the x-axis and turns around (creating a maximum or minimum - see below). For example, the cubic function f(x) = (x-2)^{2}(x+5) has a double root at x = 2 and a single root at x = -5. |

Triple root |
A solution of f(x) = 0 where the graph crosses the x-axis and the curvature changes sign. See the graph below. For example, the cubic function f(x) = x^{3} has a triple root at x = 0. |

Inflection point |
The name of the point that is a triple root of a polynomial function. The curvature of the graph changes sign at an inflection point between concave-upward and concave-downward. Not all inflection points are located at triple roots (or even at roots at all), but all triple roots are inflection points located on the x-axis. |

y-intercept |
The solution to f(0); the point where a graph crosses the y-axis, usually a convenient (and very easy-to-find) point to plot when sketching a graph. |

Local maximum/minimum |
When a graph turns around (up to down or down to up), a maximum or minimum value is created. Local maxima or minima are not the highest or lowest points on a graph. |

Global maximum/minimum |
The parabola f(x) = x has a global minimum at x = 0, but no global maximum (it increases without bound). The parabola ^{2}f(x) = -x has a global maximum, but no global minimum. The graph below has a global maximum at x = 1. The highest/lowest point on a graph (one may not exist).^{2} |

End behavior |
When x is large, either positive or negative, we are concerned with whether the function increases or decreases without bound (it will do one or the other). |

*This proof uses calculus. If you don't know how to apply differential calculus in this way, don't worry about it. Just take the conclusion that a double root means a "bounce" off of the x-axis for granted.*

Let $f(x) = (x - a)(x - a) = x^2 - 2ax + a^2,$ then the first derivative is $2x + 2a.$

If we set that equal to zero, we get the location of the single critical point, $2x - 2a = 0$ or $x = a.$

Now check the slope of $f(x)$ on the right and left of $x = a$ by letting c be a small, positive number:

$$ \begin{align} f'(a - c) &= 2(a - c) - 2a \\[4pt] &= 2a - c - 2a \lt 0 \phantom{000} \color{#E90F89}{\text{and}} \\[6 pt] f'(a + c) &= 2(a + c) - 2a \\[4pt] &= 2a + c - 2a \gt 0 \end{align}$$

The fact that the slope changes sign across the critical point, a, and that f(a) = 0 show that this is a point where the function touches the axis and "bounces" off.

*This proof uses calculus. If you don't know how to apply differential calculus in this way, don't worry about it. Just take the conclusion that a double root means a "bounce" off of the x-axis for granted.*

Let $f(x) = (x - a)(x - a)(x - a)$ $= x^3 - 3ax^2 - 3a^2x = a^3,$ then the first and second derivatives are:

$$ \begin{align} f'(x) &= 3x^2 - 6ax - 3a^2 \\[4pt] f''(x) &= 6x - 6a \end{align}$$

Now if we set $f''(x) = 0,$ we find the inflection point, $x = a.$ We can check to make sure that the curvature changes by letting c be a small, positive number:

$$ \begin{align} f''(a - c) &= 6(a - c) - 6a \\[4pt] &= 6a - 6c - 6a \lt 0 \phantom{000} \color{#E90F89}{\text{and}} \\[6 pt] f''(a + c) &= 6(a + c) - 6a \\[4pt] &= 6a + 6c - 6a \gt 0, \end{align}$$

The second derivative changes sign across $x = a,$ so $a$ is indeed an inflection point.

Here are the graphs of two cubic polynomials. they differ only in the sign of the leading coefficient.

The leading term of any polynomial function dominates its behavior. When that term has an odd power of the independent variable (x), negative values of x will yield (for large enough |x|) a negative function value, and positive x a positive value.

That is, for large enough | x |,

f(x > 0) > 0

f(x < 0) < 0

The opposite is true when the coefficient of the leading power of x is negative.

Note also in these figures and the ones below that a cubic polynomial (degree = 3) can have two turning points, points where the slope of the curve turns from positive to negative, or negative to positive. The quartic polynomial (below) has three turning points.

In general, we say that the graph of an nth degree polynomial has (at most) n-1 turning points. It may have fewer, however.

When the degree of a polynomial is even, negative and positive values of the independent variable will yield a positive leading term, unless its coefficient is negative. Negative numbers raised to an even power multiply to a positive result:

(-2)(-2) = 4

(-2)(-2)(-2) = -8

(-2)(-2)(-2)(-2) = 16, and so on.

The result for the graphs of polynomial functions of even degree is that their ends point in the same direction for large | x |:

upwhen the coefficient of the leading term ispositive,

downwhen the coefficient isnegative.

Notice that these quartic functions (left) have up to three turning points. A quartic function need not have all three, however. The graph of f(x) = x^{4} is U-shaped (not a parabola!), with only one turning point and one global minimum.

The table below summarizes some of these properties of polynomial graphs.

Very often, we are faced with finding the solution to an equation like this:

$$4x^4 - 3x^3 + 6x^2 = x + 12$$

Such an equation can always be rearranged by moving all of the terms to the left side, leaving zero on the right side:

$$4x^4 - 3x^3 + 6x^2 - x - 12 = 0$$

Now the solutions to this equation are just the roots or zeros of the polynomial function $f(x) = 4x^4 - 3x^3 + 6x^2 - x - 12.$ They are the points at which the graph of f(x) crosses (or touches) the x-axis. Our task now is to explore how to solve polynomial functions with degree greater than two. We already know how to solve quadratic functions of all kinds. First, a little bit of formalism:

Every non-zero polynomial function of degree *n* has exactly *n* complex roots.

The **fundamental theorem of algebra** tells us that a quadratic function has two roots (numbers that will make the value of the function zero), that a cubic has three, a quartic four, and so forth. The number of roots will equal the degree of the polynomial.

But there's a catch: They don't all have to be real numbers. The theorem says they're complex, and we know that real numbers are complex numbers with a zero imaginary part.

Further, when a polynomial function *does* have a complex root with an imaginary part, it always has a partner, its complex conjugate.

**Pro tip**: When a polynomial function has a complex root of the form * a + bi*,

When faced with finding roots of a polynomial function, the first thing to check is whether there is something that can be factored away from all of its terms. Here are some examples:

$x^4 - 3x^3 + x^2 - 7x = 0$ $x(x^3 - 3x^2 + x - 7) = 0$ |
Factor out an |

$3x^3 - 27x + 81 = 0$ $x^3 - 9x + 27 = 0$ |
All terms are divisible by three, so get rid of it. Note that the zero on the right makes this very convenient ... the 3 just "disappears". We haven't simplified our polynomial in degree, but it's nice not to carry around large coefficients. |

$11x^3 - 121x^2 + 3x - 2 = 0$ |
Tantalizing when you look at the x's, and the 11 and 121, but there is no GCF here. |

**Pro tip**: Always look for a **greatest common factor** first when working with any polynomial function. Finding one can make things a lot easier.

Find all roots of these polynomial functions by finding the greatest common factor (GCF).

1. |
$f(x) = 5x^4 + 10x^3 - 75x^2$ ## SolutionThe greatest common factor (GCF) in all terms is ^{2} out of each term, we get$$f(x) = 5x^2(x^2 + 2x - 15)$$ The quadratic part turns out to be factorable, too (always check for this, just in case), thus we can further simplify to: $$f(x) = 5x^2(x + 5)(x - 3)$$ Now the zeros or roots of the function (the places where the graph crosses the x-axis) are obvious. They occur when 5x $$x = 0, \, 0, \, -5, \, 3$$ Notice that zero is a double root |

2. |
$f(x) = -3x^5 - 12x^4 - 12x^3$ ## SolutionThe greatest common factor (GCF) in all terms is ^{3} out of each term, we get$$f(x) = -3x^3(x^2 + 4x + 4)$$ The quadratic part turns out to be factorable, too (always check for this, just in case), thus we can further simplify to: $$f(x) = -3x^3(x + 2)(x + 2)$$ Now the zeros or roots of the function occur when -3x $$x = 0, \, 0, \, 0, \, -2, \, -2$$ Notice that zero is a triple root and -2 is a double root. For a polynomial function like this, the former means an inflection point and the latter a point of tangency with the x-axis. |

3. |
$f(x) = -4x^6 - 16x^4$ ## SolutionThe greatest common factor (GCF) in all terms is ^{4} out of each term, we get$$f(x) = -4x^4(x^2 - 4)$$ The quadratic part turns out to be factorable, $$f(x) = -4x^4(x + 2)(x - 2)$$ Now the zeros or roots of the function are: $$x = 0, \, 0, \, 0, \, 0, \, -2, \, 2$$ |

4. |
$f(x) = 4x^4 + 12x^3 - 112x^2$ ## SolutionThe greatest common factor (GCF) in all terms is ^{2} out of each term, we get$$f(x) = 4x^2(x^2 + 3x - 28)$$ The quadratic part is factorable, $$f(x) = 4x^2(x + 7)(x - 4)$$ Now the zeros or roots of the function are: $$x = 0, \, 0, \, -7, \, 4$$ |

5. |
$f(x) = 7x^4 - 42x^3 + 63x^2$ ## SolutionThe greatest common factor (GCF) in all terms is ^{2} out of each term, we get$$f(x) = 7x^2(x^2 - 6x + 9)$$ The quadratic part is factorable, $$f(x) = 4x^2(x - 3)(x - 3)$$ Now the zeros or roots of the function are: $$x = 0, \, 0, \, 3, \, 3$$ |

6. |
$f(x) = 2x^5 - 26x^3 + 72x$ ## SolutionThe greatest common factor (GCF) in all terms is $$f(x) = 2x(x^4 - 13x^2 + 36)$$ The term in parentheses has the form of a quadratic and can be factored like this: $$f(x) = 2x(x^2 - 9)(x^2 - 4)$$ Each of the parentheses is a difference of perfect squares, so they can be factored, too: $$f(x) = 2x(x + 3)(x - 3)(x + 2)(x - 2)$$ The roots are $$x = 0, \, -3, \, 3, \, -2, \, 2$$ |

Sometimes **factoring by grouping** works. The first thing you'll need to check is whether you've got an **even number of terms**. If it's odd, move on to another method; grouping won't work.

The example below shows how grouping works. First find common factors of subsets of the full polynomial, say two or three terms, and move that out as a common factor.

If what's been left behind is common to all of the groups you started with, it can also be factored away, leaving a product of binomials that are simpler and easier to solve for roots.

The trickiest part of this for students to understand is the second factoring. Look at the example. Between the second and third steps. The binomial (x + 3) is just treated as any other number or variable. It appears in both added terms of the second step, therefore it can be factored out.

Find all roots of these polynomial functions by factoring by grouping.

1. |
$f(x) = 2x^4 - 6x^3 - 4x + 12$ ## Solution2 x 3 = 6 and 4 x 3 = 12. These patterns are present in this function and suggest pulling 4 out of the second two terms and 2x $$f(x) = 2x^3(x - 3) - 4(x - 3)$$ It takes some practice to get the signs right, but this does the trick. Now factor out the (x - 3), which is common to both terms: $$f(x) = (x - 3)(2x^3 - 4)$$ Finally, we can take a 2 out of the last term to get the factored form: $$f(x) = 2(x - 3)(x^3 - 2)$$ The roots are x = 3, $2^{1/3}$, and two imaginary roots. |

2. |
$f(x) = x^4 - 2x^3 - 8x + 16$ ## SolutionLet's try grouping the 1 $$f(x) = x(x^3 - 8) - 2(x^3 - 8)$$ It takes some practice to get the signs right, but this does the trick. Now factor out the (x^3 - 8), which is common to both terms: $$f(x) = (x^3 - 8)(x - 2)$$ Now the roots can be found by solving x - 2 = 0 and x x = 2, 2, plus two imaginary roots. |

3. |
$f(x) = x^4 - 9x^3 - 4x^2 + 36x$ ## SolutionLet's try grouping the 1 $$f(x) = x^2(x^2 - 4) - 9x(x^2 - 4)$$ It takes some practice to get the signs right, but this does the trick. Now factor out the (x $$f(x) = (x^2 - 4)(x^2 - 9x)$$ Now we can factor an x out of the second term, and recognize that the first is a difference of perfect squares: $$f(x) = x(x + 2)(x - 2)(x - 9)$$ Now the roots are easy to identify: x = 0, -2, 2, 9 |

4. |
$f(x) = -3x^4 + 3x^3 - 2x^2 + 2x$ ## SolutionLet's try grouping the 1 $$f(x) = -3x^2(x^2 - 1) - 2(x^2 - 1)$$ Now factor out the (x $$f(x) = -(x^2 - 1)(3x^2 + 2)$$ Now the roots can be found: $$x = ±1, ±i \sqrt{2/3}$$ |

5. |
$f(x) = -8x^3 + 56x^2 + x - 7$ ## Solution$$ \begin{align} f(x) &= -8x^3 + 56x^2 + x - 7 \\ &= -8x^2 (x - 7) + (x - 7) \\ &= (x - 7)(1 - 8x^2) \\ \\ x &= 7, \, ± \frac{1}{2} \sqrt{2} \end{align}$$ |

6. |
$f(x) = 3x^4 - 12x^3 - 2x^2 + 8x$ ## Solution$$ \begin{align} f(x) &= 3x^4 - 12x^3 - 2x^2 + 8x \\ &= 3x^3 (x - 4) - 2x(x - 4) \\ &= (x - 4)(3x^3 - 2x) \\ &= x(x - 4)(3x^2 - 2) \\ x &= 0, \, 4, \, ± \sqrt{\frac{2}{3}} \end{align}$$ |

7. |
$f(x) = 7x^3 + 28x^2 + x + 4$ ## Solution$$ \begin{align} f(x) &= 7x^3 + 28x^2 + x + 4 \\ &= 7x^2 (x + 4) + (x + 4) \\ &= (x + 4)(7x^2 + 1) \\ x &= 4, \, ± i\sqrt{\frac{4}{7}} \end{align}$$ |

8. |
$f(x) = x^6 + 2x^5 - 4x^2 - 8x$ ## Solution$$ \begin{align} f(x) &= x^6 + 2x^5 - 4x^2 - 8x \\ &= x^5 (x + 2) - 4x(x + 2) \\ &= (x + 2)(x^5 - 4x) \\ &= x(x + 2)(x^4 - 4) \\ x &= 0, \, -2, \, ± 4^{1/4} \end{align}$$ |

You have worked with quadratic equations enough to recognize their basic form:

$$f(x) = Ax^2 + Bx + c$$

In this form, there is a constant term, and the first term has twice the degree as the middle term. Now consider equations of the form

$$ \begin{align} 4x^4 - 3x^2 + 2 &= 0 \; \; \text{or}\\ \\ x^6 - 5x^3 + 6 &= 0 \end{align}$$

Notice that each of those equations has the same pattern. All have three terms, the highest power is twice that of the middle term, and each has a constant term (if it didn't, we'd be able to find a GCF). They have the same general form as a quadratic. Here's an example:

Let's find the roots of the quartic polynomial equation,

$$x^4 - 5x^2 + 6 = 0$$

To do this, we make a simple substitution: Let **u = x ^{2}**, which means that

$$u^2 - 5u + 6 = 0$$

Now this quadratic polynomial is easily factored:

$$(u - 2)(u - 3) = 0$$

Now we can re-substitute x^{2} for u like this:

$$(x^2 - 2)(x^2 - 3) = 0$$

Finally, it's easy to solve for the roots of each binomial, giving us a total of four roots, which is what we expect.

$$x = ±\sqrt{2} \; \; \text{and} \; \; x = ±\sqrt{3}$$

Doing these by substitution can be helpful, especially when you're just learning this technique for this special group of polynomials, but you will eventually just be able to factor them directly, bypassing the substitution.

Each of these functions has the form of a quadratic function. Find the roots of each.

1. |
$f(x) = x^4 - 5x^2 - 14$ ## SolutionMethod 1: substitution Let u = x $$ \begin{align} f(u) &= u^2 - 5u - 14 \\ &= (u - 7)(u + 2) \\ u &= -2, \, 7, \; \text{ so} \\ x^2 &= -2, \, 7 \\ x &= ±i\sqrt{2}, \; ±\sqrt{7} \end{align}$$ Method 2: factor directly $$ \begin{align} f(x) &= (x^2 - 7)(x^2 + 2) \\ x^2 &= -2, \, 7 \\ x &= ±i\sqrt{2}, \; ±\sqrt{7} \end{align}$$ Use either method that suits you. Substitution is a good method to learn for other kinds of problems, too. |

2. |
$f(x) = x^6 - 7x^3 + 10$ ## SolutionMethod 1: substitution Let u = x $$ \begin{align} f(u) &= u^2 - 7u + 10 \\ &= (u - 5)(u - 2) \\ u &= 2, \, 5, \; \text{ so} \\ x^3 &= 2, \, 5 \\ x &= 5^{1/3}, \, 2^{1/3} \end{align}$$ plus two imaginary roots for each of those. Method 2: factor directly $$ \begin{align} f(x) &= (x^3 - 5)(x^3 + 2) \\ x^3 &= 2, \, 5 \; \dots \end{align}$$ Note that every real number has three cube-roots, one purely real and two imaginary roots that are complex conjugates. |

3. |
$f(x) = x^4 - x^2 - 110$ ## SolutionMethod 1: substitution Let u = x $$ \begin{align} f(u) &= u^2 - u - 10 \\ &= (u - 11)(u + 10) \\ u &= -10, \, 11, \; \text{ so} \\ x^2 &= -10, \, 11 \\ x &= ±i\sqrt{10}, \, ±sqrt{11} \end{align}$$ plus two imaginary roots for each of those. Method 2: factor directly $$ \begin{align} f(x) &= (x^2 - 11)(x^2 + 10) \\ x^2 &= -10, \, 11 \; \dots \end{align}$$ |

4. |
$f(x) = 2x^4 + 4x^2 - 3$ ## SolutionLet u = x $$f(u) = 2u^2 + 4u - 3$$ This function isn't factorable, so we have to complete the square or use the quadratic equation (same thing) to get: $$ \begin{align} u &= -1 ± \sqrt{\frac{5}{2}} \\ x &= ± \sqrt{-1 ± \sqrt{\frac{5}{2}}} \end{align}$$ |

While this method of finding roots isn't used all that often, it's a huge time saver when it *can* be used. You don't have to memorize these formulae (you can always look them up), but use them in situations where your polynomial equation is a sum or difference of cubes, such as

$$ \begin{matrix} f(x) = x^3 - 8 & \color{#E90F89}{= x^3 - 2^3} \\[5pt] f(x) = x^6 - 27 & \color{#E90F89}{= (x^2)^3 - 3^3} \\[5pt] f(x) = 8x^3 + 125 & \color{#E90F89}{= (2x)^3 + 5} \end{matrix}$$

$$ \begin{align} x^3 + y^3 = (x + y)(x^2 - xy + y^2) \\[5pt] x^3 - y^3 = (x - y)(x^2 + xy + y^2) \end{align}$$

Use the sum/difference of perfect cubes formulae (box above) to find all of the roots (zeros) of these functions:

1. |
$f(x) = 27x^3 - 8$ ## Solution |

2. |
$f(x) = x^3 + 64$ ## Solution |

3. |
$f(x) = -343x^3 + 512$ ## Solution |

4. |
$f(x) = 8x^3 - 216$ ## Solution |

The **rational root theorem** is not a way to find the roots of polynomial equations directly, but * if* a polynomial function does have any

The important thing to keep in mind about the rational root theorem is that any given polynomial may not even *have* any rational roots. In those cases, we have to resort to estimating roots using a computer, using methods you will learn in calculus.

Consider a polynomial equation of the form

$$Ax^n + Bx^{n -1} + \dots + Z = 0$$

where **A** is the coefficient of the leading term and **Z** is the constant term. Now let **p** = the set of all possible integer factors of **Z**, and their negatives, and let **q** = the set of all possible integer factors of **A**, and their negatives. The rational root theorem says that* if there are* any **rational** roots of the equation (there may not be), then they will have the form **p/q**. That is, any rational root of the equation will be one of the **p**'s divided by one of the **q**'s.

Given a polynomial function **Ax ^{n} + Bx^{n-1} + Cx^{n-2} + ... + Z**, where

For example, given the polynomial function

$$f(x) = 6x^4 - 4x^3 + 9x^2 + 3.$$

The set $q = ±\{1, 2, 3, 6\},$ the integer factors of 6, and the set $p = ±\{1, 3\},$ the integer factors of 3.

Now we can construct the complete list of all possible rational roots of f(x):

$$\frac{p}{q} = ±1, \; ± \; 3, \; ±\frac{1}{2}, \; ±\frac{1}{3}, \; ±\frac{1}{6}, \; ±\frac{3}{2}$$

Now it's very important that you understand just what the rational root theorem says. It gives us a list of all *possible* rational roots, and we need to plug those each, in turn, into the function to test whether they are indeed roots. Not all of them can be, and it's entirely possible that *none* are.

What remains is to test them. Before we do that, we'll take a brief detour and discuss a very easy way to do that, **synthetic substitution**.

Here's a step-by-step example of how synthetic substitution works.

Sometimes (erroneously) called synthetic *division*, this procedure is illustrated by this example. It's a quick and easy method to test whether a value of the independent variable is a root.

The method starts with writing the coefficients of the polynomial in decreasing order of the power of x that they multiply, left to right. **It's important to include a zero if a power of x is missing**. In the example, if there had been no linear term, we'd put a 0 in the top line instead of a 1 in the first step.

The number to be substituted for **x** is written in the square bracket on the left, and the first coefficient is written below the line (second step). That's the setup. Now it's just a matter of doing the same thing to the end.

The number in the bracket is multiplied by the first number below the line. The result becomes the next number in the second row, above the line. The numbers now aligned in the first and second row are added to become the next number under the line. Repeat until you're finished. The last number below the line is the result of substituting the value in the bracket into f(x).

In our example, -1 is a root because it makes the function zero. The binomial (x+1) must then be a factor of f(x).

*possibilities* of rational roots, if any exist. Now synthetic substitution gives us a quick method to check whether those possibilities are actually roots. Using the rational root theorem is a trial-and-error procedure, and it's important to remember that any given polynomial function may not actually have any rational roots. Its roots might be irrational (repeating decimals) or imaginary.

Find the four solutions to the equation $x^4 + 4x^3 + 2x^2 - 4x - 3 = 0$

We begin by identifying the p's and q's. For this function it's pretty easy. The constant term is 3, so its integer factors are **p = 1, 3**. The coefficient of the highest degree term (**x ^{4}**), is one, so its only integer factor is

$$\frac{p}{q} = ±1, \; ±3$$

Now we test to see if any of these is a root. For work in math class, here's a hint: always try the smallest integer candidates first. This is just a matter of practicality; some of these problems can take a while and I wouldn't want you to spend an inordinate amount of time on any one, so I'll usually make at least the first root a pretty easy one. Here we try one and see that it's a root because the value of the function is zero. Notice that the coefficients of the new polynomial, with the degree dropped from 4 to 3, are right there in the bottom row of the synthetic substitution grid.

Now we don't want to try another positive root because the coefficients of the new cubic polynomial are all positive. There's no way that a positive value for x will ever make the function equal zero. We'll try the next-easiest candidate, x = -1:

That worked, and now we're left with a quadratic function multiplied by our two factors. That's good news because we know how to deal with quadratics. This one is easily factorable:

$$x^2 + 4x + 3 = (x + 3)(x + 1)$$

So that's the whole problem. The complete factorization is:

$$x^4 + 4x^3 + 2x^2 - 4x - 3 = (x + 3)(x - 1)(x + 1)^2$$

and the roots are

$$x = -1,\; -1, \; 1,\; -3 $$

x = -1 is a double root

Use the rational root theorem to find all of the roots (zeros) of these functions:

Note: For some of the solutions to these problems, I've skipped some of the trial-and-error parts just to save space and keep the solutions simple. Sometimes there's a lot of trial-and-error — and failure — involved in these problems. Keep in mind that *all* of the possible rational roots might fail.

1. |
$f(x) = x^4 + 3x^3 - 3x^2 - 7x + 6$ ## Solution |

2. |
$f(x) = 2x^3 - x^2 - 7x + 6$ ## Solution |

3. |
$f(x) = 3x^4 + 3x^3 - 8x^2 - 2x + 4$ ## Solution |

4. |
$f(x) = 8x^5 + 56x^4 + 80x^3 - x^2 - 7x - 10$ ## Solution |

Sometimes you won't find a GCF, grouping won't work, it's not a sum or difference of cubes and it doesn't look like a quadratic, . . . and it doesn't have any rational roots. What to do? Well, you're stuck, and you'll have to resort to numerical methods to find the roots of your function.

That means graphing the function on a calculator and estimating x-axis crossings or using a numerical root-finding algorithm. Some calculators and many computer programs can do this. You'll also learn about Newton's method of finding roots in calculus.

This function has an odd number of terms, so it's not group-able, and there's no greatest common factor (GCF), so it's a good candidate for using the** rational root theorem** with the set of possible rational roots: {±1, ±2}. If none of those work, **f(x)** has no rational roots (this one does, though).

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