Next Generation Advanced Algebra and Functions Study Guide for the ACCUPLACER Test
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Polynomial Equations
A polynomial equation equates an expression containing multiple terms to a constant value, such as:
\[3x^2 + 4y^2 + 4x^2y - xy^2 + 9 = 0\]For an expression to be a polynomial, it must be written in terms that only contain positive exponents. The terms can be combined through addition, subtraction, or multiplication. Polynomial equations cannot include non-reducible division operations.
Polynomial equations can be used to relate situations involving multiple variables.
Creating
Consider the following example: The length of a cube is increased by twice the side length. The width of the cube is decreased by half the side length. The height is increased by \(15\) meters. The new volume is \(1250 \;m^3\). Write an equation representing the new volume.
Recall that the formula for the volume of a cube is \(lwh\), where \(l\) is the length, \(w\) is the width, and \(h\) is the height:
\[l = w = h\]Use the given alterations to the variables to rewrite the volume formula:
\[1250 = (s + 2s)(s - \frac{s}{2})(s + 15)\]where \(s\) is the original side length.
Distribute each set of parentheses to create a polynomial expression:
which, if necessary, can be solved using any appropriate factoring method.
Now an example including \(2\) variables:
A cylinder’s radius is increased by three times its height, and its height is increased by \(4\) times its radius. What is the new formula for the cylinder’s volume?
Recall that the volume of a cylinder is:
\(V = \pi r^2 h\), where \(r\) is the radius and \(h\) is the height.
Apply the given alterations to generate the new volume formula:
\[V = \pi (r + 3h)^2 (h + 4r)\]Apply the exponent and distribute the resulting expression to find the new volume formula:
\[V = r^2h + 4r^3 + 6rh^2 + 24r^2h + 9h^3 + 36rh^2\]and combine like terms to find:
\[V = 4r^3 + 25r^2h + 42rh^2 + 9h^3\]Solving
Solving a polynomial equation means finding the values for the variable, which, when substituted into the equation yield a true statement. This is also stated as finding the roots of the equation.
In the \(xy\)-plane, the roots are the locations at which the curve passes through the \(x\)-axis. Solving polynomial equations is exactly like solving quadratic and cubic equations (each of these is considered a polynomial equation). As such, the same methods of rewriting the equation into a more manageable form, primarily through factoring (but also using any other valid algebraic manipulation) are used to find the solutions.
Let’s solve this equation:
\[x^3 - 3x^2 + x - 3 = 0\]When presented with a polynomial equation, the first consideration is for a greatest common factor. There is no GCF for this example, so we must find a different method of factoring. You should notice that the equation is cubic in nature, but it is not a sum or difference of cubes. Another possible method of factoring a cubic is by grouping. Applying the grouping method:
\[(x^3 - 3x^2) + (x - 3) = 0\]which enables a GCF to be factored from the first expression:
\[x^2(x - 3) + 1(x - 3) = 0\]Because the \(x - 3\) is common to both terms, it can also be factored out to form:
\[(x - 3)(x^2 + 1) = 0\]Setting both left side terms equal to \(0\):
\(x - 3 = 0\) and \(x^2 + 1 = 0\)
results in:
\(x = 3\) and \(x = \sqrt{-1} = \pm i\)
The polynomial has one real, valued root, \(3\), and two imaginary roots, \(\pm i\), and it graphs as the following:
Notice that the graph shows the curve passing through the \(x\)-axis in only one location. This is representative of the one real, valued root.
Graphing
As always, any function can be graphed by supplying an adequate number of input values and computing their corresponding output values. However, to save time, it is far more efficient to develop an intuition for graphing these types of functions. Rather than arbitrarily choosing input values, it is recommended to find some of the critical points and features of the function in question.
These critical points will be locations at which the function passes through the \(x\) or \(y\) axis or points that serve as a vertex. Because polynomial functions are continuous, there is no need to search for asymptotes, but it is still wise to consider the end behavior of the function—does it increase or decrease as \(x\) becomes very large or very small?
Polynomials containing an even valued integer as its largest exponent will have both ends of the function pointing in the same direction. Odd valued polynomials will have ends pointing in opposite directions.
Let’s practice graphing with a couple examples. First, \(f(x) = x^4 - 16\)
Right away, notice that when \(x = 0\), \(f(x) = -16\), indicating that the \(y\)-intercept is \((0, -16)\).
And, even though the function is a \(4\)th degree polynomial, it should be recognized as a difference of squares, and as a result, can be rewritten into a more manageable form:
\[f(x) = (x^2 - 4)(x^2 + 4)\]which also contains a difference of squares, so:
\[f(x) = (x + 2)(x - 2)(x^2 + 4)\]Setting \(f(x) = 0\) enables the roots, or \(x\)-intercepts, to be found:
\[0 = (x + 2)(x - 2)(x^2 + 4)\]so:
\(x + 2 = 0\) and \(x - 2 = 0\) and \(x^2 + 4 = 0\)
which results in:
\(x = \pm 2\) and \(x = \sqrt{-4} = \pm 2i\)
Each of these values corresponds with a point on the curve: \((-2, 0)\), \((2, 0)\) as the real valued intercepts, and \((-2i, 0)\), \((2i, 0)\) as the imaginary valued intercepts.
The end behavior can be determined by assessing the largest exponent and its coefficient. A positive coefficient and an even valued exponent indicate that the function will be positive for very large and very small \(x\) values. From what we’ve uncovered so far, a general shape for the curve can be drawn, but we should get an idea of what’s happening between the two roots and \(x = 0\).
Evaluating \(f(-1)\) and \(f(1)\), we find:
\(f(-1) = (-1)^4 - 16 = -15\) and \(f(1) = -15\), which gives the points \((-1, -15)\) and \((1, -15)\). And, visualizing the points collected thus far, the general shape can be graphed:
Let’s graph one more: \(f(x) = -x^3 - x^2 + 2x\).
As before, consider any common factors. Here, an \(x\) can be factored from each expression:
\(f(x) = x(-x^2 - x + 2)\).
Evaluating \(f(0)\) to find a \(y\)-intercept of 0 and the point \((0, 0)\).
Setting the expression in parentheses equal to 0 and solving for \(x\):
\(-x^2 - x + 2 = 0\), factoring out a \(-1\):
\(-1(x^2 + x - 2) = 0\), and factoring the quadratic:
\(-1(x + 2)(x - 1) = 0\), to give:
\(x = - 2\) and \(x = 1\), or as points \((-2, 0)\) and \((1, 0)\).
To determine end behavior, consider a very negative value of \(x\), \(-100\) for example:
\(f(-100) = (-100)(-(-100)^2 - (-100) + 2) = 989,800\), stated otherwise: as \(x\) decreases, \(f(x)\) increases. And testing a large value of \(x\):
\(f(100) = (100)(-(100)^2 - 100 + 2) = -1,009,800\), stated otherwise: as \(x\) increases, \(f(x)\) decreases.
Now, consider the \(x\) values between each of the roots (\((-2, 0), (0, 0), (1 ,0)\)). Test \(x = -1\) and \(x = \frac{1}{2}\)
\[f(-1) = -1(-1 + 2)(-1 - 1) = 2\]and
\[f(\frac{1}{2}) = -(\frac{1}{2})^3 -(\frac{1}{2})^2 + 2(\frac{1}{2}) = \frac{5}{8}\]which gives the points \((-1, 2)\) and \((\frac{1}{2}, \frac{5}{8})\).
Plotting these points, a general shape of the graph can be produced:
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