# Step By Step Calculus » 2.7 - Polynomial Functions and Algebra

Synopsis
Polynomial: A polynomial (in the variable xx) is a function that can be written as a_0+a_1x+a_2x^2+\cdots+a_nx^na_0+a_1x+a_2x^2+\cdots+a_nx^n for real constants a_0, a_1, \cdots a_n a_0, a_1, \cdots a_n .
• The degree of a polynomial is the highest power of xx which appears in its formula.
• A linear polynomial has degree 11 (e.g. a_0+a_1xa_0+a_1x). A quadratic polynomial has degree 22 (e.g. a_0+a_1x+a_2x^2a_0+a_1x+a_2x^2). A cubic function has degree 33 (e.g. a_0+a_1x+a_2x^2+a_3x^3a_0+a_1x+a_2x^2+a_3x^3). And so on ...
• The zeros (or roots) of a polynomial f(x)f(x) are the values x=cx=c for which f(c)=0f(c)=0.
• Factor Theorem: A number cc is a zero of a polynomial f(x)f(x) if and only if x-cx-c is a factor of f(x)f(x).
• For a quadratic ax^2+bx+cax^2+bx+c, the zeros are
and so ax^2+bx+c=a(x-c_1)(x-c_2)ax^2+bx+c=a(x-c_1)(x-c_2).
Factoring an expression:
• The name Factor refers to one of the quantities that are multiplied together in a product, while the name term refers to one of the quantities that are added together in a sum.
• To factor an expression means to change it from a sum of terms to an equivalent product of factors. It is useful in various computations.
• There are many ways of factoring an expression, depending on its structure. Some basic methods are: (i) factoring by collecting a common factor, e.g. x^2-3x=x(x-3)x^2-3x=x(x-3), (ii) factoring by grouping, e.g. \underline{x^3-2x^2}-\underline{3x+6}=x^2\underline{(x-2)}-3\underline{(x-2)}=(x-2)(x^2-3)\underline{x^3-2x^2}-\underline{3x+6}=x^2\underline{(x-2)}-3\underline{(x-2)}=(x-2)(x^2-3), (iii) factoring by using formulae for special products, i.e.
\begin{tabular}{l|l}\hline Difference of squares & $x^2-a^2=(x-a)(x+a)$\\ \hline Difference of cubes & $x^3-a^3=(x-a)(x^2+ax+a^2)$ \\ \hline Sum of cubes & $x^3+a^3=(x+a)(x^2-ax+a^2)$\\ \hline Difference of fourth powers & $x^4-a^4=(x-a)(x+a)(x^2+a^2)$\\ \hline Square of a sum & $x^2+2ax+a^2=(x+a)^2$\\ \hline Square of a difference & $x^2-2ax+a^2=(x-a)^2$\\ \hline \end{tabular}
Completing the square: We use a procedural approach to change a quadratic expression from its standard form ax^2+bx+cax^2+bx+c to the perfect square form a(x+h)^2+ka(x+h)^2+k. The procedure is as follows.
\begin{tabular}{ll|l}\hline \textbf{Step 1:} & Organize the expression as a quadratic polynomial. & $ax^2+bx+c$ \\ \hline \textbf{Step 2:} & Group terms with $x$'' together, and if $a\ne1$, &$(ax^2+bx)+c$\\ &factor it out from the group by dividing &$=a\left(\frac{ax^2}{a}+\frac{bx}{a}\right)+c$\\ & each term by $a$. & $=a\left(x^2+\frac{b}{a}x\right)+c$\\\hline \textbf{Step 3:} & Determine the square of one-half of the coefficient of $x$. & \\ & Since the coefficient of $x$ is $\frac{b}{a}$, divide this number by $2$ & \\ & and then square it $\left(\frac{b}{2a}\right)^2=\frac{b^2}{4a^2}$. Next, add and subtract&$\displaystyle a\left(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}\right)+c$\\ & this number inside the bracket if $a\ne 1$. & \\\hline \textbf{Step 4:} & Separate the first three terms inside the bracket, since & \\ & they make up the perfect square. Be careful with how &$\displaystyle a\left(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}\right)-\frac{b^2}{4a}+c$ \\ &you take the fourth term out of the bracket. & \\ \hline \textbf{Step 5:} & Rewrite the bracket as a perfect square and & $\displaystyle a\left(x+\frac{b}{2a}\right)^2+\left(c-\frac{b^2}{4a}\right)$\\ & combine the remaining constants if possible. & \\ \hline \end{tabular}