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Section 7.2 Graphing Polynomial Functions

Subsection 1. Factor polynomials

Factoring can help us analyze a polynomial. In particular, for polynomials in one variable, factoring may help us find the \(x\)-intercepts of the graph.

Subsubsection Examples

Example 7.13.

Factor completely \(~4x^3y-12x^2y-40xy\)

Solution

First, factor out the common factor, \(~4xy\text{.}\)

\begin{equation*} 4x^3y-12x^2y-40xy = 4xy(x^2-3x-10) \end{equation*}

Now factor the quadratic trinomial. We need two numbers \(p\) and \(q\) that satisfy

\begin{equation*} ~pq=-10~~~~ \text{and}~~~~ p+q=-3~ \end{equation*}

By trial and error we find \(~p=-5~\) and \(~q=2~\text{,}\) so

\begin{equation*} x^2-3x-10 = (x-5)(x+2) \end{equation*}

and thus

\begin{equation*} 4x^3y-12x^2y-40xy = 4xy(x-5)(x+2) \end{equation*}
Example 7.14.

Factor the quadratic trinomial \(~8x^2-21x-9\)

Solution

We want to find two factors so that \(~8x^2-21x-9 = (ax+b)(cx+d)\text{.}\) Now

\begin{equation*} ~(ax+b)(cx+d) = adx^2 +(ad+bc)x + bd~ \end{equation*}

so \(~(ad)(bc)=8(-9)=-72~\) and \(~ad+bc=-21.~\)

To simplify the calculations, we let \(~p=ad~\) and \(~q=bc~\text{.}\) We need to find the two numbers \(p\) and \(q\) that satisfy

\begin{equation*} ~pq=-72~~~~ \text{and}~~~~ p+q=-21~ \end{equation*}

By trying different factors of \(-72\text{,}\) we find that \(~p=-24~\) and \(~q=3~\text{.}\)

Finally, we write the trinomial as \(~8x^2-24x+3x-9~\text{,}\) and factor by grouping:

\begin{align*} (8x^2-24x)+(3x-9) \amp\amp\amp \begin{array}{l} \blert{\text{Factor out the greatest common factor }}\\ \blert{\text{from each group.}} \end{array}\\ 8x\alert{(x-3)}+3\alert{(x-3)}\amp\amp\amp \blert{\text{Factor out the common binomial.}}\\ (x-3)(8x+3) \end{align*}

Thus, \(~8x^2-21x-9 = (x-3)(8x+3).\)

Subsubsection Exercises

Factor completely \(~18a^2b-9ab-27b\)

Answer
\(9b(2a-3)(a+1)\)

Factor completely \(~4x^3+12x^2y+8xy^2\)

Answer
\(4x(x+2y)(x+y)\)

Factor completely \(~9x^3y+9x^2y^2-18xy^3\)

Answer
\(9xy(x+2y)(x-y)\)

Factor completely \(~12b^3y^2+15b^2y+3b\)

Answer
\(3b(4by+1)(by+1)\)

Factor completely \(~5x^2-14x-24\)

Answer
\((5x+6)(x-4)\)

Factor completely \(~12t^2-10t-50\)

Answer
\(2(2t-5)(3t+5)\)

Subsection 2. Factor special products

Here is more practice using formulas to factor the special quadratic and cubic polynomials.

Quadratic Polynomials.
  1. \(\displaystyle (a+b)^2=a^2+2ab+b^2\)
  2. \(\displaystyle (a-b)^2=a^2-2ab+b^2\)
  3. \(\displaystyle (a+b)(a-b)=a^2-b^2\)
  4. \(a^2+b^2~\) cannot be factored
Cubic Polynomials.
  1. \(\displaystyle (x+y)^3=x^3+3x^2y+3xy^2+y^3\)
  2. \(\displaystyle (x-y)^3=x^3-3x^2y+3xy^2-y^3\)
  3. \(\displaystyle x^3+y^3=(x+y)(x^2-xy+y^2)\)
  4. \(\displaystyle x^3-y^3=(x-y)(x^2+xy+y^2)\)

Subsubsection Examples

Example 7.21.

Factor \(~x^4-24x^2+144\)

Solution

From the square terms we see that \(a=x^2\) and \(b=12\text{.}\) We check that the middle term is \(-2ab\text{.}\)

\begin{equation*} -2ab=-2(x^2)(12)=-24x^2 \end{equation*}

The polynomial fits the pattern for \((a-b)^2\text{,}\) so

\begin{equation*} x^4-24x^2+144 = (x^2-12)^2 \end{equation*}
Example 7.22.

Factor \(~27t^3+8v^2\)

Solution

The polynomial is the sum of two cubes, with \(x=3t\) and \(y=2v\text{.}\) We substitute these values into the formula.

\begin{align*} (x+y)(x^2-xy+y^2) \amp = (3t+2v)((3t)^2-(3t)(2v)+(2v)^2)\\ \amp =(3t+2v)(9t^2-6tv+4v^2) \end{align*}

Thus,

\begin{equation*} 27t^3+8v^2= (3t+2v)(9t^2-6tv+4v^2) \end{equation*}

Subsubsection Exercises

Factor \(~a^6-4a^3b+4b^2\)

Answer
\((a^3-2b)^2\)

Factor \(~m^2+30m+225\)

Answer
\((m+15)^2\)

Factor \(~125a^{12}+1\)

Answer
\((5a^4+1)(25a^8-5a^4+1)\)

Factor \(~64p^3-q^6\)

Answer
\((4p-q^2)(16p^2+4pq^2+q^4)\)

Subsection 3. Divide polynomials

If a polynomial cannot be factored, we can use a process like long division to write it as a quotient plus a remainder.

Subsubsection Example

Example 7.27.

Divide \(~\dfrac{2x^2+x-7}{x+3}\)

Solution

First write

\begin{gather*} \require{enclose}x+3 \enclose{longdiv}{2x^2+x-7}\kern-.2ex \end{gather*}

and divide \(2x^2\) (the first term of the numerator) by \(x\) (the first term of the denominator) to obtain \(x\text{.}\) (It may be helpful to write down the division: \(\dfrac{2x^2}{x}=2x\text{.}\)) Write \(\alert{2x}\) above the quotient bar as the first term of the quotient, as shown below.

Next, multiply \(x+3\) by \(2x\) to obtain \(2x^2+6x\text{,}\) and subtract this product from \(2x^2+x-7\text{.}\)

\begin{align*} \require{enclose} \begin{array}[t]{rll} \alert{2x} \hphantom{1+x-7}&& \\[-3pt] x+3 \enclose{longdiv}{2x^2+x-7}\kern-.2ex \\[-3pt] \underline{-(2x^2+6x)\phantom{11}} && \blert{\text{Multiply } x+3 \text{ by } 2x \text{, and}}\\[-3pt] -5x-7 && \blert{\text{subtract the result}.} \end{array} \end{align*}

Repeating the process, divide \(-5x\) by \(x\) to obtain \(-5\text{.}\) Write \(-5\) as the second term of the quotient. Then multiply \(x+3\) by \(-5\) to obtain \(-5x-15\text{,}\) and subtract:

\begin{align*} \require{enclose} \begin{array}[t]{rll} 2x\alert{-5} \hphantom{x-11}&& \\[-3pt] x+3 \enclose{longdiv}{2x^2+x-7}\kern-.2ex \phantom{11}&& \\[-3pt] \underline{-(2x^2+6x)\phantom{11}} \phantom{11}&& \\[-3pt] -5x-7\phantom{11}\\[-3pt] \underline{-(-5x-15)} && \blert{\text{Multiply } x+3 \text{ by }{-5} \text{, and}} \\[-3pt] 8\phantom{11} &&\blert{\text{subtract the result}.} \end{array} \end{align*}

The remainder is 8. Because the degree of 8 is less than the degree of \(x+3\text{,}\) the division is finished. The quotient is \(2x-5\text{,}\) with a remainder of 8. We write the remainder as a fraction to obtain

\begin{equation*} \dfrac{2x^2+x-7}{x+3} = 2x-5 + \dfrac{8}{x+3} \end{equation*}

Subsubsection Exercises

Divide \(~\dfrac{4x^2+12x+7}{2x+1}\)

Answer
\(2x+5+\dfrac{2}{2x+1}\)

Divide \(~\dfrac{8z^4+4z^2+5z+3}{2z+1}\)

Answer
\(4z^3-2z^2+3z+1+\dfrac{2}{2z+1}\)