2.3: Polynomial Equations
 Page ID
 130842
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{\!\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\ #1 \}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\ #1 \}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{\!\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{\!\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
By the end of this section, you will be able to:
 Understand what it means to solve a polynomial equation
 Solve certain polynomial equations with one variable.
Before you get started, take this readiness quiz.
 Solve \((x2)(3x7)=0\).
 Divide \((x^36x^2+11x6)\) by \((x1)\).
 Solve \(2x^25x1=0\).
Polynomial equations with one variable have features that are similar to the quadratic equations, the kind of polynomial equations we treated in the previous section.
For example, consider the polynomial equation
$$ 0=2(x1)(2x3)(x+7).\nonumber$$
Note that the right hand side of the equation is a polynomial of degree 3 with leading coefficient \(4\) which we can see by imagining distributing the product. We see that if we replace \(x\) with 1 we arrive at the true statement \(0=0\) so that \(1\) is a solution. Also, when we replace \(x\) with \(\dfrac32\), the factor \((2x3)\) has the value \(0\) so that the right hand side is also \(0\) and we have again the true statement \(0=0\). Similarly, if we replace \(x\) with \(7\) the factor \((x+7)\) becomes \(0\) and again we arrive at \(0=0\).
So, \(1,\dfrac32\), and \(7\) are solutions. For other values of \(x\), none of the factors are zero, so the product on the right hand side is not zero. It follows that these are the only solutions.
A number \(c\) is a root of a polynomial of a single variable if when we substitute \(c\) for that variable the result is equal to 0.
So, in the above example, the roots of \(2(x1)(2x3)(x+7)\) are \(1, \dfrac32\) and \(7\). Note that, then, \(1, \dfrac32\) and \(7\) are solutions to the equation \(2(x1)(2x3)(x+7)=0.\)
We can also easily write down a polynomial with prescribed roots. For example, suppose we want to write down a polynomial of degree 3 with roots 3, 2, and 4 and leading coefficient 7. We simply try to arrange the situation that we have above. We attempt to find factors so that one of the factors is zero when we substitute 3, another is zero when we substitute \(2\) and yet another is zero when we substitute \(4\). We see that \((x3)(x+2)(x4)\) is a polynomial with the appropriate zeros. But the leading coefficient is 1. To arrange the leading coefficient to be 7, we need only multiply (so that \((x3)\), \((x+2)\) and \((x4)\) are still factors) by 7. So,
$$7(x3)(x+2)(x4)$$
is a polynomial of degree 7 with leading coefficient 7 and roots \(4,2,\) and \(4\).
More generally, we have the following
If \(c\) is a root of a polynomial, then \((xc)\) is a factor of that polynomial, i.e., the polynomial is equivalent to \((xc)\text{(some polynomial)}\).
Now suppose we want to solve
$$0=x^3+2x^211x12.$$
The right hand side is not quadratic and in general when the degree is 3 or larger the equation can be very difficult to solve.
However, in this case, we can observe (via trial and error or with the help of a calculator) that
$$0=(1)^3+2(1)^211(1)12.$$
It follows that \((x+1)\) is a factor! What is the other factor? We can use long division to find
$$x^3+2x^211x12=(x+1)(x^2+x12).$$
This is a product of a linear expression and a quadratic expression (which can then be factored!!).
We see then that
$$x^3+2x^211x12=(x+1)(x+4)(x3).$$
So,
$$0=x^3+2x^211x12$$
is equivalent to
$$0=(x+1)(x+4)(x3).$$
So we can see the solutions are \(1, 4\) and \(3\).
Solve \(2x^3+x^25x+2=0\).
 Solution

To begin solving this equation, we note that by guess and checking some simple values for \(x\) , perhaps with the help of a calculator,
\(2(1)^3+(1)^25(1)+2=0\)
so, \(1\) is a root of \(2x^3+x^25x+2\)! It follows that \((x1)\) is a factor, i.e.,
\(2x^3+x^25x+2=(x1)(\text{some polynomial})\).
We can proceed by guessing and checking or by dividing \(2x^3+x^25x+2\) by \((x1)\). By guessing at coefficients we can easily determine that that mystery polynomial has a leading term of \(2x^2\) and a constant term of \(2\). So, this mystery polynomial looks like \(2x^2+bx2\). But, the coefficient of \(x\) when we multiply \((x1)(2x^2+bx2)\) is \(b2\) but, being equal to \(2x^3+x^25x+2\) must also be \(5\) so \(b\) must be 3 and
\[2x^3+x^25x+2=(x1)(2x^23x2).\]
This should be checked by multiplying the right hand side.
We can now hope to factor the quadratic factor by guessing and checking, the AC method, or by finding roots (and therefore corresponding factors) by using the quadratic formula as we will see in a later section. Here we see
\[2x^3+x^25x+2=(x1)(2x^23x2)=(x1)(2x+1)(x2)\]
so the equation
\[2x^3+x^25x+2=0\]
is equivalent to
\[(x1)(2x+1)(x2)=0.\]
We can use the zeroproductproperty then to find that the solutions are \(1, \dfrac12\) and \(2\).
We may check that substituting any of the three values lead to a true statement in the original equation.
Solve \(y^36y^2y+30=0\).
 Answer

\(y= 2, 3\) or \(5.\)
Solve \(2x^35x^2=x6\).
 Answer

\(x=1,\dfrac32,\) or \(2\)
 Explain the relationships between roots and factors of a polynomial.
 What can you say about the degree of a polynomial which has roots \(1,0\) and \(2\)?
 If the height at time \(t\) is a polynomial with variable \(t\), what do the roots represent?
 Give a polynomial of degree 4 with roots \(2,3, 1\) and \(0\).
 Find all solutions to the polynomial equation \(x^3+3x^25x=6\).
Key Concepts
roots of a polynomial
Factors of a polynomial