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When presented with a polynomial, it is possible to determine the number of roots by using the Fundamental Theorem of Algebra. However, this theorem does not give any detailed information about the roots — whether they are integer, rational, real, or imaginary. This lesson will explore methods to find these amazing inner workings of a root.
### Catch-Up and Review

**Here are a few recommended readings before getting started with this lesson.**

Explore

Recalling the Fundamental Theorem of Algebra, a polynomial function of degree greater than $0$ always has a root. These roots can be found using diverse methods like the quadratic formula for quadratic polynomial functions.

How can the roots of the polynomial be related to the constant term of a polynomial function?

$ax_{2}+bx+c=0⇓x=2a-b±b_{2}−4ac $

However, factoring a polynomial function of greater degrees can be tough, as there are not easy to use formulas to do so. The good news is that looking at the terms of a polynomial function can be useful to find more characteristics of a given function. Consider the following functions and their roots. Polynomial Function | Roots |
---|---|

$p(x)=x_{3}−6x_{2}+11x−6$ | $1,$ $2,$ $3$ |

$g(x)=x_{4}−4x_{3}−19x_{2}+106x−120$ | $-5,$ $2,$ $3,$ $4$ |

$m(x)=x_{3}+6x_{2}−55x−252$ | $-9,$ $-4,$ $7$ |

Discussion

Consider a polynomial where every coefficient is an integer.
### Proof

Rational Root Theorem

$P(x)=a_{n}x_{n}+a_{n−1}x_{n−1}+⋯+a_{1}x+a_{0} $

The following properties are true for the roots of the polynomial. - Every integer root is a factor of the constant $a_{0}.$
- All rational roots must have a numerator that is a factor of $a_{0}$ and a denominator that is a factor of the leading coefficient $a_{n}.$

To prove this theorem, each of the properties will be considered individually. To start with the first property, consider a polynomial $P(x)$ with integer coefficients that has an integer root $x_{r}.$ This means that $P(x_{r})=0.$
Since the coefficients of $P(x)$ are integers and $x_{r}$ is an integer, the expression between the parentheses results in an integer. Also, $-a_{0}$ is the product of an integer and $x_{r},$ which means that $x_{r}$ is a factor of $a_{0}.$ This completes the proof of the first property.
Since the coefficients of $P(x),$ $p,$ and $q$ are all integers, the expression between parentheses results in an integer. Also, since $qp $ is written in its simplest form, $p$ and $q$ do not have a common factor, which means that $p$ and $q_{n}$ do not have a common factor either. Therefore, $p$ is a factor of $a_{0}.$
If the modification is done differently, it is possible to come to the other property.
With the same reasoning as before, it is possible to conclude that $q,$ the denominator of the root, is a factor of $a_{n},$ finishing the proof.

$a_{n}x_{r}+a_{n−1}x_{r}+⋯+a_{1}x_{r}+a_{0}=0 $

Now $a_{0}$ is subtracted from both sides of the equation.
$a_{n}x_{r}+a_{n−1}x_{r}+⋯+a_{1}x_{r}+a_{0}=0$

SubEqn

$LHS−a_{0}=RHS−a_{0}$

$a_{n}x_{r}+a_{n−1}x_{r}+⋯+a_{1}x_{r}=-a_{0}$

FactorOut

Factor out $x_{r}$

$x_{r}(a_{n}x_{r}+a_{n−1}x_{r}+⋯+a_{1})=-a_{0}$

For the second property, suppose that the polynomial $P(x)$ has a rational root $qp ,$ such that the fraction is written in its simplest form. Again, it is obtained that $P(qp )=0.$

This time the equation will be modified in a different way.$a_{n}(qp )_{n}+a_{n−1}(qp )_{n−1}+⋯+a_{1}(qp )+a_{0}=0$

PowQuot

$(ba )_{m}=b_{m}a_{m} $

$a_{n}(q_{n}p_{n} )+a_{n−1}(q_{n−1}p_{n−1} )+⋯+a_{1}(qp )+a_{0}=0$

MultEqn

$LHS⋅q_{n}=RHS⋅q_{n}$

$a_{n}(p_{n})+a_{n−1}(p_{n−1}q)+⋯+a_{1}(pq_{n−1})+a_{0}(q_{n})=0$

SubEqn

$LHS−a_{0}(q_{n})=RHS−a_{0}(q_{n})$

$a_{n}(p_{n})+a_{n−1}(p_{n−1}q)+⋯+a_{1}(pq_{n−1})=-a_{0}(q_{n})$

FactorOut

Factor out $p$

$p(a_{n}p_{n−1}+a_{n−1}p_{n−2}q+⋯+a_{1}q_{n−1})=-a_{0}q_{n}$

The numerator of the root, $p,$ is a factor of $a_{0}.$

$a_{n}(p_{n})+a_{n−1}(p_{n−1}q)+⋯+a_{1}(pq_{n−1})+a_{0}(q_{n})=0$

SubEqn

$LHS−a_{n}p_{n}=RHS−a_{n}p_{n}$

$a_{n−1}p_{n−1}q+⋯+a_{1}pq_{n−1}+a_{0}q_{n}=-a_{n}p_{n}$

FactorOut

Factor out $q$

$q(a_{n−1}p_{n−1}+⋯+a_{1}pq_{n−2}+a_{0}q_{n−1})=-a_{n}p_{n}$

$p(x)=6x_{4}−59x_{3}+94x_{2}−49x+8 $

The integer roots of the equation must be factors of the value of $a_{0},$ which in this case is $8.$ The factors of $8$ are $1,$ $2,$ $4,$ and $8.$ Each of these values is substituted for $x$ into the equation $p(x)=0$ to determine which is a root. Equation | Result |
---|---|

$6⋅1_{4}−59⋅1_{3}+94⋅1_{2}−49⋅1+8=0$ | $0=0✓$ |

$6⋅2_{4}−59⋅2_{3}+94⋅2_{2}−49⋅2+8=0$ | $-90=0×$ |

$6⋅4_{4}−59⋅4_{3}+94⋅4_{2}−49⋅4+8=0$ | $-924=0×$ |

$6⋅8_{4}−59⋅8_{3}+94⋅8_{2}−49⋅8+8=0$ | $0=0✓$ |

As determined, the integer roots are $1$ and $8.$ The rational roots have a numerator that is a factor of $a_{0}$ and a denominator that is a factor of $a_{n}.$ In this case,$a_{n}$ is $6$ with factors of $1,$ $2,$ $3,$ and $6.$ The following table displays the possible fractions that can be formed with the factors. The columns of the table are the factors of $8,$ and the rows are the factors of $6.$

$1$ | $2$ | $4$ | $8$ | |
---|---|---|---|---|

$2$ | $21 $ | $22 =1$ | $24 =2$ | $28 =4$ |

$3$ | $31 $ | $32 $ | $34 $ | $38 $ |

$6$ | $61 $ | $62 =31 $ | $64 =32 $ | $68 =34 $ |

It is important to understand that the integers do not need to be tested because the integer factors have already been found. Now, each fraction will be tested to see if they are roots of the equation.

Equation | Result |
---|---|

$6(21 )_{4}−59(21 )_{3}+94(21 )_{2}−49(21 )+8=0$ | $0=0✓$ |

$6(31 )_{4}−59(31 )_{3}+94(31 )_{2}−49(31 )+8=0$ | $0=0✓$ |

$6(32 )_{4}−59(32 )_{3}+94(32 )_{2}−49(32 )+8=0$ | $2722 =0×$ |

$6(34 )_{4}−59(34 )_{3}+94(34 )_{2}−49(34 )+8=0$ | $-9100 =0×$ |

$6(38 )_{4}−59(38 )_{3}+94(38 )_{2}−49(38 )+8=0$ | $-277280 =0×$ |

$6(61 )_{4}−59(61 )_{3}+94(61 )_{2}−49(61 )+8=0$ | $108235 =0×$ |

Example

Izabella is tutoring some of her friends for an upcoming math exam.

To study, they decided to go over some exercises that the math professor assigned for homework in the past. One polynomial function, in particular, caught Izabella's attention.$p(x)=x_{4}−x_{3}−6x_{2}+14x−12 $

Solve following exercises. a Find the integer roots of the given polynomial function.

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b Find the remaining roots of the function.

{"type":"text","form":{"type":"roots","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x"],"constants":["i"]},"hideNoSolution":false,"variable":"x"},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"texts":["1+i","1-i"]}}

a Use the Rational Root Theorem.

b Use synthetic division.

a The integer roots of a polynomial function can be found using the Rational Root Theorem. This theorem indicates that the integer roots of a polynomial are factors of the constant term $a_{0},$ which in this case is $-12.$ Below are the factors of $-12$ listed.

$Positive:Negative: 1,2,3,4,6,12-1,-2,-3,-4,-6,-12 $

To determine which of these is a root of the given polynomial function, each of the factors will be substituted into the equation $p(x)=0$ to check which results in a true statement. First, the positive factors will be substituted. Equation | Solution |
---|---|

$(1)_{4}−(1)_{3}−6(1)_{2}+14(1)−12=0$ | $-4 =0×$ |

$(2)_{4}−(2)_{3}−6(2)_{2}+14(2)−12=0$ | $0=0✓$ |

$(3)_{4}−(3)_{3}−6(3)_{2}+14(3)−12=0$ | $30 =0×$ |

$(4)_{4}−(4)_{3}−6(4)_{2}+14(4)−12=0$ | $140 =0×$ |

$(6)_{4}−(6)_{3}−6(6)_{2}+14(6)−12=0$ | $936 =0×$ |

$(12)_{4}−(12)_{3}−6(12)_{2}+14(12)−12=0$ | $18300 =0×$ |

The value of $x=2$ is a root of the function. Since the polynomial function is of degree $4,$ it is possible that one of the negative factors is also a root, so these values will be substituted next.

Equation | Solution |
---|---|

$(-1)_{4}−(-1)_{3}−6(-1)_{2}+14(-1)−12=0$ | $-30 =0×$ |

$(-2)_{4}−(-2)_{3}−6(-2)_{2}+14(-2)−12=0$ | $-40 =0×$ |

$(-3)_{4}−(-3)_{3}−6(-3)_{2}+14(-3)−12=0$ | $0=0✓$ |

$(-4)_{4}−(-4)_{3}−6(-4)_{2}+14(-4)−12=0$ | $156 =0×$ |

$(-6)_{4}−(-6)_{3}−6(-6)_{2}+14(-6)−12=0$ | $1200 =0×$ |

$(-12)_{4}−(-12)_{3}−6(-12)_{2}+14(-12)−12=0$ | $21420 =0×$ |

As can be seen in the table, $-3$ is also a root of $p(x).$ Since every integer root has to be a factor of $-12$ and every factor was tested, the only integer roots of the given function are $-3$ and $2.$

b The Factor Theorem can be used to factor the given polynomial into a polynomials of a lesser degree. Since $2$ is a root of $p(x),$ $(x−2)$ is a factor of $p(x).$ This means that multiplying a polynomial $q(x)$ by $(x−2)$ results in $p(x).$

$p(x)=(x−2)q(x) $

To find the polynomial $q(x),$ which is one degree less than $p(x),$ the polynomial $p(x)$ has to be divided by the binomial $(x−2).$
$q(x)=x−2x_{4}−x_{3}−6x_{2}+14x−12 $

This division can be performed by synthetic division. First, the coefficients are ordered and the leftmost coefficient goes down, below the horizontal line.
$2 1 -1-614-12 $

Bring down $1$

$2 1 -1 -6 14 -12 1 -1 -6 14 -12 $

$2 1 -1 -6 14 -12 1 -1 -6 14 -12 $

▼

Evaluating the division

$2⋅1=2$

$2 1 -12 -6 14 -12 1 -1 -6 14 -12 $

$-1+2=1$

$2 1 -12 -6 14 -12 1 1 -6 14 -12 $

$2⋅1=2$

$2 1 -12 -62 14 -12 1 1 -6 14 -12 $

$-6+2=-4$

$2 1 -12 -62 14 -12 1 1 -4 14 -12 $

$2⋅-4=-8$

$2 1 -12 -62 14-8 -12 1 1 -4 14 -12 $

$14−8=6$

$2 1 -12 -62 14-8 -12 1 1 -4 -6 -12 $

$2⋅6=12$

$2 1 -12 -62 14-8 -1212 1 1 -4 6 -12 $

$-12+12=0$

$2 1 -12 -62 14-8 -1212 1 1 -4 6 .0 $

$q(x)=x−2x_{4}−x_{3}−6x_{2}+14x−12 ⇓q(x)=x_{3}+x_{2}−4x+6 $

Now the given function can be rewritten factoring the binomial $(x−2).$
$p(x)=(x−2)q(x)⇓p(x)=(x−2)(x_{3}+x_{2}−4x+6) $

A similar thing can be done to rewrite $q(x).$ A root of the function $p(x)$ is also a root of $q(x).$ Therefore, the binomial $(x+3)$ can be factored from $q(x)$ using synthetic division again.
$-3 1 1-46 $

▼

Evaluate the division

Bring down $1$

$-3 1 1 -4 6 1 1 -4 6 $

$-3⋅1=-3$

$-3 1 1-3 -4 6 1 1 -4 6 $

$1−3=-2$

$-3 1 1$