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When working with polynomial equations, valuable information can be gathered just by looking at the structure of the equation itself. This lesson will present how to discover information about the solutions of a polynomial equation without having to solve it.
### Catch-Up and Review

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

It is common to be given an equation and asked for a solution or root. For example, consider the following quadratic equation.

$x_{2}−3x+2=0⇕(x−1)(x−2)=0 $

The equation is factored into two binomials. In the factored equation, it can be determined that its solutions are $x=1$ and $x=2.$ Now consider another quadratic equation.
$x_{2}−2x+1=0⇕(x−1)(x−1)=0 $

The only solution to this equation is $x=1,$ which is repeated twice. A quadratic equation can have $0,$ $1,$ or $2$ real solutions. Is it possible to determine how many solutions an equation can have without solving it? Consider a third equation.
$2x_{5}−7x_{4}+12x_{3}−12x_{2}=0 $

How many real solutions does the equation have? Is it possible to determine if the solutions are positive or negative?A theorem is used to determine the maximum number of solutions or zeros, that a polynomial equation can have. This theorem is called the Fundamental Theorem of Algebra.

For a polynomial $p(z)$ of degree $n>0,$ the equation $p(z)=0$ has at least one solution in the set of complex numbers.

For a polynomial $p(z)$ of degree $n>0,$ the equation $p(z)=0$ has exactly $n$ solutions, counting each repeating solution the number of times the solution is repeated. This means that a solution repeated two times counts as two solutions, a solution repeated three times counts as three solutions, and so on.

To justify this theorem, polynomials with low degrees will be examined.

$p_{1}(x)=ax+b $

The solution to the equation $p_{1}(x)=0$ can be found by performing inverse operations.
$p_{1}(x)=0$

Substitute

$p_{1}(x)=ax+b$

$ax+b=0$

Solve for $x$

SubEqn

$LHS−b=RHS−b$

$ax=-b$

DivEqn

$LHS/a=RHS/a$

$x=a-b $

MoveNegNumToFrac

Put minus sign in front of fraction

$x=-ab $

$p_{2}(x)=ax_{2}+bx+c $

This time, the Quadratic Formula can be used to find the solutions to $p_{2}(x)=0.$
$ax_{2}+bx+c=0⇓x_{1},x_{2}=2a-b±b_{2}−4ac $

There are three possible combinations for the resulting values of $x_{1}$ and $x_{2}.$ - The values of $x_{1}$ and $x_{2}$ can be two different real numbers.
- The values of $x_{1}$ and $x_{2}$ can be two different complex numbers.
- The values of $x_{1}$ and $x_{2}$ can be the same real number.

$p_{2}(x)=(x−x_{1})(x−x_{2}) $

$p_{3}(x)=ax_{3}+bx_{2}+cx+d $

This time, the graph of an arbitrary cubic polynomial will be examined.
The end behavior of a cubic polynomial is always down-up or up-down. Therefore, it always has at least one real solution. Let $r_{1}$ be this solution. Using the Factor Theorem, it is possible to rewrite the equation $p_{3}(x)=0$ as the product of $(x−r_{1})$ and a polynomial of degree $2.$
$ax_{3}+bx_{2}+cx+d=0⇕(x−r_{1})(a_{2}x_{2}+b_{2}x+c_{2})=0 $

It has been already established that the equation $p_{2}(x)=0,$ where $p_{2}(x)$ is a polynomial of degree $2,$ always has two solutions. This means that the equation $p_{3}(x)$ has exactly $3$ solutions: $r_{1},$ $r_{2},$ and $r_{3}.$ Therefore, the left-hand side of the equation $p_{3}(x)=0$ can be written as a product of three factors.
$p_{3}(x)=0⇕(x−r_{1})(x−r_{2})(x−r_{3})=0 $

$For Any Natural Numbernp_{n}(x)=0⇕(x−r_{1})…(x−r_{n})=0 $

It should be noted that this is an informal justification and should not be taken as a formal proof.
$9z_{4}+13z_{3}−2z+19=0 $

The degree of the polynomial is $4.$ Therefore, the above equation has $4$ solutions. Now consider another polynomial equation.
$x_{2}−6x+9=0 $

Since the equation's degree is $2,$ it has $2$ solutions. The left-hand side of this equation can also be written as the square of a binomial.
$(x−3)_{2}=0 $

By the Factor Theorem, the solution to the equation is $3.$ It is worth noting that since the factor is squared, the solution is repeated and therefore counts as two solutions.Indicate how many solutions each polynomial equation has.

For a polynomial function $p(x)$ with real coefficients written in standard form, the following properties are true.

- The number of positive real zeros is equal to the number of sign changes in $p(x)$ or less than this by an even number.
- The number of negative real zeros is equal to the number of sign changes in $p(-x)$ or less than this by an even number.

To justify this theorem, the graph of a polynomial function will be examined for non-negative values of $x.$ Consider an arbitrary polynomial function $p(x)$ of degree $n.$
*up*.
When $x=8,$ the value of $f$ returns to the original value of $24$ before continuously going up as $x$ increases. In the interval that goes from $0$ to $8,$ the term $-8x_{2}$ drives the graph down. Then the value of $x_{3}$ catches up and directs the graph upwards. Now consider a variation of the function.

$p(x)=a_{n}x_{n}+a_{n−1}x_{n−1}+…+a_{1}x+a_{0} $

Here, $a_{n}$ is the leading coefficient and $a_{0}$ the constant term. Consider the case where both $a_{n}$ and $a_{0}$ are positive. Because $a_{0}>0,$ the graph intersects the $y-$axis above the $x-$axis. Also, since $a_{n}>0,$ the end behavior of the graph for positive values of $x$ is In this case, the number of sign changes must be even. This is because every time a positive sign changes to a negative sign, it must return to positive again, making the number of sign changes a multiple of $2.$ The number of times the graph crosses the $x-$axis must also be even, since the graph begins and ends above the $x-axis.$

Without sign changes, a polynomial function only increases or decreases as $x$ increases. Therefore, each change in direction of the graph is related to a sign change in the polynomial. To explore this idea, consider the following polynomial function.$f(x)=x_{3}−8x_{2}+24 $

It has already been said that only non-negative values of $x$ are considered, so only non-negative values of $x$ will be considered for the graph of $f(x).$
The $y-$intercept of the graph is $24.$ Then, the graph goes down from $24$ for a while before going up. This happens because in the first interval, the term $-8x_{2}$ directs the graph before the leading term $x_{3}$ takes direction. For the values of this first interval, $x_{3}$ is less than $8x_{2}$ and, therefore, $x$ is less than $8.$
$x_{3}<8x_{2}⇒x<8 $

This value can be substituted into the function $f(x)$ to find its corresponding output.
$f(x)=x_{3}−8x_{2}+24$

Substitute

$x=8$

$f(8)=(8)_{3}−8(8)_{2}+24$

MultBasePow

$a⋅a_{m}=a_{1+m}$

$f(8)=8_{3}−8_{3}+24$

SubTerm

Subtract term

$f(8)=24$

$f_{2}(x)=8x_{3}−x_{2}+24 $

The same will be done to find the values where $x_{2}$ is greater than $8x_{3}.$
This time the interval is significantly shorter than in the previous case. This can also be seen in the graph.
Here, the effect of the term $-x_{2}$ is negligible compared to the effect of the term $8x_{3}.$ This indicates that a change in direction of the graph of a function is caused by a sign change, but a sign change does not always result on a change in direction of the graph. Consider the graph of the initial polynomial function again.

The function starts above the $x-$axis and has $4$ significant direction changes. Therefore, there are at least $4$ sign changes in the equation. Also, each time the graph moves in a different direction, it crosses the $x-$axis, which suggests that the number of zeros can be the same as the number of sign changes. Consider now a different graph.

Here, the number of zeros is less than the number of times the graph changes direction. Since sign changes indicate changes in direction, it is possible to have fewer zeros than sign changes. Also, since there cannot be more zeros without changing direction, there cannot be more zeros than sign changes. Two conclusions can be made from this reasoning.

- The number of sign changes and the number of zeros are both even.
- The number of zeros has to be less than or equal to the number of sign changes.

This is equivalent to what is written in the rule. The other cases can be examined on a similar way. Also, negative zeros are found by considering $p(-x)$ because this is a reflection in the $y-$axis. Note that this is an informal justification and cannot be taken as a formal proof.

$f(x)=-x_{4}+2x_{3}+7x_{2}−8x−12 $

The number of sign changes can be found by examining the polynomial. A sign change is when an addition follows a negative term or a subtraction follows a positive term. Finding the sign changes can help determine the possible number of positive zeros.
Two sign changes were identified. Therefore, the number of positive real zeros can be either $2$ or $0.$ To determine the number of negative real zeros, the polynomial function $f(-x)$ needs to be examined. Note that the sign of a term only changes if its exponent is odd.

There are two sign changes in this polynomial. This means that there are also two or zero possible negative zeros. Since the polynomial has a degree of $4,$ the polynomial has $4$ zeros by the Fundamental Theorem of Algebra. If there are less than $4$ real zeros, the remaining zeros are complex. The possible combinations of zeros are shown on the following table.

Positive Real Zeros | Negative Real Zeros | Imaginary Zeros | Total Zeros |
---|---|---|---|

$2$ | $2$ | $0$ | $4$ |

$2$ | $0$ | $2$ | $4$ |

$0$ | $2$ | $2$ | $4$ |

$0$ | $0$ | $4$ | $4$ |

$f(x)=-x_{4}+2x_{3}+7x_{2}−8x−12⇕f(x)=-(x+1)(x−2)(x+2)(x−3) $

By using the Zero Product Property, the zeros can be identified as $-2,$ $-1,$ $2,$ and $3.$ Therefore, $f(x)$ has two real positive zeros and two real negative zeros, which is one of the possibilities shown in the table above. Positive Real Zeros | Negative Real Zeros | Imaginary Zeros | Total Zeros |
---|---|---|---|

$2$ | $2$ | $0$ | $4$ |

$2$ | $0$ | $2$ | $4$ |

$0$ | $2$ | $2$ | $4$ |

$0$ | $0$ | $4$ | $4$ |

Jordan is starting to learn about polynomial functions.

She knows that it is possible to find information about the zeros of a polynomial function before solving it. The following polynomial was included on her homework.$p(x)=-2x_{5}+x_{4}−2x_{3}−3x_{2}+9x−7 $

Help Jordan answer each of the following questions. a Determine the possible numbers of positive real zeros.

b Determine the possible numbers of negative real zeros.

c Make a table that summarizes the possibilities of zeros that the function has, including the imaginary zeros.

a $4,$ $2,$ and $0$

b $1$

c

Real Positive Zeros | Real Negative Zeros | Imaginary Zeros | Total Zeros |
---|---|---|---|

$4$ | $1$ | $0$ | $5$ |

$2$ | $1$ | $2$ | $5$ |

$0$ | $1$ | $4$ | $5$ |

a Use Descartes' Rule of Signs.

b Use Descartes' Rule of Signs.

c Use the Fundamental Theorem of Algebra.

a Descartes' Rule of Signs is used to determine the possible numbers of positive real zeros of a polynomial. To find the number of zeros of the polynomial, the number of sign changes has to be determined first. Be aware that to do this, the polynomial must be written in standard form.

This polynomial function has $4$ sign changes. Descartes' Rule of Signs indicates that the polynomial can have $4$ positive real zeros, or a number that is an even number less than $4.$

Possible Number of Positive Real Zeros | ||
---|---|---|

$4$ | $4−2=2$ | $4−4=0$ |

Since there cannot be a negative number of real positive zeros, the possible numbers of positive real zeros are $4,$ $2,$ and $0.$

b Similar to what was done in Part A, Descarte's Rule of Signs will be used. However, to find the number of negative zeros of the function, the sign changes of the polynomial $p(-x)$ are counted this time. It is important to remember that since every even power of $-1$ is $1,$ the terms with even powers of $x$ will not change their sign.

The polynomial $p(-x)$ has only one sign change. Since there are no positive numbers that are an even number apart and are less than one, the only possible number of real negative zeros is $1.$

c The Fundamental Theorem of Algebra establishes that the degree of a polynomial indicates the total number of zeros of the polynomial. Therefore, the first step to finding all the possible combinations of zeros is to determine the degree of the given polynomial.

$p(x)=-2x_{5}+x_{4}−2x_{3}−3x_{2}+9x−7 $

The degree of the polynomial is $5.$ This means that the polynomial function has $5$ zeros. When considering all the possible combinations of zeros for this function, the total number of zeros has to be $5.$ Therefore, the sum of the positive, negative, and imaginary zeros must be $5.$ Real Positive Zeros | Real Negative Zeros | Imaginary Zeros | Total Zeros |
---|---|---|---|

$4$ | $1$ | $0$ | $5$ |

$2$ | $1$ | $2$ | $5$ |

$0$ | $1$ | $4$ | $5$ |

Write the correct answer for each polynomial equation.

Jordan is continuing to work through her homework on polynomial functions.

Help Jordan solve each of the following.

a Which of the following is not a possible combination of zeros for $f(x)=-x_{5}−8x_{4}$ $+6x_{3}−2x+4?$

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b Determine which function has at least one negative real zero.

{"type":"choice","form":{"alts":["<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:1em;vertical-align:-0.25em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.10764em;\">f<\/span><span class=\"mopen\">(<\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mclose\">)<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.897438em;vertical-align:-0.08333em;\"><\/span><span class=\"mord\"><span class=\"mord mathdefault\">x<\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.8141079999999999em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">4<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">\u2212<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.897438em;vertical-align:-0.08333em;\"><\/span><span class=\"mord\">7<\/span><span class=\"mord\"><span class=\"mord mathdefault\">x<\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.8141079999999999em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">3<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">+<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.897438em;vertical-align:-0.08333em;\"><\/span><span class=\"mord\">4<\/span><span class=\"mord\"><span class=\"mord mathdefault\">x<\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.8141079999999999em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">\u2212<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.72777em;vertical-align:-0.08333em;\"><\/span><span class=\"mord\">9<\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">+<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">1<\/span><span class=\"mord\">3<\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:1em;vertical-align:-0.25em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.03588em;\">g<\/span><span class=\"mopen\">(<\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mclose\">)<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.897438em;vertical-align:-0.08333em;\"><\/span><span class=\"mord\"><span class=\"mord mathdefault\">x<\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.8141079999999999em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">4<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">+<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.897438em;vertical-align:-0.08333em;\"><\/span><span class=\"mord\">7<\/span><span class=\"mord\"><span class=\"mord mathdefault\">x<\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.8141079999999999em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">+<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">1<\/span><span class=\"mord\">9<\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:1em;vertical-align:-0.25em;\"><\/span><span class=\"mord mathdefault\">h<\/span><span class=\"mopen\">(<\/span><span class=\"mord mathdefault\">x<\/span><span 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a Use Descartes' Rule of Signs to find all the possible combinations of zeros for the polynomial function.

b Determine the real negative zeros of each polynomial function.

The function has $3$ sign changes. Therefore, according to Descartes' Rule of Signs, the function can have $3$ or $1$ positive real zeros. To find the numbers of possible negative real zeros, the sign changes will be found for $f(-x).$ Note that a term does not change signs if its exponent is even.

Since there are $2$ sign changes for $f(-x),$ there are $2$ or $0$ negative real zeros. Now it is possible to make a table with all possible combinations of zeros. Since degree of the polynomial is $5,$ by the Fundamental Theorem of Algebra, the total number of zeros is $5.$ If the sum of the real roots is less than $5,$ the other zeros are imaginary numbers.

Real Positive Zeros | Real Negative Zeros | Imaginary Zeros | Total Zeros |
---|---|---|---|

$3$ | $2$ | $0$ | $5$ |

$3$ | $0$ | $2$ | $5$ |

$1$ | $2$ | $2$ | $5$ |

$1$ | $0$ | $4$ | $5$ |

Comparing these combinations with the given options, it can be concluded that the only of the given combinations that is not in the table is $1$ positive real zero, $4$ negative zeros, and $0$ imaginary zeros.

This is an impossible combination.

b For a polynomial function $f(x)$ to have at least one negative real zero, the function $f(-x)$ must have at least one sign change. The given functions will be examined, one by one. The first function $f(x)$ will be considered.

$f(x)=x_{4}−7x_{3}+4x_{2}−9x+13⇓f(-x)=x_{4}+7x_{3}+4x_{2}+9x+13 $

There are no sign changes for $f(-x).$ Therefore, $f(x)$ has no negative real zeros. Now, the function $g(x)$ will be examined.
$g(x)=x_{4}+7x_{2}+19⇓g(-x)=x_{4}+7x_{2}+19 $

Like in the previous function, there are no sign changes for $g(-x).$ This means that $g(x)$ has no negative real zeros. The function $h(x)$ will now be examined.
$h(x)=-x_{4}+9x_{3}−8x_{2}+11x−2⇓h(-x)=-x_{4}−9x_{3}−8x_{2}−11x−2 $

Since $h(-x)$ has no sign changes, $h(x)$ also has no negative real zeros. Finally, $m(x)$ will be investigated.
Looking at the function, it can be seen that there is one sign change. This means that this function has exactly one negative real zero. Therefore, the correct answer is the last polynomial function, $m(x).$

Tearrik knows that Jordan has been studying about identifying the zeros of a polynomial function. He asks for her help with an assignment.

Tearrik wants to write and sketch the graph of a polynomial function of degree $4$ for which all its zeros are real and positive.**Example Solution:**

Use the Factor Theorem to write a function, starting from the number of zeros.

By the Fundamental Theorem of Algebra, a polynomial function of degree $4$ has exactly $4$ zeros. The values of the zeros of the polynomial can be determined beforehand and be used to construct the polynomial. For simplicity, let the zeros be $1,$ $2,$ $3,$ and $4.$ By the Factor Theorem, the polynomial function $p(x)$ can be written as follows.
The third binomial can be multiplied now.
Finally, the fourth binomial can be multiplied to complete the function $p(x)$ in standard form.
To display the results, the function can be sketched using some information known about the function. Finding the $y-$intercept of the function is done by substituting $0$ for $x$ on the function.
The solution points and the $y-$intercept can be drawn on the graph.

$p(x)=(x−1)(x−2)(x−3)(x−4) $

The binomials of the function can be expanded to write the function in standard form. This can be done in steps, starting with the first two binomials.
$p(x)=(x−1)(x−2)(x−3)(x−4)$

$p(x)=(x_{2}−3x+2)(x−3)(x−4)$

$p(x)=(x_{2}−3x+2)(x−3)(x−4)$

Multiply by $(x−3)$

Distr

Distribute $(x−3)$

$p(x)=(x_{2}(x−3)−3x(x−3)+2(x−3))(x−4)$

Distr

Distribute $x_{2},-3x,&2$

$p(x)=(x_{3}−3x_{2}−3x_{2}+9x+2x−6)(x−4)$

AddSubTerms

Add and subtract terms

$p(x)=(x_{3}−6x_{2}+11x−6)(x−4)$

$p(x)=(x_{3}−6x_{2}+11x−6)(x−4)$

Multiply by $(x−4)$

Distr

Distribute $(x−4)$

$p(x)=x_{3}(x−4)−6x_{2}(x−4)+11x(x−4)−6(x−4)$

Distr

Distribute $x_{3},-6x_{2},11x,&-6$

$p(x)=x_{4}−4x_{3}−6x_{3}+24x_{2}+11x_{2}−44x−6x+24$

AddSubTerms

Add and subtract terms

$p(x)=x_{4}−10x_{3}+35x_{2}−50x+24$

$p(x)=x_{4}−10x_{3}+35x_{2}−50x+24$

Substitute

$x=0$

$p(0)=(0)_{4}−10(0)_{3}+35(0)_{2}−50(0)+24$

ZeroPropMult

Zero Property of Multiplication

$p(0)=24$

The graph of a polynomial function with no double zeros always moves in the same direction before and after the solution points. Since every solution point is already known and the polynomial has an even degree, the graph moves upwards as $x<1$ towards the $y-$axis.

Each solution point indicates a point where the graph crosses the $x-$axis. Therefore, the graph changes its vertical direction between the solution points in order to cross the axis.

Finally, since the polynomial has an even degree, the graph moves upwards as $x>4.$

Remember that this is only one example solution and that other solutions can also be correct.

Tearrik and Jordan have run into a difficult problem on their math homework.

Their teacher gave them the graph of a polynomial function of degree $5.$

Help Tearrik and Jordan to solve the following exercises.

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b Write the possible number of sign changes in the coefficients of the graphed function, supposing it is written in standard form.

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a Recall the Fundamental Theorem of Algebra.

b Use Descartes' Rule of Signs.

a The real zeros of a function correspond to the $x-$intercepts of the graph.

From the diagram, the following zeros can be identified.

- There are two positive real zeros.
- There is one negative real zero.

Next, the Fundamental Theorem of Algebra states that a polynomial function of degree $n$ has $n$ zeros. Since only $3$ real zeros are identified by looking at the graph and the given polynomial function is of degree $5,$ there must be $5−3=2$ additional complex zeros. Finally, all the zeros can be summarized in a table.

Zero Type | Number |
---|---|

Positive Real | $2$ |

Negative Real | $1$ |

Complex | $2$ |

b Descartes' Rule of Signs can be used to describe the possible sign changes of the function. This rule is presented below. Suppose that the given polynomial function is called $p(x).$

- The number of positive real zeros is equal to the number of sign changes in $p(x),$ or less than that by an even number.
- The number of negative real zeros is equal to the number of sign changes in $p(-x),$ or less than that by an even number.

$2,4,6,8,… $

However, by the Fundamental Theorem of Algebra, the function cannot have more than $5$ zeros. This means that the only possible number of sign changes for $p(x)$ are $2$ and $4.$
At the beginning of the lesson, the following polynomial equation was given.

There is something else that can be said about this equation. Notice that every term has a common factor of $x_{2}.$

$2x_{5}−7x_{4}+6x_{3}−12x_{2}=0 $

Now it is possible to determine how many solutions the equation has, how many of them are real, and how many of them are complex. The polynomial on the left-hand side is written in standard form and has $3$ sign changes.
Since the polynomial function equals zero, the zeros of the function are the solutions to the equation. Since there are $3$ sign changes, by Descartes' Rule of Signs, the equation has $3$ or $1$ possible positive real solutions. The number of potential negative real solutions can be found by replacing $x$ with $-x.$
$2(-x)_{5}−7(-x)_{4}+6(-x)_{3}−12(-x)_{2}=0⇓-2x_{5}−7x_{4}−6x_{3}−12x_{2}=0 $

Since the polynomial on the left-hand side has no sign changes, the given equation has no real negative solutions. By the Fundamental Theorem of Algebra, a polynomial of degree $5$ has $5$ total solutions, since repeating solutions are considered different solutions. With this information, it is possible to make a table for the different combinations of types of solutions. Positive Real | Negative Real | Complex | Total |
---|---|---|---|

$3$ | $0$ | $2$ | $5$ |

$1$ | $0$ | $4$ | $5$ |

$2x_{5}−7x_{4}+6x_{3}−12x_{2}=0⇓x_{2}(2x_{3}−7x_{2}+6x−12)=0 $

This indicates that $x=0$ is a solution to the equation. Also, since the term is squared, $x=0$ is repeated twice in the equation. An interesting observation now is that the expression in parentheses still has $3$ sign changes.
Another table for the possible real and complex solutions of this new equation can be written.

Positive Real | Negative Real | Complex | Total |
---|---|---|---|

$3$ | $0$ | $0$ | $3$ |

$1$ | $0$ | $2$ | $3$ |

The information found in this exercise can be used to expand Descartes' Rule of Signs.

- The $x=0$ solutions were not identified by the sign changes.
- These solutions can exist if every term of the function is multiplied by a power of $x.$