<< Chapter < Page | Chapter >> Page > |
A new bakery offers decorated sheet cakes for children’s birthday parties and other special occasions. The bakery wants the volume of a small cake to be 351 cubic inches. The cake is in the shape of a rectangular solid. They want the length of the cake to be four inches longer than the width of the cake and the height of the cake to be one-third of the width. What should the dimensions of the cake pan be?
This problem can be solved by writing a cubic function and solving a cubic equation for the volume of the cake. In this section, we will discuss a variety of tools for writing polynomial functions and solving polynomial equations.
In the last section, we learned how to divide polynomials. We can now use polynomial division to evaluate polynomials using the Remainder Theorem . If the polynomial is divided by $\text{\hspace{0.17em}}x\u2013k,\text{\hspace{0.17em}}$ the remainder may be found quickly by evaluating the polynomial function at $\text{\hspace{0.17em}}k,\text{\hspace{0.17em}}$ that is, $\text{\hspace{0.17em}}f\left(k\right)\text{\hspace{0.17em}}$ Let’s walk through the proof of the theorem.
Recall that the Division Algorithm states that, given a polynomial dividend $\text{\hspace{0.17em}}f(x)\text{\hspace{0.17em}}$ and a non-zero polynomial divisor $\text{\hspace{0.17em}}d(x)\text{\hspace{0.17em}}$ where the degree of $\text{\hspace{0.17em}}\text{\hspace{0.17em}}d(x)\text{\hspace{0.17em}}$ is less than or equal to the degree of $\text{\hspace{0.17em}}f(x),\text{\hspace{0.17em}}$ there exist unique polynomials $\text{\hspace{0.17em}}q(x)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}r(x)\text{\hspace{0.17em}}$ such that
If the divisor, $\text{\hspace{0.17em}}d(x),\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}x-k,\text{\hspace{0.17em}}$ this takes the form
Since the divisor $\text{\hspace{0.17em}}x-k\text{\hspace{0.17em}}$ is linear, the remainder will be a constant, $\text{\hspace{0.17em}}r.\text{\hspace{0.17em}}$ And, if we evaluate this for $\text{\hspace{0.17em}}x=k,\text{\hspace{0.17em}}$ we have
In other words, $\text{\hspace{0.17em}}f(k)\text{\hspace{0.17em}}$ is the remainder obtained by dividing $\text{\hspace{0.17em}}f(x)\text{\hspace{0.17em}}$ by $\text{\hspace{0.17em}}x-k.\text{\hspace{0.17em}}$
If a polynomial $\text{\hspace{0.17em}}f(x)\text{\hspace{0.17em}}$ is divided by $\text{\hspace{0.17em}}x-k,\text{\hspace{0.17em}}$ then the remainder is the value $\text{\hspace{0.17em}}f(k).\text{\hspace{0.17em}}$
Given a polynomial function $\text{\hspace{0.17em}}f,$ evaluate $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}x=k\text{\hspace{0.17em}}$ using the Remainder Theorem.
Use the Remainder Theorem to evaluate $\text{\hspace{0.17em}}f(x)=6{x}^{4}-{x}^{3}-15{x}^{2}+2x-7\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}x=2.\text{\hspace{0.17em}}$
To find the remainder using the Remainder Theorem, use synthetic division to divide the polynomial by $\text{\hspace{0.17em}}x-2.\text{\hspace{0.17em}}$
The remainder is 25. Therefore, $\text{\hspace{0.17em}}f(2)=25.\text{\hspace{0.17em}}$
Use the Remainder Theorem to evaluate $\text{\hspace{0.17em}}f(x)=2{x}^{5}-3{x}^{4}-9{x}^{3}+8{x}^{2}+2\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}x=-3.\text{\hspace{0.17em}}$
$\text{\hspace{0.17em}}f(-3)=-412\text{\hspace{0.17em}}$
The Factor Theorem is another theorem that helps us analyze polynomial equations. It tells us how the zeros of a polynomial are related to the factors. Recall that the Division Algorithm tells us
If $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ is a zero, then the remainder $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}f(k)=0\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}f(x)=(x-k)q(x)+0\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}f(x)=(x-k)q(x).\text{\hspace{0.17em}}$
Notice, written in this form, $\text{\hspace{0.17em}}x-k\text{\hspace{0.17em}}$ is a factor of $\text{\hspace{0.17em}}f(x).\text{\hspace{0.17em}}$ We can conclude if $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ is a zero of $\text{\hspace{0.17em}}f(x),\text{\hspace{0.17em}}$ then $\text{\hspace{0.17em}}x-k\text{\hspace{0.17em}}$ is a factor of $f(x).\text{\hspace{0.17em}}$
Notification Switch
Would you like to follow the 'Precalculus' conversation and receive update notifications?