Polynomial Functions

We have already discussed a particualar polynomial, the quadratic.In this journey, we are going to extend our knowledge to other types of polynomials.

The purpose here is to know how to manipulate the polynomials in our daily tasks.

In this site we will have more practice than theory.We have already seen that a monomial can be a number, a variable or a product of both.

A polynomial, on the other hand, is a monomial or a combination of monomials.An example is:

$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_{2}x^{2}+a_{2}x+a_{0}$

In this function $a_{n}\neq0$ , is the leading coefficient and $n$ is aclled the $degree$ .We also note that $a_{0}$ is the constant.
The exponents are whole numbers while the coefficients are real numbers.

Polynomial:

$\pi x^{3}-6x^{2}-\sqrt{5}$

Exponents are whole numbers and coefficients are real numbers.

Not a polynomial:

$3^{x}-3x^{2}+9$
Some exponents are not whole numbers.

Operations on Polynomials:

Direct substitution:

Example:Evaluate: $f(x)=3x^{5}-4x^{2}+8$ when $x=2$

We get: $f(2)=3\times 2^{5}-4\times 2^{2}+8$

(1) $\begin{equation*} \begin{split} f(2)&=3\times 2^{5}-4\times 2^{2}+8\\ &=3\times 32-4\times 4+8\\&=96-16+8\\ &=88 \end{split} \end{equation*}$

Adding or Subtracting Polynomials:

To add or subtract polynomials, we just add or subtract the coefficients of like terms.
Add: $2x^{3}+x^{2}+3x+2$ and $x^{6}+x^{4}+3x^{2}+2$
We add like terms:

$x^{6}+x^{4}+2x^{3}+4x^{2}+3x+2$

Multiplying polynomials

To multiply polynomials, we multiply each term of the first polynomial by each term of the other polynomial.

This reminds us of the identities seen in another course in this site.

(2) $\begin{equation*} \begin{split} (2x+5)(3x^{2}-5x+2)&=(2x)(3x^{2})+(2x)(-5x)+(2x)(2)+5(3x^{2})+5(-5x)+5(2)\\ &=6x^{3}-10x^{2}+4x+15x^{2}-25x+10\\ &=6x^{3}+(-10+15)x^{2}+(4-25)x+10\\ &=6x^{3}+5x^{2}-21x+10 \end{split} \end{equation*}$

Factoring a polynomial:

A polynomial is factored completely if it is written as a product of unfactorable polynomials with integer coefficients.
Example:

Factor $x^{3}+2x^{2}-15x$

(3) $\begin{equation*} \begin{split} x^{3}+2x^{2}-15x &=x(x^{2}+2x-15)\\ &=x(x^{2}+5x-3x-15)\\ &=x[x(x+5)-3(x+5)\\ &=x(x-3)(x+5) \end{split} \end{equation*}$

We just used factoring by grouping.

Another example:
Factor: $x^{3}-3x^{2}-16x+48$

(4) $\begin{equation*} \begin{split} x^{3}-3x^{2}-16x+48&=x^{2}(x-3)-16(x-3)\\ &=(x^{2}-16)(x-3)\\ &=(x+4)(x-4)(x-3) \end{split} \end{equation*}$

Power function:

The power function has the form: $f(x)=ax^{n}$

In this function, the coefficient is a real number $a\neq 0$ , and $n>0$ is an integer.

The power function has some properties worth noting.

If $n>0$ and $n$ is even we have:-The function is said to be $even$ . This means the graph is symmetric with respect to the y-axis.-The domain is the set of all real numbers with a range being the set of nonnegative numbers.

-This graph contains the main points $(0,0)$ , $(-1,1)$ and $(1,1)$
If $n>0$ and $n$ is odd we have:-The function is said to be $odd$ .

This means the graph is symmetric with respect to the origin.-The domain and range are all the set of all real numbers.

-This graph contains the main points $(0,0)$ , $(-1,1)$ and $(1,1)$
We will cover later the variation of functions.

Solving polynomials, finding zeros

REAL ZEROS:

A polynomial of degree $n$ cannot have more a number of real zeros more than $n$

Example: $f(x)=x^{4}-3x^{3}+x^{2}-8$

cannot have more than 4 real zeros.

Rational zeros theorem:

For:

$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ if $a_{n}\neq 0$ and $a_{0}\neq 0$

with all integer coefficients, when $\frac{p}{q}$ is a rational zero of $f$ then $p$ is a factor of $a_{0}$ and $q$ is a factor of $a_{n}$ .

Example:

$f(x)=2x^{4}-7x^{3}+x-8$

We have $a_{0}=-8$ the $p$ factors are $\pm 1$ , $\pm 2$ , $\pm 4$ , $\pm 8$

$a_{4}=2$ the $q$ factors are $\pm 1$ , $\pm 2$

We check all the possible $\frac{p}{q}$

$\pm 1$ , $\pm \frac{1}{2}$ , $\pm 2$ , $\pm 4$ , $\pm 8$

THEOREM: A polynomial function of odd degree that has real coefficients has at least one real zero.

THEOREM: For a polynomial function $f$ . If $a<b$ and if $f(a)$ and $f(b)$ are of opposite sign, then there is at least one real zero of $f$ between $a$ and $b$ .

Complex Zeros of a polynomial:A complex polynomial function $f$ of degree $n\geq 0$ has at least one complex coefficient.
A polynomial function $f$ of odd degree with real coefficients has at least one real zero.

Third degree or cubic function solutions: Girolamo Cardano formula

We are going to study the third degree equation and find an easy way to find the solutions.

The Cubic formula we are going to use will have variations, according to the situation.

Trigonometric Solutions:

Given:

$ax^{3}+bx^{2}+cx+d=0$

Let’s depress the equation.

$x=y-\frac{b}{3a}$

We apply the substitution

$a(y-\frac{b}{3a})^{3}+b(y-\frac{b}{3a})^{2}+c(y-\frac{b}{3a})+d=0$

$ay^{3}+(c-\frac{b^{2}}{3a})y+(d+\frac{2b^{3}}{27a^{2}}-\frac{bc}{3a})=0$

We get the deprecated function.

$y^{3}+py+q=0$

Please note that when $q=0$ , we can simply deduct the 3 values of $x$ from those of $y$ .

First case: when $p<0$ and $27q^{2}+4p^{3}<0$

We proceed as follows:

$y=\sqrt{-\frac{4p}{3}}\cos \theta$

When we replace $z$ in our equation:

$\left(\sqrt{-\frac{4p}{3}}\right)^{3}\cos^{3} \theta+p\sqrt{-\frac{4p}{3}}\cos \theta+q=0$

$\sqrt{16\frac{p^{2}}{9}(-\frac{4p}{3}})\cos^{3} \theta+p\sqrt{-\frac{4p}{3}}\cos \theta=-q$

$4\sqrt{-\frac{4p^{3}}{27}}\cos^{3} \theta+\sqrt{-\frac{4p^3}{3}}\cos \theta=-q$

$4\sqrt{-\frac{4p^{3}}{27}}\cos^{3} \theta+3\sqrt{-\frac{4p^2}{27}}\cos \theta=-q$

By simple division we get:

$4\cos^{3} \theta-3\cos \theta=-q\sqrt{-\frac{27}{4p^{3}}}$

Let’s simplify the left side:

(5) $\begin{equation*} \begin{split} 4\cos^{3} \theta-3\cos \theta &=\cos^{3} \theta+3\cos^{3} \theta-3\cos \theta\\ &=\cos^{2} \theta \cos \theta+3\cos \theta(\cos^{2} \theta-1)\\ &=\cos^{2} \theta \cos \theta-3\cos \theta(1-\cos^{2} \theta)\\ &=\cos^{2} \theta \cos \theta-3\cos \theta \sin^{2} \theta\\ &=\cos \theta (\cos^{2} \theta-3\sin^{2} \theta)\\ &=\cos \theta (\cos^{2} \theta- \sin^{2} \theta-2\sin^{2} \theta)\\ &=\cos \theta (\cos (2\theta)-2\sin^{2} \theta)\\ &=\cos (2\theta) \cos \theta-2\sin^{2} \theta \cos \theta\\ &=\cos (2\theta) \cos \theta-2\sin \theta \cos \theta \sin \theta\\ &=\cos (2\theta) \cos \theta-\sin (2\theta) \sin \theta\\ &=\cos (2\theta +\theta)\\ &=\cos (3\theta) \end{split} \end{equation*}$

The simplified Equation is:

$\cos (3\theta)=-q\sqrt{-\frac{27}{4p^{3}}}$

We can now find $theta$ , $y$ and $x$ .

Second case: when $p<0$ and $27q^{2}+4p^{3}>0$

Here we say:

$y=\sqrt{-\frac{4p}{3}}\cosh \theta$

The same transformations yield:

$4\cosh^{3} \theta-3\cosh \theta=-q\sqrt{-\frac{27}{4p^{3}}}$

And:

$\cosh (3\theta)=-q\sqrt{-\frac{27}{4p^{3}}}$

Third case: when $p>0$

Here we say:

$y=\sqrt{-\frac{4p}{3}}\sinh \theta$

The same transformations yield:

$4\sinh^{3} \theta+3\sinh \theta=-q\sqrt{-\frac{27}{4p^{3}}}$

And:

$\sinh (3\theta)=-q\sqrt{-\frac{27}{4p^{3}}}$

Trigonometric method for complex conjugates:

From

$y^{3}+py+q=0$

With:

$x=y-\frac{b}{3a}$

First case: when $p<0$

Let’s calculate $\theta$ and $\phi$

$\sin \theta=-\frac{2p}{3q}\sqrt{-\frac{p}{3}}$

Now:

$\tan \phi=\sqrt[3]{\tan (\frac{\theta}{2})}$

The $y$ values:

$\begin{cases}y1=-2\sqrt{-\frac{p}{3}}\frac{1}{\sin (2\phi)}\\y2=\sqrt{-\frac{p}{3}}\frac{1}{\sin (2\phi)}-i\frac{\sqrt{-p}}{\tan (2\phi)}\\y3=\sqrt{-\frac{p}{3}}\frac{1}{\sin (2\phi)}+i\frac{\sqrt{-p}}{\tan (2\phi)}\end{cases}$

For $x$ values we just apply $x=y-\frac{b}{3a}$

Second case: when $p>0$

Let’s calculate $\theta$ and $\phi$

$\tan \theta=\frac{2p}{3q}\sqrt{\frac{p}{3}}$

Now:

$\tan \phi=\sqrt[3]{\tan (\frac{\theta}{2})}$

The $y$ values:

$\begin{cases}y1=-2\sqrt{\frac{p}{3}}\frac{1}{\tan (2\phi)}\\y2=\sqrt{\frac{p}{3}}\frac{1}{\tan (2\phi)}-i\frac{\sqrt{p}}{\sin (2\phi)}\\y3=\sqrt{\frac{p}{3}}\frac{1}{\tan (2\phi)}+i\frac{\sqrt{p}}{\sin (2\phi)}\end{cases}$

For $x$ values we just apply $x=y-\frac{b}{3a}$

Other methods:

In this case we consider the same steps but with more details

$x^{3}+ax^{2}+bx+c=0$

The same steps above will be used

We get:

$x^{3}+(b-\frac{a^{2}}{3})x+\frac{a}{27}(2a^{2}-9b)+c=0$

This is of the form:

$x^{3}+px+q=0$

So we have to calculata $p$ and $q$

$p=b-\frac{a^{2}}{3}$

and

$q=\frac{a}{27}(2a^{2}-9b)+c$

Next we calculate $u$ and $v$

$u=\sqrt[3]{-\frac{q}{2}+\sqrt{(\frac{q}{2})^{2}+(\frac{p}{3})^{3}}}$

$v=\sqrt[3]{-\frac{q}{2}-\sqrt{(\frac{q}{2})^{2}+(\frac{p}{3})^{3}}}$

We can now deduct the 3 roots:

$x_{1}=u+v-\frac{a}{3}$

$x_{2}=(-\frac{u}{2}-\frac{v}{2}-\frac{a}{3})+ \frac{1}{2}i\sqrt{3}(u-v)$

$x_{3}=(-\frac{u}{2}-\frac{v}{2}-\frac{a}{3})+ \frac{1}{2}i\sqrt{3}(u-v)$

Let D be:

$D=4p^{3}+27q^2$

If $D>0$ , the equation has one real root and 2 complex roots

If $D=0$ , the equation has one root $\frac{3q}{p}$ and a double root $\frac{-3q}{p}$

If $D<0$ , the equation has three real roots

Worked examples:

1.Find the solutions of the equation: $4x^{5}-40x^{3}+36x=0$

Solution

Factoring by $4x$

We get: $4x(x^{4}-10x^{2}+9)=0$

Now let’s factor the quantity: $x^{4}-10x^{2}+9$

(6) $\begin{equation*} \begin{split} x^{4}-10x^{2}+9&=x^{4}-x^{2}-9x^{2}+9\\ &=x^{2}(x^{2}-1)-9(x^{2}-1)\\ &=(x^{2}-9)(x^{2}-1)\\ &=(x+3)(x-3)(x+1)(x-1) \end{split} \end{equation*}$

Now put together: $4x(x^{4}-10x^{2}+9)=0$

Means: $4x((x^{4}-10x^{2}+9)=0$ $4x(x+3)(x-3)(x+1)(x-1)=0$

Each of the factors can verify the relationships.

$4x=0 \Leftrightarrow x=0$

$x+3=0 \Leftrightarrow x=-3$

$x-3=0 \Leftrightarrow x=3$

$x+1=0 \Leftrightarrow x=-1$

$x-1=0 \Leftrightarrow x=1$

We have now the five solutions:

Answer: $\{-3,-1,0,1,+3\}$

Signs of zeros of a polynomial

Descartes’ Rule of signs theorem

If $f$ is a standard polynomial function:

-The number of positive real zeros of $f$ is either equals the number of variations in the sign of the non-zero coefficients of $f(x)$ or else that number less than an even integer.

-The number of negative real zeros of $f$ is either equals the number of variations in the sign of the non-zero coefficients of $f(-x)$ or else that number less than an even integer.

Example:

$f(x)=x^{5}-5x^{4}+12x^{3}-24x^{2}+32x-16$

-At least 5, 3 or 1 positive zeros.(5 changes in the sign)

$f(x)=-x^{5}-5x^{4}-12x^{3}-24x^{2}-32x-16$

-No sign variation means no negative root.

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Polynomial Functions