3. Polynomials

Lesson

If we start with the most basic form of a polynomial, $y=ax^n$`y`=`a``x``n`, we can immediately see a pattern emerge for odd and even powers. Experiment with the following applet, where $a=1$`a`=1.

Odd degree polynomials move in opposite directions at the extremities and even degree polynomials move in the same direction at the extremities! This is because the degree of a polynomial, $n$`n`, together with the leading coefficient, $a$`a`, dictate the overall shape and behavior of the function at the extremities:

$n$n |
$a>0$a>0 |
$a<0$a<0 |
---|---|---|

Even |
||

Odd |

But what happens between the extremities?

A polynomial of degree $n$`n` can have up to $n$`n` $x$`x`-intercepts, with those of odd degree having at least one.

A polynomial of degree $n$`n` can have up to $n-1$`n`−1 turning points, with those of even degree having at least one.

Details of key features can be found using a table of values and technology or from the factored forms of the function. Let's focus on what we can learn from factored forms of polynomials.

Recall the zero product property from when we solved quadratic equations previously:

Zero product property

If $ab=0$`a``b`=0, then $a=0$`a`=0 or$b=0$`b`=0

For example:

$\left(x+3\right)\left(x-4\right)=0$(`x`+3)(`x`−4)=0

$x+3=0$`x`+3=0 or $x-4=0$`x`−4=0

$x=-3$`x`=−3 or $x=4$`x`=4

We will extend this concept to polynomials of any degree, so if we can fully factor a polynomial, then we can quickly find the $x$`x`-intercepts, solutions or roots.

Certain polynomials of degree $n$`n` can be factored into up to $n$`n` linear factors over the real number field. For example the $4$4th degree polynomial $P\left(x\right)=2x^4-x^3-17x^2+16x+12$`P`(`x`)=2`x`4−`x`3−17`x`2+16`x`+12 can be expressed as $P\left(x\right)=\left(x+3\right)\left(2x+1\right)\left(x-2\right)^2$`P`(`x`)=(`x`+3)(2`x`+1)(`x`−2)2. Note that there are two distinct factors, the $\left(x+3\right)$(`x`+3) and $\left(2x+1\right)$(2`x`+1) and two equal factors, the $\left(x-2\right)$(`x`−2) appears twice. But we can immediately identify that there are roots at $x=-3,-\frac{1}{2}$`x`=−3,−12 and $x=2$`x`=2.

A root is said to have multiplicity $k$`k` if a linear factor occurs $k$`k` times in the polynomial function. For example the function of degree $6$6 given by $y=\left(x-1\right)\left(x+1\right)^2\left(x-2\right)^3$`y`=(`x`−1)(`x`+1)2(`x`−2)3 is said to have a root $x=1$`x`=1 of multiplicity $1$1 (a single root), another root of $x=-1$`x`=−1 of multiplicity $2$2 (a double root or two equal roots) and a root $x=2$`x`=2 of multiplicity $3$3 (a triple root or three equal roots). In effect that is $6$6 roots altogether.

The key to understanding how a root's multiplicity effects the graph is given in the following statement:

**The curve's shape near a root depends on the root's multiplicity****. **

Look closely at the sketch of $y=\left(x-1\right)\left(x+1\right)^2\left(x-2\right)^3$`y`=(`x`−1)(`x`+1)2(`x`−2)3. What do you notice about the different roots?

**Think:** The curve approaching a root of multiplicity $1$1 will behave like a linear function across that root (see position B on graph). The curve approaching a root of multiplicity $2$2 will behave like a quadratic function across that root (see position A on graph). The curve approaching a root of multiplicity $3$3 will behave like a cubic function across that root (see position C on graph).

Did you know?

In general, the curve approaching a root of multiplicity $k$`k` will behave like the function $y=x^k$`y`=`x``k` across that root.

There are also two other considerations that will help with the graphing of the function.

- Firstly, note that this function is an even degree function, so at the left and right extremes it will move away from the $x$
`x`- axis in the same direction. - Secondly, because the coefficient of the greatest power of the function is positive, the curve moves away from the $x$
`x`- axis in the positive direction.

Finally, always check the position of the $y$`y`- intercept. For example, in the above sketch, at $x=0$`x`=0, $y=\left(0-1\right)\left(0+1\right)^2\left(0-2\right)^3=8$`y`=(0−1)(0+1)2(0−2)3=8 and this confirms the direction of the curve downward toward the first positive root.

We will eventually have more mathematical tools to determine the positions of local turning points. But for now, the important principles here allow us to understand the function's basic shape.

Describe the multiplicity of the roots of $y=-\left(x-1\right)\left(x+2\right)^2$`y`=−(`x`−1)(`x`+2)2 and then sketch the function.

**Think:** This polynomial has a root of multiplicity $1$1 at $x=1$`x`=1 and a root of multiplicity $2$2 ($2$2 equal roots) at $x=-2$`x`=−2. The $y$`y` - intercept is $-\left(-1\right)\left(2\right)^2=4$−(−1)(2)2=4. The function is of odd degree, so its ends move off in different directions. The leading coefficient will be negative, so for large positive values of $x$`x`, the curve becomes very negative (overall shape of down to the right).

**Do: **

Which of the following is the graph of the function $f\left(x\right)=x\left(x+3\right)\left(x-3\right)$`f`(`x`)=`x`(`x`+3)(`x`−3)?

- Loading Graph...ALoading Graph...BLoading Graph...C