1.04 Polynomial equations

Power functions

Functions with a form like $y=x^2$y=x2, $y=5x^3$y=5x3 and $y=ax^n$y=axn are known as power functions.

The following applet lets you see the general shape of power functions where $a=1$a=1 and the powers are positive integers.

• What happens at the extremities of the graph? Is it different for odd and even powers? 
• Also look carefully at the shape of the curves as they come close to the origin.
• Do you notice anything else about the curves?

Created with Geogebra

 

Odd degree power functions all have the following properties:

• The graph has $180$180$^\circ$° rotational symmetry about the origin, that is $f(-x)=-f(x)$f(−x)=−f(x) for all $x$x, 
• The graph moves in opposite directions at the extremities
• They all pass through $\left(0,0\right)$(0,0), $\left(-1,-1\right)$(−1,−1) and $\left(1,1\right)$(1,1)
• Domain and range are both $\left(-\infty,\infty\right)$(−∞,∞)
• As $n$n increases, within the interval $\left[-1,1\right]$[−1,1] the graphs lie closer to the $x$x-axis, 'flatter' about the point of inflection
• As $n$n increases and $x$x extends beyond the interval $\left[-1,1\right]$[−1,1], the functions values grow further away from the $x$x-axis than those of lower powers    

Even degree power functions all have the following properties:

• The graph is symmetric about the y-axis, that is $f(-x)=f(x)$f(−x)=f(x) for all $x$x 
• The graph moves in same directions at the extremities
• They all pass through $\left(0,0\right)$(0,0), $\left(-1,1\right)$(−1,1) and $\left(1,1\right)$(1,1)
• Domain is and $\left(-\infty,\infty\right)$(−∞,∞)and the range is$\left[0,\infty\right)$[0,∞)
• As $n$n increases, within the interval $\left[-1,1\right]$[−1,1] the graphs lie closer to the $x$x-axis, 'flatter' about the turning point
• As $n$n increases and $x$x extends beyond the interval $\left[-1,1\right]$[−1,1], the functions values grow further away from the $x$x-axis than those of lower powers        

 

Polynomials of degree n

Let's look at polynomials more generally, that is, functions of the form :

 $P(x)=ax^n+bx^{n-1}+cx^{n-2}+...+d$P(x)=axn+bxn−1+cxn−2+...+d, where the powers are positive integers.

Recall the degree of the polynomial is the highest power of $x$x in the expression. The degree of the polynomial, $n$n, together with the leading coefficient, $a$a, dictate the overall shape and behaviour 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 calculus and technology, or from factored forms of the function.  Let's focus on what can be learned from the factored forms of polynomials.

 

Factored forms of polynomials

Certain polynomials of degree $n$n can be factorised into $n$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)=2x4−x3−17x2+16x+12 can be re-expressed as $P\left(x\right)=\left(x+3\right)\left(2x+1\right)\left(x-2\right)^2$P(x)=(x+3)(2x+1)(x−2)2. Note that there are two distinct factors, the $\left(x+3\right)$(x+3) and $\left(2x+1\right)$(2x+1) and two equal factors, the $\left(x-2\right)$(x−2) appears twice.

A polynomial function $y=P\left(x\right)$y=P(x) that can be completely factorised into its linear factors can be roughly sketched using the axes intercepts and general shape. The function $y=\left(x+3\right)\left(2x+1\right)\left(x-2\right)^2$y=(x+3)(2x+1)(x−2)2 has roots immediately identifiable as $x=-3,-\frac{1}{2}$x=−3,−12​ and $x=2$x=2.

 

Multiplicity

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 affects the graph is given in the following statement:

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

Worked examples

Example 1

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

 

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).

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

There are a few 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 highest power of the function is positive, the curve moves away from the $x$x- axis in the positive direction.

Thirdly, 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.

Although more mathematical tools are needed to determine the positions of local turning points, the important principles here allow us to understand the function's basic shape.

Example 2

The function $y=-\left(x-1\right)\left(x+2\right)^2$y=−(x−1)(x+2)2 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. Note that for large positive values of $x$x, the curve becomes very negative (See graph below).

 

EXAMPLE 3

The function $y=\frac{1}{2}\left(x-3\right)^2\left(x+1\right)^3$y=12​(x−3)2(x+1)3 is of degree $5$5, with a double root at $x=3$x=3 and a triple root at $x=-1$x=−1. The y intercept is $4.5$4.5. The coefficient $\frac{1}{2}$12​ at the front simply halves the size of every $y$y value, but note that there is no change at all in the position of the roots because half of zero is still zero (see graph below).

Practice questions

Question 1

Question 2

 

Finding factored form

Since factored form tells us about key features in a graph and allows us to roughly sketch a graph, let's practice finding the factored form of polynomials of higher degree. From factoring quadratics and cubic equations, recall our toolbox of strategies:

• Look for the highest common factor
• Looking for special patterns, such as difference of two squares, perfect squares, sum or difference of two cubics
• Factorise by grouping in pairs
• Factorise by completing the square
• Given one linear factor, find other factors by polynomial division

And if we are not given a linear factor, try to find one using the remainder theorem as with cubics.

Recall the following guess and check strategy:

• Guess a root of the expression, say $a$a
• Try evaluating the expression for this value, that is find $P(a)$P(a)
• If $P(a)=0$P(a)=0, we have found a factor $\left(x-a\right)$(x−a)

Good candidates for guesses for a root of any polynomial are factors of the constant term divided by factors of the leading coefficient, $a$a. This comes from the rational root theorem.

 

Worked example

Example 4

A polynomial $g(x)=x^2+kx-28$g(x)=x2+kx−28 has a zero at $x=4$x=4. Find the other zero.

Think: If $x=4$x=4 is a zero of the polynomial $g(x)$g(x), when substituting $x=4$x=4 into the polynomial the answer will be zero. Let's see what happens when we do that here.

$g(x)$g(x)	$=$=	$x^2+kx-28$x2+kx−28
$g(4)$g(4)	$=$=	$4^2+k\times4-28$42+k×4−28
	$=$=	$16+4k-28$16+4k−28
	$=$=	$-12+4k$−12+4k

At this stage we can solve for $k$k, because we know that $x=4$x=4 is a zero and thus $g(4)=0$g(4)=0.

Therefore $-12+4k=0$−12+4k=0 . Hence, $k=3$k=3.

We can now express the polynomial as $g(x)=x^2+3x-28$g(x)=x2+3x−28, but we are yet to solve for the second zero.

How can we do that? We know that to solve a quadratic we can factorise fully. Once factorised, each of the linear factors present us with a solution. In this case, we already know that $x=4$x=4 is a solution, this means that $x-4$x−4 is one of the linear factors.

Factorising $g(x)$g(x), we get $g(x)=(x-4)(x+7)$g(x)=(x−4)(x+7), which means that the other zero is at $x+7=0$x+7=0, which is $x=-7$x=−7.

We have now found the value of $k$k to get the complete polynomial, and then were able to factorise and solve to find the other zero.

 

Practice questions

Question 3

Question 4