Factored Form and Factor Theorem

Given the polynomial in factored form, it is possible to calculate x intercepts, y intercept and degree. With that information we can draw a reasonably accurate sketch of the graph.

Curve Sketching using the Factored Form

Consider the polynomial p(x)=(x+5)(x+1)(x-1)^2.

The x intercepts occur when y=0. We solve 0=(x+5)(x+1)(x-1)^2. There are three distinct solutions, x=-5, x=-1 and x=1. Note that x=1 is a double solution.

The y intercept occurs when x=0. We calculate p(0)=(0+5)(0+1)(0-1)^2=(5)(1)(-1)^2=5.

Plotting the intercepts:

Multiplying the x term in each bracket, we see that the degree of the polynomial is 4, and the leading coefficient is +1. The first and last terms of the expansion of the regular factored form give us degree and y-intercept:

    \[p(x)=(x+5)(x+1)(x-1)(x-1)=x^4 + \dots + 5\]

Using the intercepts we can sit a ‘positive quartic’ shape onto the points calculated, bearing in mind the various forms a positive quartic (anywhere between a symmetric ‘w’ to asymmetric ‘w’ to simple ‘u’ shape) can take. Note that there is a ‘double zero’ at x=1. Also note that for the curve to be smooth, larger gaps between zeros produce larger ‘valleys’ or ‘mountains’. You may wish to ‘spot check’ a value of x in a large gap to get an idea of how far to dip.

 

The Factor Theorem

The factor theorem allows us to find the factors of a polynomial if the factors are  integers.

Given polynomial p(x), if p(c)=0 then (x-c) is a factor.

Remember that the final term in an expansion of factors is the product of all the zeros (\pm). Therefore, we need only consider factors of the constant as possible zeros of the polynomial.

Example

    \[p(x)=x^4-x^3-7x^2+x+6\]

We need only consider the factors of 6 which are \pm1, \pm2,\pm 3, \pm6.

    \[p(1)=1^4-1^3-7(1)^2+1+6=0\checkmark\]

Therefore (x-1) is a factor.

    \[p(x)=x^4-x^3-7x^2+x+6=(x-1)(???)\]

To determine what goes in the second bracket, we may use one of the following methods:

  1. polynomial long division,
  2. equating coefficients
  3. juggling method.
  4. repeat the factor theorem until all values c where p(c)=0 have been found.

Equating Coefficients

    \[p(x)=x^4-x^3-7x^2+x+6=(x-1)(???)\]

Since (x-1) is a factor of p(x), and p(x) is a polynomial of degree 4, we know that the unknown polynomial is a cubic.

    \[x^4-x^3-7x^2+x+6=(x-1)(ax^3+bx^2+cx+d)\]

Multiplying out and gathering like terms gives us:

    \begin{align*}x^4-x^3-7x^2+x+6=&(x-1)(ax^3+bx^2+cx+d)\\[10 pt]=&ax^4+bx^3+cx^2+dx-ax^3-bx^2-cx-d\\[10 pt]=&ax^4+(b-a)x^3+(c-b)x^2+(d-c)x-d\end{align}

We now equate coefficients of each term with the corresponding term in p(x)

    \begin{align*}x^4: \quad &a=1\\[10 pt]x^3: \quad &b-a=-1\quad\Rightarrow b-1=-1 \quad\Rightarrow b=0\\[10 pt] x^2: \quad &c-b=-7\quad\Rightarrow c-0 =-7 \quad\Rightarrow c=-7\\[10 pt]x: \quad &d-c=1\quad\Rightarrow d-(-7)=1 \quad\Rightarrow d+7=1 \quad\Rightarrow d=-6\end{align}

We now have

    \[x^4-x^3-7x^2+x+6=(x-1)(x^3-7x-6)\]

We begin the process again, considering the factors of -6 in the smaller polynomial q(x)=x^3-7x-6.

Try q(1):   q(1)=1^3-7(1)-6=-12. Therefore (x-1) is not a factor.

Try q(-1): q(-1)=(-1)^3-7(-1)-6=-1+7-6=0. Therefore (x+1) is a factor.

    \[x^3-7x-6=(x+1)(ax^2+bx+c)\]

.

(Here we use a, b and c again for convenience, however don’t mix these up with the a, b and c we already used.) Multiply out, gather like terms and equate coefficients gives us:

    \begin{align*}x^3-7x-6&=(x+1)(ax^2+bx+c)\\[10 pt]&=ax^3+bx^2+cx+ax^2+bx+c\\[10 pt]&=ax^3+(b+a)x^2+(c+b)x+c\\[20 pt]&x^3:\quad a=1\\[10 pt]&x^2: \quad b+a=0 \quad \Rightarrow b+1 = 0 \quad \Rightarrow b=-1\\[10 pt]&x: \quad c+b=-7 \quad \Rightarrow c-1 = -7 \quad \Rightarrow c=-6\end{align}

That is,

    \[x^3-7x-6=(x+1)(x^2-x-6)\]

and

    \[x^4-x^3-7x^2+x+6=(x-1)(x+1)(x^2-x-6)\]

We are now down to a quadratic which is straightforward to factor:

    \[x^2-x-6=(x-3)(x+2)\]

Gives us:

    \[x^4-x^3-7x^2+x+6=(x-1)(x+1)(x-3)(x+2)\]

And our factoring is done.

Juggling

Juggling is a condensed version of the ‘equate coefficients’ method. Here, we calculate one term and ‘deal with the consequences’ ( or, correct the following term).

Let’s take a new example:

    \[p(x)=x^4+8x^3+26x^2+43x+30\]

Using the factor theorem, let’s try factors of 30 until we have a value x where p(x)=0.

We find that p(-2)=0. Therefore (x+2) is a factor.

    \[x^4+8x^3+26x^2+43x+30=(x+2)(x^3 + \dots )\]

The first term in the second bracket must be x^3 in order to multiply to x^4.

Multiplying (x+2) with x^3 we have

    \begin{align*}x^4+8x^3+26x^2+43x+30&=(x+2)(x^3 +\dots)\\[10 pt]&=x^4+2x^3 +\dots\end{align}

The first term is correct, we calculated that term. However we need a total of 8x^3, which is 6 more than we have. Let’s make the next term in the bracket 6x^2 and multiply out:

    \begin{align*}x^4+8x^3+26x^2+43x+30&=(x+2)(x^3 + 6x^2+\dots)\\[10 pt]&=x^4+2x^3 +6x^3+12x^2\dots\end{align}

We now have the correct coefficient for x^4 and x^3, but we need to ‘correct’ the x^2 term. Currently, we have 12x^2 but we need 26 which is 14 more. Let’s make the next term 14x.

    \begin{align*}x^4+8x^3+26x^2+43x+30&=(x+2)(x^3 + 6x^2+14 x\dots)\\[10 pt]&=x^4+2x^3 +6x^3+12x^2+14x^2+28x\dots\\[10 pt]&=x^4+8x^3+26x^2+28x\dots\end{align}

Now, all terms are correct except the x term. We currently have 28x but we need 43x which is 15 more.

    \begin{align*}x^4+8x^3+26x^2+43x+30&=(x+2)(x^3 + 6x^2+14 x+15)\\[10 pt]&=x^4+2x^3 +6x^3+12x^2+14x^2+28x+15x+30\dots\\[10 pt]&=x^4+8x^3+26x^2+43x+30\end{align}

Now because (x+2) is a factor, the last term in the bracket works out perfectly.

    \[x^4+8x^3+26x^2+43x+30=(x+2)(x^3 + 6x^2+14 x+15)\]

Now we repeat the process with the cubic x^3+6x^2+14x+15. By some examination we find that (-3)^3+6(-3)^2+14(-3)+15=0, so (x+3) is a factor.

    \begin{align*}x^3 + 6x^2+14 x+15&=(x+3)(x^2 \dots)\\[10 pt]&=x^3+3x^2 ...\end{align}

We need another 3x^2 so let’s add the term 3x, which will multiply out to 3x^2.

    \begin{align*}x^3 + 6x^2+14 x+15&=(x+3)(x^2 +3x\dots)\\[10 pt]&=x^3+3x^2 +3x^2+9x...\end{align}

We need another 5x so let’s add the term 5 which will multiply out to 5x.

    \begin{align*}x^3 + 6x^2+14 x+15&=(x+3)(x^2 +3x+5)\\[10 pt]&=x^3+3x^2 +3x^2+9x+5x+15\end{align}

We are now down to a quadratic:

    \begin{align*}x^4+8x^3+26x^2+43x+30&=(x+2)(x+3)(x^2 +3x+5)\\[10 pt]&=(x+2)(x+3)(x+?)(x+?)\end{align}

After a little investigation, we see that x^2 +3x+5 doesn’t factor. When graphed, this quadratic does not cross the x axis, so there are no real zeros. This is called an ‘irreducible quadratic’. Therefore,

    \[x^4+8x^3+26x^2+43x+30=(x+2)(x+3)(x^2+3x+5)\]

is the fully factored form of this quartic. There are only two real zeros which occur at x=-2 and x=-3. We still know the end behaviour of this graph, however we don’t have enough information yet to determine if there is one or two turning points.

Using technology, we find that there is one turning point. The graph:

Practice

All of the polynomials in this applet have at least one value c where c \in \mathbb{Z} and p(c)=0.

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