Factors of quadratic polynomials and zeroes
Quadratic functions are examples of polynomials, which have a pleasant arithmetic much like that of numbers, but with some additional aspects. The question of factoring polynomials is particularly interesting, and challenging. Descartes found an important relation between the zeroes of a polynomial, and linear factors of that polynomial.
In this step, we will

review basics of polynomial arithmetic

relate zeroes and factors of quadratics via Descartes theorem

give a proof of Descartes theorem.
Polynomials and their arithmetic
A polynomial is a general algebraic expression made from powers of \(\normalsize{x}\) with arbitrary coefficients, such as
\(\normalsize{p(x)=x^24x+3}\) or \(\normalsize{q(x)=x+2}.\)
The degree of a polynomial is the highest power of \(\normalsize{x}\) that appears, so that \(\normalsize{p}\) has degree \(\normalsize{2}\) and \(\normalsize{q}\) has degree \(\normalsize{1}\).
There is an arithmetic with polynomials: they can be added, subtracted or multiplied just like numbers, so that for example
\[\Large{(p+q)(x)=x^23x+5}\] \[\Large{(pq)(x)=x^25x+1}\] \[\Large{(pq)(x)=x^32x^25x+6}.\]Q1 (E): If \(\normalsize{r(x)=x+3}\) and \(\normalsize{s(x)=2x1}\) then determine the polynomials i) \(\normalsize{r+s}\) ii) \(\normalsize{rs}\) and iii) \(\normalsize{rs}\).
Q2 (E): If \(\normalsize{u(x)=x^2+x+1}\) and \(\normalsize{v(x)=x^31}\) then determine the polynomials i) \(\normalsize{u+v}\) ii) \(\normalsize{uv}\) and iii) \(\normalsize{uv}\).
Factoring polynomials
Sometimes polynomials can also be divided: just like numbers, this occurs most smoothly when one polynomial is a multiple of the other.
For example you will all know the factorization
\[\Large{x^21=(x+1)(x1)}.\]In this case we say that \(\normalsize{x^21}\) is a multiple of \(\normalsize{x+1}\), and that equivalently \(\normalsize{x+1}\) is a factor of \(\normalsize{x^21}\).
It is nice to know that just as natural numbers factor uniquely into primes, for example \(\normalsize{1428=2\times2\times3\times7\times17}\), so too a polynomial factors uniquely (up to rearrangement) into irreducible polynomials that do not have factors.
Given any two linear polynomials \(\normalsize{u(x)=ax+b}\) and \(\normalsize{v(x)=cx+d}\) their product is the quadratic polynomial
\[\Large{(uv)(x)=(ax+b)(cx+d)=acx^2+(ad+bc)x+bd.}\]Please check that you understand completely how the distributive law works here! A rather deep and interesting problem is the reverse: given a quadratic polynomial, how can you write it as a product of linear factors.
Q3 (M): I have multiplied two linear factors together to get the quadratic polynomial
\[\Large{30x^228x240}.\]What were my two linear polynomials?
Factoring challenge
This is actually a fun activity. Please take two linear factors, and multiply them, and challenge your fellow participants to find the factors. And also try to find the factors of at least one of your fellow course mates quadratic.
Zeroes of a quadratic polynomial
If \(\normalsize{p(x)=x^212x+35}\) is a polynomial, then a zero of \(\normalsize{p(x)}\) is a number \(\normalsize{r}\) with the property that \(\normalsize{p(r)=0}\). In other words, we are solving the equation \(\normalsize{r^212r+35=0}\). The basic technique for solving this kind of quadratic equation goes back to the ancient Hindus.
If we look at a table of values for \(\normalsize{p(x)}\), we can find such zeros:
\(\normalsize{x}\)  \(\normalsize{p(x)}\) 

3  80 
2  63 
1  48 
0  35 
1  24 
2  15 
3  8 
4  3 
5  0 
6  1 
7  0 
8  3 
9  8 
10  15 
We see that the zeroes of \(\normalsize{p(x)}\) are \(\normalsize{5}\) and \(\normalsize{7}\). However we have to consider ourselves lucky here: not every quadratic has zeroes that can be found so easily.
We could also find the zeroes graphically, at least approximately, by plotting the corresponding polynomial function
\[\Large{y=p(x)=x^212x+35}:\]The connection with factors
We say that the polynomial \(\normalsize{p(x)}\) has a linear factor \(\normalsize{(xr)}\) when \(\normalsize{p(x)=(xr)q(x)}\) for some other polynomial \(\normalsize{q(x)}\). Descartes realized that the zeroes of \(\normalsize{p(x)}\) were intimately connected with its linear factors. Note that we can write
\[\Large{x^212x+35=(x5)(x7)}\]and this explains why this polynomial has zeroes \(\normalsize{5}\) and \(\normalsize{7}\).
Q4 (E): What are the linear factors of \(\normalsize{x^2+8x+15}\)?
Now we come to a major theoretical insight of Descartes.
Theorem (Descartes’ Factor Theorem) If \(\normalsize{p(x)}\) is a polynomial, then \(\normalsize{p(x)}\) has a zero \(\normalsize{r}\) precisely when \(\normalsize{(xr)}\) is a factor of \(\normalsize{p(x)}\).
Note that we are using the term ‘precisely when’ to mean that there is a twoway implication here:
 if \(\normalsize{p(x)}\) has a zero \(\normalsize{r}\) then \(\normalsize{(xr)}\) is a factor, and
 if \(\normalsize{(xr)}\) is a factor then \(\normalsize{p(x)}\) has a zero \(\normalsize{r}\).
We are getting two statements for the price of one!
Examples with Descartes’ Factor Theorem
To illustrate the Theorem, let’s look at some examples. The polynomial \(\normalsize{x^212x+35}\) factors has zeroes \(\normalsize{5}\) and \(\normalsize{7}\), and factors as
\[\Large{x^212x+35=(x5)(x7)}.\]Another example: the polynomial \(\normalsize{x^2+3x28}\) factors as
\[\Large{x^2+3x28=(x4)(x+7)}\]and so it has zeroes \(\normalsize{4}\) and \(\normalsize{7}\).
Although we like to give examples with integers, fractions are also possible.
Q5 (M): What are the linear factors of \(\normalsize{x^2+\frac{7}{2}x+3}\)?
It is not always the case that a polynomial has zeroes, and hence factors. The equation \(\normalsize{x^2+2=0}\) has no real solutions. Hence the polynomial \(\normalsize{x^2+2}\) has no linear factors.
Q6 (C): How many linear factors does the polynomial have whose graph is shown in the image to this step?
A proof of Descartes theorem (advanced)
Why is Descartes theorem true?? If you have some stronger background, we invite you to follow the argument for quadratic polynomials. In fact the same proof applies more generally.
Proof: If \(\normalsize{(xr)}\) is a factor of \(\normalsize{p(x)=ax^2+bx+c}\), it means that \(\normalsize{p(x)=(xr)q(x)}\) for some other polynomial \(\normalsize{q(x)}\). In that case if we substitute \(\normalsize{r}\) we have \(\normalsize{p(r)=(rr)q(r)}\) which means that \(\normalsize{p(r)=0}\) so that yes, \(\normalsize{r}\) is a zero of \(\normalsize{p(x)}\).
Now suppose that \(\normalsize{r}\) is a zero of \(\normalsize{p(x)}\), so that \(\normalsize{p(r)=ar^2+br+c=0}\) then
\[\Large{\begin{align*} p(x) &= p(x)  p(r)\\ & = ax^2+bx+c(ar^2+br+c)\\ &=a(x^2r^2)+b(xr)\\ &=a(xr)(x+r) + b(xr) \\ & = (xr) (a(x+r) + b) \end{align*}}\]So \(\normalsize{p(x) = (xr) q(x) }\), which means that \(\normalsize{(xr)}\) is a factor of \(\normalsize{p(x)}\). QED.
Answers
A1.
i) The sum is \(\normalsize{(r+s)(x)=x+3+2x1=3x+2}\).
ii) The difference is \(\normalsize{(rs)(x)=x+3(2x1)=x+4}\).
iii) The product is \(\normalsize{(rs)(x)=(x+3)\times(2x1)=2x^2+5x3}\).
A2.
i) The sum is \(\normalsize{(u+v)(x)=x^2+x+1+x^31=x^3+x^2+x}\).
ii) The difference is \(\normalsize{(uv)(x)=x^2+x+1(x^31)=x^3+x^2+x+2}\).
iii) The product is \(\normalsize{\begin{align}(uv)(x)&=(x^2+x+1)\times(x^31)\\&=x^5x^2+x^4x+x^31 \\&=x^5+x^4+x^3x^2x1. \end{align}}\)
A3. The factorization is:
\[\Large{30x^228x240=(5x+12)(6x20)}.\]How could we get that? We need to find two numbers that multiply to \(\normalsize{30}\), and two other numbers that multiply to \(\normalsize{240}\), such that the sum of products of two of them with the other two is \(\normalsize{28}\). It certainly appears that trial and error are needed!
A4. The linear factors of \(\normalsize{x^2+8x+15}\) are \((x+3)\) and \((x+5)\).
A5. The linear factors of \(\normalsize{x^2+\frac{7}{2} x+3}\) are \(\normalsize{(x+\frac{3}{2})}\) and \(\normalsize{(x+2)}\).
A6. The polynomial visually has zeros at \(\normalsize{x=3,2,1,0,1,2}\) and \(\normalsize{3}\). So according to Descartes’ theorem, it will have factors \(\normalsize{(x+3),(x+2),(x+1),x,(x1),(x2)}\) and \(\normalsize{(x3)}\). So a good guess for the polynomial might be:
\[\Large{p(x)= (x+3)(x+2)(x+1)x(x1)(x2)(x3)}.\]
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