Sums and Products of Roots (quadratic, cubic and quartic)

The zeros of a polynomial $P(x)$P(x) are the values of $x$x which make the polynomial equal to zero. That is, they are the roots of the corresponding polynomial equation $P(x)=0$P(x)=0.

If $a$a is a zero of a polynomial $P(x)$P(x), then $x-a$x−a is a factor of $P(x)$P(x). We can write $P(x)=(x-a)Q_1(x)$P(x)=(x−a)Q1​(x) where $Q_1$Q1​ is a polynomial of degree one less than $P$P.

If $Q_1$Q1​ has a zero $b$b, we can continue this process and write $P(x)=(x-a)(x-b)Q_2(x)$P(x)=(x−a)(x−b)Q2​(x). The factorisation process ends when the quotient polynomial $Q_i$Qi​ has degree $1$1 or when it is irreducible in the sense that it cannot be further factored.

For example, the quartic polynomial $x^4-2x^3-x+2$x4−2x3−x+2 has zeros $x=1$x=1 and $x=2$x=2 so that it has linear factors $(x-1)$(x−1) and $(x-2)$(x−2). When these factors are taken out the polynomial is written $(x-1)(x-2)(x^2+x+1)$(x−1)(x−2)(x2+x+1). (This can be verified by expanding the brackets.) The quadratic factor $(x^2+x+1)$(x2+x+1) is said to be irreducible in the field of real numbers in the sense that $x^2+x+1=0$x2+x+1=0 has no real roots and hence cannot be factored.

On the other hand, the quartic polynomial $x^4-2x^3-2x^2+5x-2$x4−2x3−2x2+5x−2 has zeros $x=1$x=1, $x=2$x=2, $x=-\frac{1+\sqrt{5}}{2}$x=−1+√52​ and $x=-\frac{1-\sqrt{5}}{2}$x=−1−√52​. In factored form it is seen to have four linear factors:$(x-1)(x-2)\left(x+\frac{1+\sqrt{5}}{2}\right)\left(1+\frac{1-\sqrt{5}}{2}\right)$(x−1)(x−2)(x+1+√52​)(1+1−√52​)

 

Here, we consider polynomials that can be factored completely into linear factors involving just the real numbers. (We mention, in passing, that every polynomial can be written as the product of linear factors if we agree to use complex numbers and this is essentially the content of the fundamental theorem of algebra which was proved by Karl Friedrich Gauss in 1799.)

 

From the discussion above, it is apparent that a polynomial of degree $n$n can have at most $n$n real zeros. This is because the polynomial can have at most $n$n linear factors.

Consider a quartic with four zeros $a$a, $b$b, $c$c and $d$d. It can be expressed as $P(x)=A(x-a)(x-b)(x-c)(x-d)$P(x)=A(x−a)(x−b)(x−c)(x−d). In expanded form, this is $P(x)=A\left[x^4-(a+b+c+d)x^3+(ab+ac+ad+bc+bd+cd)x^2-(abc+abd+acd+bcd)x+abcd\right]$P(x)=A[x4−(a+b+c+d)x3+(ab+ac+ad+bc+bd+cd)x2−(abc+abd+acd+bcd)x+abcd]. 

If, for convenience, we restrict our attention to monic polynomials (those with $A=1$A=1) we see that in this case, the coefficient of the cubic term is the negative of the sum of the roots, and the constant term is the product of the roots.

In general, for monic polynomials, the coefficient of the second highest power term is the negative sum of the zeros. If the degree of the polynomial is even, then the product of the zeros is equal to the constant term, while if the degree is odd, then the product of the zeros is equal to the negative of the constant term.

Let's look at a few examples. Consider the polynomial equation $x^6-3x^5+8x^4+2x^3-x^2+7x+9=0$x6−3x5+8x4+2x3−x2+7x+9=0. It looks quite foreboding, but using the rules above we know that to find the sum of the roots we look at the second highest power term: $-3x^5$−3x5. The negative of the coefficient of this term is $-\left(-3\right)=3$−(−3)=3. Next, to find the product of the roots we look at the constant term: $9$9. Since this polynomial is of degree $6$6, which is even, the product of the roots is simply this constant term, $9$9.

Next, consider the polynomial equation $x^3+5x^2-11x+6=0$x3+5x2−11x+6=0. Using the same approach, we find that the sum of the roots is the negative of the coefficient of the term $5x^2$5x2, which is $-5$−5. Secondly, the product of the roots is the negative of the constant term, which is $-6$−6, since the degree of this polynomial is odd.

Now let's think about the process of factoring a quadratic of the form $x^2+bx+c$x2+bx+c. We look for two numbers, $\alpha$α and $\beta$β, such that $\alpha+\beta=b$α+β=b and $\alpha\beta=c$αβ=c. We can then write $x^2+bx+c=(x+\alpha)(x+\beta)$x2+bx+c=(x+α)(x+β) and the zeros are $-\alpha$−α and $-\beta$−β.

 

Example 1

It was asserted above that the polynomial equation $x^2+x+1=0$x2+x+1=0 has no real roots. This can be seen by observing that its discriminant is negative: $1^2-4\times1\times1=-3$12−4×1×1=−3.

Another way to see that there can be no real roots is to look for two numbers, $\alpha$α and $\beta$β, such that $\alpha+\beta=1$α+β=1 and $\alpha\times\beta=1$α×β=1. We must have both numbers positive or both negative (since their product is positive). We assume both are positive, as a similar argument will work if both are negative. Clearly, $\alpha=1$α=1 and $\beta=1$β=1 is impossible since we would then have $\alpha+\beta=2$α+β=2. In all other cases, in order for $\alpha\times\beta=1$α×β=1, we must have one of these numbers greater than $1$1 and the other less. So, the sum must be greater than $1$1. Hence, the requirement cannot be met and we conclude that there can be no real roots.

 

Example 2

How many monic cubic polynomials can be constructed that have integer zeros when the constant term is $17$17?

The general form of the polynomial is $x^3+bx^2+cx+17$x3+bx2+cx+17 and there can be at most three zeros: $\alpha$α, $\beta$β and $\gamma$γ, all of them integers.

Now, $\alpha\times\beta\times\gamma=-17$α×β×γ=−17 and since $17$17 is a prime number, we have the following possibilities (apart from relabelling the zeros).

$\alpha$α	$\beta$β	$\gamma$γ
$-17$−17	$-1$−1	$-1$−1
$-17$−17	$1$1	$1$1
$17$17	$1$1	$-1$−1

So, there are three polynomials. We can go further and produce them specifically. They are

$P_1(x)=x^3+19x^2+35x+17=(x+1)(x+1)(x+17)$P1​(x)=x3+19x2+35x+17=(x+1)(x+1)(x+17)
$P_2(x)=x^3-17x^2-x+17=(x+1)(x-1)(x-17)$P2​(x)=x3−17x2−x+17=(x+1)(x−1)(x−17)
$P_3(x)=x^3+15x^2-33x+17=(x-1)(x-1)(x+17)$P3​(x)=x3+15x2−33x+17=(x−1)(x−1)(x+17)

The coefficient of the linear term in each case is the sum of the products of the zeros taken two at a time. 

 

Worked examples

Question 1

Question 2

Question 3