Polynomials: Sums and Products of Roots

Roots of a Polynomial

A "root" (or "zero") is where the polynomial is equal to zero:

Graph of Inequality

Put simply: a root is the x-value where the y-value equals zero.

General Polynomial

If we have a general polynomial like this:

f(x) = axn + bxn-1 + cxn-2 + ... + z


Which can sometimes help us solve things.

How does this magic work? Let's find out ...


We can take a polynomial, such as:

f(x) = axn + bxn-1 + cxn-2 + ... + z

And then factor it like this:

f(x) = a(x−p)(x−q)(x−r)...

Then p, q, r, etc are the roots (where the polynomial equals zero)


Let's try this with a Quadratic (where the variable's biggest exponent is 2):

ax2 + bx + c

When the roots are p and q, the same quadratic becomes:


Is there a relationship between a,b,c and p,q ?

Let's expand a(x−p)(x−q):

= a( x2 − px − qx + pq )
= ax2 − a(p+q)x + apq

Now let us compare:
Quadratic: ax2 +bx +c
Expanded Factors: ax2 −a(p+q)x +apq

We can now see that −a(p+q)x = bx, so:

−a(p+q) = b
p+q = −b/a

And apq = c, so:

pq = c/a

And we get this result:

This can help us answer questions.

Example: What is an equation whose roots are 5 + √2 and 5 − √2

The sum of the roots is (5 + √2)  + (5 − √2) = 10
The product of the roots is (5 + √2) (5 − √2) = 25 − 2 = 23

And we want an equation like:

ax2 + bx + c = 0


When a=1 we can work out that:

Which gives us this result

x2 − (sum of the roots)x + (product of the roots) = 0

The sum of the roots is 10, and product of the roots is 23, so we get:

x2 − 10x + 23 = 0

And here is its plot:

polynomial roots

(Question: what happens if we choose a=−1 ?)


Now let us look at a Cubic (one degree higher than Quadratic):

ax3 + bx2 + cx + d

As with the Quadratic, let us expand the factors:

= ax3 − a(p+q+r)x2 + a(pq+pr+qr)x − a(pqr)

And we get:

Cubic: ax3 +bx2 +cx +d
Expanded Factors: ax3 −a(p+q+r)x2 +a(pq+pr+qr)x −apqr

We can now see that −a(p+q+r)x2 = bx2, so:

−a(p+q+r) = b
p+q+r = −b/a

And −apqr = d, so:

pqr = −d/a

This is interesting ... we get the same sort of thing:

(We also get pq+pr+qr = c/a, which can itself be useful.)

Higher Polynomials

The same pattern continues with higher polynomials.

In General: