# Polynomials: Sums and Products of Roots

## Roots of a Polynomial

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

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) = ax^{n} + bx^{n-1} + cx^{n-2} + ... + z

Then:

**Adding**the roots gives**−b/a****Multiplying**the roots gives (where "z" is the constant at the end):**z/a**(for even degree polynomials like quadratics)**−z/a**(for odd degree polynomials like cubics)

It works on Linear, Quadratic, Cubic and Higher!

It can sometimes help us solve things.

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

## Factors

We can take a polynomial, such as:

f(x) = ax^{n} + bx^{n-1} + cx^{n-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)

## Quadratic

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

ax^{2} + bx + c

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

a(x−p)(x−q)

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

Let's expand a(x−p)(x−q):

a(x−p)(x−q)

= a( x^{2} − px − qx + pq )

= ax^{2} − a(p+q)x + apq

Quadratic: | ax^{2} |
+bx | +c |

Expanded Factors: | ax^{2} |
−a(p+q)x | +apq |

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

And apq = c, so:

And we get this result:

- Adding the roots gives
**−b/a** - Multiplying the roots gives
**c/a**

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:

ax^{2} + bx + c = 0

When **a=1** we can work out that:

- Sum of the roots =
**−b/a**=**-b** - Product of the roots =
**c/a**=**c**

Which gives us this result

x^{2} − (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:

x^{2} − 10x + 23 = 0

And here is its plot:

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

## Cubic

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

ax^{3} + bx^{2} + cx + d

As with the Quadratic, let us expand the factors:

a(x−p)(x−q)(x−r)

= ax^{3} − a(p+q+r)x^{2} + a(pq+pr+qr)x − a(pqr)

And we get:

Cubic: | ax^{3} |
+bx^{2} |
+cx | +d |

Expanded Factors: | ax^{3} |
−a(p+q+r)x^{2} |
+a(pq+pr+qr)x | −apqr |

We can now see that −a(p+q+r)x^{2} = bx^{2}, so:

And −apqr = d, so:

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

- Adding the roots gives
**−b/a**(exactly the same as the Quadratic) - Multiplying the roots gives
**−d/a**(similar to +c/a for the Quadratic)

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

## Higher Polynomials

The same pattern continues with higher polynomials.

In General:

- Adding the roots gives
**−b/a** - Multiplying the roots gives (where "z" is the constant at the end):
**z/a**(for even degree polynomials like quadratics)**−z/a**(for odd degree polynomials like cubics)