Polynomials: Sums and Products of Roots

Roots of a Polynomial

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

Graph of Inequality

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

General Polynomial

If nosotros have a full general polynomial like this:

f(10) = axnorth + bxnorthward-1 + cxdue north-2 + ... + z

Then:

  • Calculation the roots gives −b/a
  • Multiplying the roots gives:
    • z/a (for even degree polynomials like quadratics)
    • −z/a (for odd degree polynomials like cubics)

Which can sometimes help united states solve things.

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

Factors

We tin take a polynomial, such as:

f(ten) = axdue north + bxn-1 + cxn-2 + ... + z

And so gene information technology like this:

f(x) = a(x−p)(ten−q)(x−r)...

So 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 ii):

axtwo + bx + c

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

a(10−p)(ten−q)

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

Let'south expand a(x−p)(x−q):

a(x−p)(10−q)
= a( x2 − px − qx + pq )
= axtwo − a(p+q)x + apq

At present let us compare:

Quadratic: axtwo +bx +c
Expanded Factors: ax2 −a(p+q)10 +apq

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

−a(p+q) = b

p+q = −b/a

And apq = c, so:

pq = c/a

And we get this result:

  • Adding the roots gives −b/a
  • Multiplying the roots gives c/a

This can assistance united states reply questions.

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

The sum of the roots is (v + √2)  + (v − √two) = ten
The product of the roots is (5 + √two) (5 − √ii) = 25 − 2 = 23

And nosotros desire an equation like:

ax2 + bx + c = 0

When a=one we can work out that:

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

Which gives us this event

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

The sum of the roots is x, 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 ?)

Cubic

Now permit united states of america wait at a Cubic (one degree college than Quadratic):

ax3 + bx2 + cx + d

As with the Quadratic, permit us aggrandize the factors:

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

And nosotros go:

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

Nosotros can now run into that −a(p+q+r)x2 = bx2 , and then:

−a(p+q+r) = b

p+q+r = −b/a

And −apqr = d, and so:

pqr = −d/a

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

  • Adding the roots gives −b/a (exactly the same equally 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.)

College Polynomials

The same pattern continues with college polynomials.

In Full general:

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