How do you find the quadratic polynomial when sum of zeros and product of zeros are given?

Solution

Let the quadratic polynomial be
ax^2+bx+c
and it's zero be alpha and beta

Then
α+β=3
α*β=2
we know that
α+β= -b/a
=3/1
α*β=c/a
=2/1

From this
a=1
b=-3
c=2
Therefore the required polynomial
=1*x^2+[-3x]+2
=x^2-3x+2

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] = axn + bxn-1 + cxn-2 + ... + z

Then:

• Adding 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 us solve things.

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

Factors

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]

Quadratic

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:

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[ x2 − px − qx + pq ]
= ax2 − a[p+q]x + apq

Now let us compare:

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:

• 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:

ax2 + 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

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:

[Question: what happens if we choose a=−1 ?]

Cubic

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:

a[x−p][x−q][x−r]
= ax3 − a[p+q+r]x2 + a[pq+pr+qr]x − a[pqr]

And we get:

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:

• 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]

Sum of zeroes = α + β =√2

Product of zeroes = α β = 1/3

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x2–[α+β]x +αβ = 0

x2 –[√2]x + [1/3] = 0

3x2-3√2x+1 = 0

Thus, 3x2-3√2x+1 is the quadratic polynomial.

How do you find a quadratic polynomial when its zeros are given?

What is the quadratic polynomial whose sum and the product of zeros?