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?