Website owner:  James Miller

Prove. Prove the following assertion:

To obtain an equation, each of whose roots is h less than the corresponding root of the equation

1)        G(x) = a₀xⁿ + a₁x^n-1 + a₂x^n-2 + ... + a_n-1x + a_n = 0 ,

compute the following R₁, R₂, .... , R_n , employing synthetic division.

            G(x) = (x - h)∙Q₁(x) + R₁

            Q₁(x) = (x - h)∙Q₂(x) + R₂

2)        .........................................

            ........................................

            Q_n-2(x) = (x - h)∙Q_n-1(x) + R_n-1

              Q_n-1(x) = (x - h)∙Q_n(x) + R_n

where each R is a constant and Q_n(x) = a₀. Then the roots of

3)        g(y) = a₀yⁿ + R_ny^n-1 + R_n-1y^n-2 + ... + R₂y + R₁ = 0

are the roots of G(x) diminished by h.

Proof. Let r be any root of G(x) = 0. Since G(r) = 0, we can write equations 2) above as

            R₁ = - (r - h)∙Q₁(r)

            R₂ = Q₁(r) - (r - h)∙Q₂(r)

4)        .....................................

            .....................................

            R_n-1 = Q_n-2(r) - (r - h)∙Q_n-1(r)

            R_n = Q_n-1(r) - (r - h)∙a₀

Let us now substitute these R’s into 3) above.

5)        g(y) = a₀yⁿ + [Q_n-1(r) - (r - h)∙a₀] y^n-1 + [Q_n-2(r) - (r - h)∙Q_n-1(r)] y^n-2 + .......

                                                                        ....... + [Q₁(r) - (r - h)∙Q₂(r)] y - (r - h)∙Q₁(r)

This equation 5) can be rewritten as

6)        g(y) = a₀[y - (r - h)] y^n-1 + Q_n-1(r) [y - (r - h)] y^n-2 + ....... + Q₁(r) [y - (r - h)] = 0

From 6) we can see that r - h is a root of 3) as was to be proved.