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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 + a1xn-1 + a2xn-2 + ... + an-1x + an = 0 ,

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

G(x) = (x - h)∙Q1(x) + R1

Q1(x) = (x - h)∙Q2(x) + R2

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

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

Qn-2(x) = (x - h)∙Qn-1(x) + Rn-1

Qn-1(x) = (x - h)∙Qn(x) + Rn

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

3)        g(y) = a0yn + Rnyn-1 + Rn-1yn-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)        .....................................

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

Rn-1 = Qn-2(r) - (r - h)∙Qn-1(r)

Rn = Qn-1(r) - (r - h)∙a0

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

5)        g(y) = a0yn + [Qn-1(r) - (r - h)∙a0] yn-1 + [Qn-2(r) - (r - h)∙Qn-1(r)] yn-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)] yn-1 + Qn-1(r) [y - (r - h)] yn-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.

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