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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) = a_{0}x^{n} + a_{1}x^{n-1} + a_{2}x^{n-2} + ... + a_{n-1}x + a_{n} = 0 ,

compute the following R_{1}, R_{2}, .... , R_{n} , employing synthetic division.

G(x) = (x - h)∙Q_{1}(x) + R_{1}

Q_{1}(x) = (x - h)∙Q_{2}(x) + R_{2}

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) = a_{0}. Then the roots of

3) g(y) = a_{0}y^{n} + R_{n}y^{n-1} + R_{n-1}y^{n-2} + ... + R_{2}y + R_{1} = 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

R_{1} = - (r - h)∙Q_{1}(r)

R_{2} = Q_{1}(r) - (r - h)∙Q_{2}(r)

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

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

R_{n-1} = Q_{n-2}(r) - (r - h)∙Q_{n-1}(r)

R_{n} = Q_{n-1}(r) - (r - h)∙a_{0}

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

5) g(y) = a_{0}y^{n} + [Q_{n-1}(r) - (r - h)∙a_{0}] y^{n-1} + [Q_{n-2}(r) - (r - h)∙Q_{n-1}(r)] y^{n-2} + .......

....... + [Q_{1}(r) - (r - h)∙Q_{2}(r)] y - (r - h)∙Q_{1}(r)

This equation 5) can be rewritten as

6) g(y) = a_{0}[y - (r - h)] y^{n-1} + Q_{n-1}(r) [y - (r - h)] y^{n-2} + ....... + Q_{1}(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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