than those of the given equation by 2. Substituting y-2 for r in the above equation, we have:

"And this is the transformed equation whose roots are increased by 2; that is, are equal to 4 and 5. Thus,

( y −4) × ( y −5)=y2 −9y+20=0.

"The old Pine will see that in the equation

y2—4y+4—5y+10+6=0,

which results from susbstituting y-2 for x in the given equation, the absolute term 6 of the given equation remains. The first two terms of the given equation (x2-5x) are changed, becoming y2-97+14, the 14 together with the 6 making the absolute term of the transformed equation.

"If then we divide this transformed equation by y-2 we will have as a remainder the absolute term 6 of the given equation. Thus,

And so, if we divide the quotient y-7 by y-2, we will get the second coefficient of the given equation. For that coefficient is formed from the sum of the roots, and this sum by the hypothesis is less by 4. But in dividing by y-2 twice, from the nature of the construction we diminish the sum of the roots of the transformed equation by 4, when it becomes equal to the sum of the roots or second coefficient of the given equation. Awfully simple. And for the same or similar reasons, if the equation is a more extended one, by continuing to divide the last quotient by the same divisor we shall continue to have a remainder that will be the next preceding coefficient of the original equation.

XXXIV.

DERIVED POLYNOMIALS.

LLEN will again build up equations. Given the equa

"ELLE

tion:

(x−1)(x-2)=x2−3x+2=0.

(1)

"The first derived polynomial, by rule, is 2x-3=(x−1)+ (x-2), or is once the sum (of the products) of the factors taken one in a set.

"The second derived polynomial, by rule, is 2, or is equal to two times (1X2) the sum of the products of the factors taken no times in a set, 2.

"Given the equation:

(x−1)(x−2)(x−3)=x3 −6x2+11x-6=0.

(2)

"The first derived polynomial, by rule, is 3x2-12x+1=

(x−1)(x-2)=x2−3x+2

(x-1)(x-3)=x2-4x+3=3x2-12x+11,

(r−2)(r−3)=x-5+6

or is equal to once the sum of the products of the factors taken two in a set.

"The second derived polynomial is 6x-12=(x−1)+(x− 2)+(x−1)+(x−3)+(x−2)+(x−3), or is equal to two times (1X2) the sum of (the products of) the factors taken one in a set.