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= x2 + xy + y2, .'. x3 — y3 = (x − y) (x2 + xy + y2)

= x2 - xy + y2, .'. x3 + y3 = (x + y) (x2 — xy + y2).

It follows from 1 that:

The difference between the cubes of two quantities is equal to the product of two factors; the first is the difference between the quantities and the second is the square of the first term plus the product of the first by the second term, plus the square of the second term of the first factor.

It follows from 2 that:

The sum of two cubes is equal to the product of two factors; the first is the sum of the quantities, and the second is the square of the first term minus the product of the first term by the second term, plus the square of the second term of the first factor.

EXAMPLE 1.-Factor

- y3.

x·¤ — y3 — (x2)3 — y3 = (x2 — y) (.x1 + x2y + y2).

EXAMPLE 2. -Factor 8 a3 2763.

Sa3-27b3 =

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(2 a)3 — (31)3 — (2 a — 3b) (4 a2 + 6 ab + 9 b2).

EXAMPLE 3.-Factor m + 8 m3.

m2 + 8n3 = (m2)3 + (2 n)3 = (m2 + 2 n) (m* — 2 m2n + 4 n2).

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98. Case V.-When the Trinomial has the Form x2+ px + q.

It frequently happens that the product of two binomials is a trinomial (286). Conversely, some trinomials can be separated into the product of two factors, thus:

1. Find the factors of x2+9x+20.

The first term of each of the binomial, factors must be x; then, if a and b are the other terms of the factors,

or

(x + a)(x+b) must be the same as x2+9x + 20

x2 + (a+b)x+ab must be the same as 2 + 9x+20. Hence, the second terms a and b of the two binomial factors must be two numbers

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The only two numbers whose sum is 9 and whose product is 20 are 4 and 5.

..x2+9x+20= (x+4)(x+5).

2. Find the factors of x2+7xy + 12 y2.

Here, as in example 1,

x2 + (a+b)x+ab must be the same as x2 + (7 y)x + 12 y2.

The second terms a and b of the two binomial factors must be two numbers

and

whose sum is 7y

whose product is 12 y.

The only two numbers whose sum is 7y and whose product is 12y are 3y and 4y.

... x2+7xy +12 y2 = (x + 3y)(x+4y).

3. Factor the trinomial x2 + 3 x 54.

Here

x2 + (a+b)x+ab must be the same as 2 + 3x -- 54. Hence, the second terms a and b must be two numbers

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If the product of ab is -54, the factors a and b must have opposite signs and the greater must be positive, since their sum is +3. The only two numbers whose sum is +3 and whose product 54 are 9 and —- 6.

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The only two numbers, a and b, whose sum is 7 and whose product is 44 are 11 and +4.

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NOTE. The student, in examples of this kind, should always verify the results, by forming mentally the product of the factors he has chosen (¿89, IV).

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90+9x − x2= (x2- 9 x90)= −(x+6) (x — 15) — (15 — x) (x+6).

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99. Case VI. A polynomial of more than three terms, if it has factors, can often be factored readily by the principles already explained. Thus, it is seen at a glance that the expression

x3-3x2y+3xy3 — y3

fulfills, in respect both to exponents and to coefficients, the laws for expanding a power of a binomial, stated in 89, VIII; and it follows at once that

x3-3x2y + 3 xy2 — y3 = (x — y)3

In the discussion of the solution of quadratic equations, it will be shown how to find the factors of expressions of the following forms:

(a) ax2 + bx + c,

(b) Ax2+2 Bey+by+2 Bx + 2 Ey + F,

(c) LxMx2y2 + Ny1;

which in general can not be factored by inspection.

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100. Remainder Theorem.-If a rational integral expression in

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PROOF. Divide the polynomial by x − r and continue the process of division until x no longer appears in the partial dividend. The quotient will be an integral polynomial of the degree n − 1 in x, and the remainder independent of x, and therefore a constant. Hence, if is the quotient and R the remainder, it follows from the definition of division (270, formula) that

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a relation which holds, whatever may be; hence, in equation (1) give the value r.

Then

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here rr = 0, and (Dx-r

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is the value the expression Q has when

x is replaced by r. Hence it follows from (2) that

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EXAMPLE. What will the remainder be on dividing

x3+2x2-4x+9 by x-3?

By rule, remainder

=

(3)3+2(3)4(3) +942.

101. The Factor Theorem.-If a rational integral expression in x vanishes, that is, becomes equal to 0, when r is substituted for x, then X r is an exact divisor of the expression.

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By (2), the first member of (4) is 0; and also r-r0, therefore from (4)

0 = Q0+ R
.. R = 0.

Hence, when we divide the expression, (1), by x-r there is no remainder, that is, expression (1) is exactly divisible by x-r.

NOTE. If expression (1) is exactly divisible by a-r, then pn is exactly divisible by r, because the product of the last term in the divisor by the last term in the quotient is equal to the last term in the dividend. Hence, in searching for a numerical value of æ which will make the given expression vanish, only exact divisors of the last term of the expression need be tried.

EXAMPLE 1.

Find the factors of a3+52 + 7 x + 2.

The factors of 2 are +1, +2, 1, - 2; of these four numbers -2 only will make the expression 0, thus,

(-2)+5(-2)2 — 14 + 2 = −8+2014+ 2 = 0.

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x-(-2) = x+2 is a factor of 3+5+7x+2.

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