Imágenes de páginas
PDF
EPUB

and

Hence from the above we have

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

x2+3xy2 – 18y1 = (x +6y2) (x − 3y2),
(a+b)2-7 (a+b) x + 10x2 = (a+b−5x) (a+b−2x),

[blocks in formation]
[merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small]

80. Factors of general quadratic expression. We proceed to shew how to find the factors of any expression of the second degree in a particular letter, x suppose.

The most general quadratic expression in x is ax2 + bx + c, where a, b and c do not contain x.

The problem before us is to find two factors which are rational and integral with respect to x, and are therefore each of the first degree in x, but which are not necessarily, and not generally, rational and integral with respect to arithmetical numbers or to any other letters which may be involved in the expression.

The factors of ax2 + bx + c can be found by changing it into an equivalent expression which is the difference of two squares.

We first note that since x2+2ax + a2 is a perfect square, in order to complete the trinomial square of which a2 and 2ax are the first two terms, we must add the square of a, that is, we must add the square of half the coefficient of x.

For example, x2+57 is made a perfect square, namely (a +5)*,

[blocks in formation]

81. To find the factors of ax2 + bx + c.

b

[blocks in formation]

2

Now a2 + - a is made a perfect square, namely (a+2),

a

[blocks in formation]
[ocr errors][merged small][ocr errors][ocr errors][merged small][merged small][merged small][ocr errors][subsumed][subsumed][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][ocr errors]

Hence as the difference of any two squares is equal to the product of their sum and difference, we have

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small]

x2+4x+3= x2+4x+4-4+3=(x+2)-1=(x+2+1)(x+2-1)
=(x+3)(x+1).

Ex. 2. To find the factors of x2-5x+3.

[merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][subsumed][subsumed][merged small][ocr errors][merged small][merged small][merged small][ocr errors][ocr errors][ocr errors][ocr errors][ocr errors][subsumed][ocr errors][ocr errors][merged small][merged small][subsumed][subsumed][ocr errors][subsumed][merged small][merged small][ocr errors][merged small][subsumed][subsumed][subsumed][subsumed][ocr errors][merged small][merged small][ocr errors][merged small]

Ex. 4. To find the factors of x2+2ax-b2-2ab.

(x-1).

x2+2ax-b2 - 2ab= x2+2ax+ a2 — a2 – b2 – 2ab = (x + a)2 − (a + b)2

={x+a+(a+b)} {x+a− (a+b) } = (x+2a+b) (x − b).

82. Instead of working out every example from the beginning, we may use the formula

[blocks in formation]

and we should then only have to substitute for a, b and c their values in the particular case under consideration.

Thus to find the factors of 3x2-4x+1.

[blocks in formation]

Here a=3, b= −4, c=1.

[blocks in formation]

=

√16

[merged small][ocr errors][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small]

Now, for particular values of a, b, c,

positive, zero, or negative.

I. Let

b2-4ac
4a2

[merged small][merged small][merged small][ocr errors]

√ b2 - 4ac).

b2-4ac

4a2 may be

be positive. Then the two factors of ax2 + bx + c will be rational or irrational according as

b2-4ac

is or is not a perfect square.

4a2

[blocks in formation]

III. Let

b2-4ac
4a2

(a

in

X, if b2-4ac0.

be negative. Then no positive or nega

tive quantity can be found whose square will be equal to b2-4ac

; for all squares, whether of positive or of negative 4a2 quantities, are positive.

Expressions of the form √-a, where a is positive, are called imaginary, and positive or negative quantities are distinguished from them by being called real.

We shall consider imaginary quantities at length in a subsequent chapter: for our present purpose it is sufficient to observe that they obey all the fundamental laws of Algebra; and this being the case, the formula of Art. 81 will hold good when b2-4ac is negative.

Note. For some purposes for which the factors of expressions are required, the only useful factors are those which are altogether rational: on this account irrational and imaginary factors are often not shewn. Thus, for example, the factorisation of 3-8 is for many purposes complete in the form (x-2) (x2 + 2x + 4), the imaginary factors of x2 + 2x + 4, namely

x+1+√3 and x + 1 √ − 3,

not being shewn.

84. We have in Art. 81 shewn how to resolve any expression of the second degree in a particular letter into two factors (real or imaginary) of the first degree in that letter.

It should be noted that the factors of the most general expression of the third degree, or of the fourth degree, can be found, although the methods are beyond the range of this book; expressions of higher degree than the fourth cannot however, except in a few special cases, be resolved into factors.

85. Factors found by re-arrangement and grouping of terms. The factors of many expressions can be found by a suitable re-arrangement and grouping of the terms.

For example

1+ax − x2 − ax3 = 1 + ax)− x2 (1 + ax) = (1+ax) (1 − x2)

=(1+ax) (1+x) (1 − x) ;

« AnteriorContinuar »