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The pairs of negative numbers whose product is 10 are 10 and · 1, and — 5 and - 2; and the sum of the last pair is 7. Hence x2 - 7x+10= (x − 5) (x − 2).

Again, to find the factors of x2+3x-18, we have to find two numbers whose product is - 18, and whose sum is 3. Since the product is - 18, one of the numbers is positive and the other negative. The pairs of numbers whose product is

and 1, -9 and 2, 6 and 3,

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18; and of these pairs the

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18 are 18

3 and 6,2 and 9, and 1 and sum of 6 and - 3 is 3.

Hence

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Find the factors of each of the following expressions:

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(ax+b) (cx + d) = acx2 + (ad + bc)x+bd.

Hence conversely, if an expression of the form px2+qx+r be the product of the two factors ax + b and cx+d, the given expression must be the same as

acx2+(ad+bc)x + bd; we must therefore have ac= = P, bd =r, and ad + bc = q. We can in simple cases find by trial the values of a, b, c, d which satisfy these relations.

5.

For example, to find the factors of 3x2 - 16x + 5. The 3x2 can only be given by the multiplication of 3 x and x. The 5 could only be given by the multiplication of 5 and 1, or - 5 and — 1; and, since the middle term is negative, the latter pair must be taken. We have now only a choice between (3x-5) (x − 1) and (3 x − 1)(x − 5), and it is at once seen that the latter must be taken in order that the coefficient of x in the product may be - 16. Thus the factors required are 3 x 1 and xAgain, to find the factors of 5x2+32 x be given by the multiplication of 5x and x. can be the product of -21 and 1, -7 and 3, 21. The possible pairs of factors, so far as the two end terms are concerned, are therefore (5 x ‡ 21) (x + 1), (5 x ‡7)(x ± 3), (5x3)(x+7), and (5x F 1)(x+21). It will be found that of these pairs the one which will give 32 x for the middle term is

(5 x − 3)(x+7).

21, the 5 x2 can only The end term - 21 -3 and 7, or -1 and

119. The factors of x2- 5 xy+4y2 can be found in the same way as the factors of x2-5x + 4. For we must find two quantities whose product is 4 y, and whose

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4y and -y. Hence

x2 − 5 xy + 4 y2 = (x — 4 y) (x − y).

16 xy+5y can be found in

So also the factors of 3x2
the same way as the factors of 3x2

factors of either can be written down
tors of the other are known.

165; and the as soon as the fac

For example, if we know that 5 x2 + 32 x 21 = (5 x − 3) (x + 7), we must have 5 x2 + 32xy - 21 y2 (5x-3y) (x + 7y).

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EXAMPLES XXXII.

Find the factors of each of the following expressions:

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120. It is clear that the method of finding by trial the factors of an expression of the form pa2 + qx+r, where p, q, r are known numbers, would be very tedious if there were many pairs of numbers whose product was p, and many pairs whose product was r, for there would then be very many pairs of factors which would agree with the given expression so far as the end terms were concerned, and out of these the single pair which would give the correct middle term would have to be

sought. It would, for example, be almost impossible to find the factors of 2310 a2 - 2419 x 9009 in this way.

Again, we not unfrequently meet with such an expression as x2+6x+7 which cannot be written as the product of two factors altogether rational, and in such a case it would be impracticable to try to guess the factors.

We therefore need some method of finding the factors of a quadratic expression which is applicable to all This method we proceed to investigate.

cases.

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

For example, x2 + 6x is made a perfect square by the addition of (9); and the square is x2+6x+9, which is (x + 3)2.

So also, x2+5x is made a perfect square by the addition of (5)2; and the square is x2 + 5x+(25), which is (x + 5)2.

And x2

that is of 49; and the square is x2

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7x is made a perfect square by the addition of (− 1)2, 7x+49, which is (x − )2. Whatever a may be, x2+ax is made a perfect square by the

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x2 + ax + (~1) 3 is ( x + 2)2

122. We will now find the factors of ax2 + bx + c by a method which is applicable in all cases.

The problem before us is to find two factors which are to be rational and integral with respect to x, and are therefore of the first degree in a, but which are not necessarily

rational and integral with respect to arithmetical numbers or any other letters which may be involved. We first apply the method to some examples.

Ex. 1. Find the factors of

x2+6x+8.

As we have seen in the last article, x2 + 6x is made a perfect square by the addition of 32; hence, adding and subtracting 32, we have

x2+6x+8= x2 + 6 x + 9 −9+8

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The last form of the expression is the difference of two squares, and the factors are therefore x + 3+1 and x + 3 1, that is x + 4 and x2. Hence the factors of x2+6x+8 are x + 4 and x + 2.

Ex. 2. Find the factors of x2 3x-28.

x2

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3 x is made a perfect square by the addition of (− 3)2, that is of; hence, adding and subtracting, we have

x23x28= x2 - 3x + 2 - - 28

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Thus the required factors are x + 4 and x − 7.

Ex. 3. To find the factors of x2 + 6x + 7.

3+ and

x2+6x is made a perfect square by the addition of 32; hence we write

x2+6x+7= x2 + 6x +9 −9+7

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Now the factors of the last expression, and therefore of x2+6x+7, are x + 3 + √2 and x + 3 − √2.

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