| XXX | (112) | XXX |
B A. R in A L G E from the root already found, and the quotient ffiall manner, gives the refalt ~o, be the root of the pro- dace give an equation of a degree lower than the propofed; pofed equation. roots will give the remaining roots required. For example, fuppofe this 1equation is to be refolved, whofe As for example, the root+3, firft found, gave x=3, 5 or x—3=0, whence dividing thus. x ——2a b') - bx'+SabX 5 -°' where the laft term is 2a1 b, whofe fimple literal divifors are a, b, 2a, 2b, each of which may be taken either pofitively or negatively: but as here we find there are variations of figns in the equation, we need only take them pofitively. Suppofe x—a the firft of the divifors, and fubftituting a for x, the equation becomes So that, the whole vanifiiing, it follows, that a is one of the roots of the equation. After the fame manner, if you fubftitute b in place of x, the equation is z b*—2ab --2a*b—2a* b~) 0 —b'+zab* 5" ' which vanifhing, (hews b to be another root of the equation. Again, if you fubftitute 2a for x, you will find all the terms deftroy one another fb as to make the furh =0. For it will then be 8«5—I2aj+4<J15—2azb') —b--(>a. b y °* Whence we find, that 2a is the third root of the equation. Which, after the firft two (+?, +£,) had been found, might have been colledted from this, that the laft term being the produft of the three roots, -r«, being known, the third muft neceflarily be equal to the laft term divided by the prodqdt ab, that is, =^-7-^=2<2. ab Let5the roots of the cubic equation x —2xl—33x-i-90=o be required. And firft the divifors of 90 are found to be 1, 2, 3, 5, 6, 9, to, 15, 18, 30, 4J, 90. If you fubftitute 1 for x, you will find x3—ax1—33X+90—56; fo that 1 is not a root of the equation. If you fubftitute 2 for x, the3 refultwill be 24: but, putting x=3, you have x —2x*——18—99+90—117—117=0. So that three is one of the roots of the propofed equation. The other affirmative root is +5 ; nnd after you find it, as it is manifeft from- the equatioiiv that the other root is negative, you are not to try any more divifors taken pofitively, but to fubftitute them, negatively taken, for x: and thus you find, that —6 is the third root. For putting x=—-6, you have x3—2x1—33x4-90=—216—72+198-1-90=0. This laft root might have been found by dividing the laft term 90, having its fign changed by 15, the product of the two roots already found, , When one of the roots of an equation is found, in order to find the reft with lefs trouble, divide the propofed equation by the fimple equation which you are to de-
The quotient (hall give a quadratic equation x’+x—30 =0, which mult be the produdt of the other two fimple equations from which the cubic is generated, and whofe roots therefore muft be two of the roots of that cubic. Now the roots of that quadratic equation are eafily found, by chap. 12. to be + 5 and—6. For, x1+v=30 add £ . . x1 +x+i=30 -l;|= and . . x=~r-x^-—^=+5 or —6. After the fame manner, if the biquadratic x4—2*3— 25x1+26x+i20=0 is to be refolved; by fubftituting the divifors of 12a for x, you will find, that +3, one of thofe divifors, is one of the roots; the fubftitutioa of 3 for x giving 81—54—225+78+120=279—279=0. And therefore, dividing the propofed equation by x—3, you muft inquire for the roots of the cubic xJ+**—2 ax —40=0, and finding that +5, one of the divifors of 40, is one of the roots, you divide that cubic by x—5, and the quotient gives the quadratic x I+6x+8=o, whofe two roots are —2, —4. So that the four roots of the. biquadratic are +3, +5, —2, —4. This rule fuppofes that you can find all the divifors of "the laft term; which you may always do thus. “ If it is a fimple quantity, divide it by its lead divi“ for that exceeds unit, and the quotient again by its “ leaft divifor, proceeding thus till you have a quotient “ that is not divifible by any number greater than unit.. “ This quotient, with thefe divifors, are the firft or “ fmple divifors of the quantity. And tire products of “ the multiplication of any 2, 3, 4, <bc. of them are “ the eompound divifors.” As to find the divifors of 60; firft I divide by,2, and the quotient 30 again by 2, then the the next quo15 by 3, and the quotient of this divifion 5 is not farther divifible by any integer above units; fo that the finiple divifors are, 2, 2, 3. 5; The produdls of two, 4, 6, xo, 15. The produdls of three, 12, 20, 30. The produdt of all four, ........ 60. The divifors of 90 are found after the fame manner; Simple
