Therefore the factors of a3 +5 x2+7 x + 2 are NOTE. An expression can sometimes be resolved into three or more factors. EXAMPLE 2. Factor 35 x2-2x-24. The factors of 24 are 1, 2, 4, 3, 6, 8, 12, 1, 2,4,3, — 6, 8, 12. Hence, x+5x-2x-24 is divisible by x-2; the quotient is +7x+12, whose factors are (x+3) (x+4). x2+5 x2-2x-24 (x-2) (x+3) (x+4). = EXERCISE XXII Show by means of the factor theorem, that +7 is a factor of the following expressions: 7. x2+ y2 −x1y-ry is divisible by (1) x-y, (2) x+y. 8. 23—122+27x+40 is divisible by - 5. x 9. æ3―J œ2 —a+9 is divisible by a+ 3. 10. a3+ a2-43a+20 is divisible by a- 3. 11. a-5ab+11 c3b2 — 14 a2b3 +9 ab1 — 2 65 is divisible by a-b, 18. When 3-5+10+11 x -6 is divided by x-3. 19. When 3+5x2+3x+2 is divided by x- — 1. ---- 20. When a3 − 4 xa + 7 x23 − 11-13 is divided by x −5. 21. When ao + 3 x2 15+2 is divided by x- 2. 22. What must be the value of the coefficient of in+ar2 - 10x+113 in order that the expression may be divisible by x+4? 23. For what values of a in No2+7 ac2+16 a2x-4a2 will the expression be divisible by x— 1? x 102. Case VII.-Compound Expressions which have a Linear Binomial Factor. A compound expression involving x and y is divisible, according to the factor theorem (101) by x -y if the expression vanishes when +y is substituted for x; and is divisible by x+y if the expression vanishes when y is substituted for x. Put 1. xn - yn " is divisible by x y, whether n is odd or even. y for x in " — y", then " — y" — y" — y" — 0 whether n is odd or even. Since y" — y” = 0, x" — y" is divisible by x-y, whether n is odd or Since y"y" = 0, x" +y" is divisible by x+y, if n is odd. 4. "y" is in no case divisible by x y. Put y for x in "+y"; then "+y" = y"+y" = 2y". It follows from these four cases, that: x” — y" — (.x — y)(xn−1 + xn−2y + x2¬3y2+ · -3 iii. For all positive odd integral values of n, iv. +xy"-2+y"−1). x"+y"=(x+y) (x"−1—‚x"¬2y+x2-3 y2 —, +, · · · —xy"-2+y^~1). n-3.2 —,+, "y" is never divisible by x-y, and is not divisible by x+y if n is even. NOTE. If the terms of the expression have a common monomial factor it should first be removed before applying the preceding rules. The proofs of the formulae in 1, 2, 3, 4 can be established by division, thus: The law of formation of the terms of the quotient is readily observed. All the terms are positive; the exponents of a decrease continually by unity, and those of a increase continually by unity. Since the remainder from the division is exact, whatever the integral exponent of m is, it follows, therefore, xm am = xm-1+axm−2+ a3xm−3+ x-a +am¬3x+am−1, This result has many applications in Algebra. When the difference of two squares is involved in an example, the principle of 94 should always be used. EXAMPLES.-1. Factor + a3. By 102, 3, x3 + a3 = (x+a) (x1 — x3a + x2a2 — xa3 + a1). 2. Factor 32 b5 + 243 c3. 326+243 c (2b)+(3c)5 =(2b+3c)[(2b)*—(2b)3 (3c)+(2b)2(3c)2−(2b) (3c)3+(3c)*] =(2b+3c) (16 b-24 b3c +36 b2c-54bc3+81 c1). 103. Case VIII. To Factor Expressions containing Four Terms. When a polynomial contains four terms which can be arranged in pairs so that each pair of terms contains the same binomial factor, the polynomial may be factored as follows: Divide the polynomial by the common binomial factor; then the divisor will be one factor and the quotient the other. Thus: Notice that the first and the second terms contain the common factor X, and the third and the fourth terms have the common factor b; therefore, the first two terms and the last two can be factored by Case I, and the result is x(x − a) + b (x − a); divide by x — and the quotient x+b, the other factor, is obtained. Thus, x2 — ax + bx — ab = x(x − a) +b(x − a) = (x + b)(x − a). (2) — SOLUTION OF EQUATIONS BY FACTORING 104. The ability the student has acquired in factoring rational integral expressions enables him to solve an extensive class of equations which are of a degree higher than the first. Suppose that it is desired to solve the equation x2-9x+20 Factoring the first member, (1) - : 0. x2 9x+20 (μ· 5) (x-4)= 0 Any value of x which makes the first member of equation (1) vanish is called a root of the equation. To solve equation (1) is to answer the question, what values of will make the first member of this equation equal to zero? The values of c which will make either of the factors 5 or x -4 4) equal to equal to zero will make the first member (x — 5) (x — zero. Therefore the roots of (1) are the values of x obtained from x3 − 2 x2 9x18 (x − 2) (x — 3) (x +3) 2. Solve the equation +4ax +4a2b2 = 0. 3. Solve the equation + 2x2+5x2+10x2-6x-12= 0. Here x+2x1+5.x3+10.c2-6x-12 _ _ (x+2) (x1+5x2—6) [101] [298, 3] V 6 can not be found, because there is no number which multiplied by itself will give -6. V 6 is called an imaginary number. |