127. + 1. · a √ x2 + a + √x2 - a { (a - b)a + 4a (a + b) — 4a3 }. { ao ( 1–26) √ {(a x + y \(x − y)2 + x − y)° 2x2 (x2+y3) 4x3 (x − y)* 132. √√√4(√a+√b)°—16 √ab+√16{a°+b*+4√ab(a+b)+6ab} Ja first by rationalizing the numerator, and then by rationalizing the denominator; and show that the sum of the results is, as it ought to be, the double of the original expression. 135. Multiply 7xa - 8x3- 5x + 3 by 3x-11x-9, using detached coefficients; and divide the result by 3x-9x-6 by the same method. 136. Write down the coefficient of x in the product of ax + bx3- cx + ex2 + fx-h by kx* + lx2 + mn - n without actual multiplication; and state how the result is obtained. 137. Multiply together √√xx2+x"y" + 3"", √x" -y", √x2-x"y"+y" and "+y". 138. Multiply together 1 1 ' (aa +ab+b")', (a−b)', (a−b)" and (a2 + ab + b2)". 139. Multiply 4a3b - 2a3b3 - 6ab3 + 3b* by ab3 - 2a-1b, using detached coefficients. 140. Square the expression -1+√√√√-1. 141. Cube the expression a√x-bay. 142. Extract the square root of 145. In any equation x+y=a+b, which involves rational quantities and quadratic surds, show that the rational parts on each side are equal, and also the irrational parts: and extract in the form of a binomial surd the square root of 146. Extract the square root of 3/3-2/6, without using a formula: show that 2+3 is the reciprocal of 2-√3, and state generally what must be the connexion between the two terms w and √x, so that their sum may be the reciprocal of their difference. Find the square roots of the following binomial surds, without using a formula: 153. Divide x + (a3 — 2b2) x − (a* - b1) x2 - ao - 2a1b3 — a2b1 by x-a-b, by Horner's or the "synthetic" method. 154. Arrange 6(x3+y3)+(18xy−4)(x+y) — 8 (x2+y3)—16xy-120 and x2 + y2+ 2x (1+ y) + 2y +6 according to the powers of (x + y), and divide the former by the latter, by the synthetic method. by 155. Divide 10a3xy-60axy3 + 6a1x3- 9a2x*y* 3x*y* - 5axy3 + a3x1y, by the synthetic method, giving six terms of the quotient, arranged according to the ascending powers of y. 4 ха 156. Divide 3x3- 9x+ by 4x2+8, using the synthetic me thod, and giving the quotient as far as the term involving x 157. Divide x3-3x3-31x2+25x+3x3-15x1-8x3 +19x2+3x+10 by 3x-21x+9x-6, by the synthetic method; showing the "final remainder," and then carrying on the quotient to twelve terms. 158. Expand synthetic division. 159. Expand 1-x and 3x*- 8x in series by x+1 х 1 in a series to five terms, by Horner's mode of division; and show the arithmetical equality of the fraction and resulting series, when x= = 3. by x-4x+2x-2, showing the final remainder, by the synthetic method. by 161. Find the remainder left after dividing 5у1 — 37 yo — 5y3 — 27y* + 7y3 — 10y – 6 4y* — 28y3 — 16y3 – 24y + 8, synthetically; the quotient involving only positive powers of y. 162. Divide 2a-1b+10b5-4ab1-32a2b3+16ab2 - 37ab+2a3 +66ab-1- 8a7b-2 by 2a-'+4a-b1-6a-3b28a-26-3, by Horner's method, showing the remainder. 163. Divide 4x1y-23x y3z-36x3y3z2+26x*y*z3— 52x3y3z*—30x3y°z5+16xy ̃z®+35y8z7 by 4x3yz ̄2 - 3xy3z1-7y3 to such extent that the quotient may involve no negative powers of x, and give the remainder, 164. Divide 2a3b-2 +9a1b ̄1 — 9a3 - &c., the remaining coefficients being - 21+ 69–74+ 68 − 15 - 36 +29, by 2a ̄1b+3a ̄2b3 — 8a ̄3b® + 4a ̄1b7, giving the remainder whose first term involves a-2b5. 165. Divide 2x1y ̄2 – 14x2 + 22xy – 22y3 + 16x ̄1y3 + 9x ̄2y1 — 12x ̄3y3 by 2x ̄1y1 — 4x ̄3y3 - 2x ̄y1o, by Horner's method, and give the remainder. 166. 10 -2.2 11 -8 Divide 3x2yz1+4x+6x'y ̄'z-21x y z2 + 16xy ̃31⁄23 -11xyz-8x3y-z - 5x3y-z-3Sxy+33x"1y- by 3x1y− 2x3z ̄2 + xoy ̄1z1 — 5x1y-2, and give the fiual remainder. -3 -3 -3 167. Divide 18x ̄*y*+6x-y+20x-y2+31xy+2x+6x-2y-1 + 39x ̄1y ̄2 — 64y ̃3 +20xy ̄1 by 3x ̄1y ̄2 — x ̄3y ̄3 + 5x ̄3y ̄* — 2x ̄1y ̄3, by Horner's method, and give the remainder. 168. Divide 8a-5.68+17a-4b7 + a ̄3bε — 9a-2b5+ 8a-1b* — 763 +6ab2 – 5a3b + 4a3 — 4a1b ̃1 + a3b-2 by a3 + 3a1b ̄' + 2a3b ̄2 — ao.b¬3 by Horner's method. --- 169. Divide 3ax ̄1y + 7y6 — 5a ̄1xy3 +35α ̄2x2y1 — 22a ̄3ñ3y3 +18a ̄*x*y2+30a ̄3x3y − 60a ̄εx®+18a ̄x1y ̄1+5a ̄x3y¬3 -4 -3 by 3x-3y3-2α ̄1x ̄ˆy+7a ̄2x ̄3— 5a ̄3x ̄2у ́ ̄ — a ̄x ̄1y ̃3, by Horner's method; giving the remainder whose first term involves a ̄x. metical equality of the fraction and its expansion, when x=1; taking four terms of the series. |