Find the quotient using long division. (I have to show work.) (8m^3 + 38m^2 - 6m + 20)/(m + 5) Answers being: (A)8m^2+2m+4 (B)m^2+2m+8 (C)8m^2-2m+4 (D) m^2+3m+4 Second question: Start with this: ---------------------- m + 5) 8m3 + 38m2 - 6m + 20 Divide m into 8m3, which amounts to 8m3/m or 8m2. Place that above the line above 38m2: 8m2 ---------------------- m + 5) 8m3 + 38m2 - 6m + 20 Multiply 8m2 by the 5, getting 40m2 and place it under the 38m2 8m2 ---------------------- m + 5) 8m3 + 38m2 - 6m + 20 + 40m2 Multiply 8m2 by the m, getting 8m3 and place this under the other 8m3 8m2 ---------------------- m + 5) 8m3 + 38m2 - 6m + 20 8m3 + 40m2 Draw a line underneath 8m2 ---------------------- m + 5) 8m3 + 38m2 - 6m + 20 8m3 + 40m2 ---------- Subtract vertically by mentally changing the sign of the 8m3 and the 40m2 and adding, getting 0 and -2m2. Write only the -2m2 8m2 ---------------------- m + 5) 8m3 + 38m2 - 6m + 20 8m3 + 40m2 ---------- -2m2 Bring down the -6m 8m2 ---------------------- m + 5) 8m3 + 38m2 - 6m + 20 8m3 + 40m2 ---------- -2m2 - 6m Divide m into -2m2, which amounts to -2m2/m or -2m. Place that above the line above " - 6m ": 8m2 - 2m ---------------------- m + 5) 8m3 + 38m2 - 6m + 20 8m3 + 40m2 ---------- -2m2 - 6m Multiply -2m by the 5, getting -10m and place it under the "- 6m" 8m2 - 2m ----------------------- m + 5) 8m3 + 38m2 - 6m + 20 8m3 + 40m2 ---------- -2m2 - 6m - 10m Multiply -2m by the m, getting -2m2 and place this under the other -2m2 8m2 - 2m ----------------------- m + 5) 8m3 + 38m2 - 6m + 20 8m3 + 40m2 ---------- -2m2 - 6m -2m2 - 10m Draw a line underneath: 8m2 - 2m ----------------------- m + 5) 8m3 + 38m2 - 6m + 20 8m3 + 40m2 ---------- -2m2 - 6m -2m2 - 10m ---------- Subtract vertically by mentally changing the sign of the -2m2 and the -10m and adding, getting 0 and 4m. Write only the 4m. 8m2 - 2m ----------------------- m + 5) 8m3 + 38m2 - 6m + 20 8m3 + 40m2 ---------- -2m2 - 6m -2m2 - 10m ---------- 4m Bring down the +20 8m2 - 2m ----------------------- m + 5) 8m3 + 38m2 - 6m + 20 8m3 + 40m2 ---------- -2m2 - 6m -2m2 - 10m ---------- 4m + 20 Divide m into 4m, which amounts to 4m/m or "+ 4". Place that above the line above "+ 20": 8m2 - 2m + 4 ----------------------- m + 5) 8m3 + 38m2 - 6m + 20 8m3 + 40m2 ---------- -2m2 - 6m -2m2 - 10m ---------- 4m + 20 Multiply 4 by the 5, getting +20, and place it under the other + 20 8m2 - 2m + 4 ----------------------- m + 5) 8m3 + 38m2 - 6m + 20 8m3 + 40m2 ---------- -2m2 - 6m -2m2 - 10m ---------- 4m + 20 + 20 Multiply 4 by the m, getting 4m and place this under the other 4m 8m2 - 2m + 4 ----------------------- m + 5) 8m3 + 38m2 - 6m + 20 8m3 + 40m2 ---------- -2m2 - 6m -2m2 - 10m ---------- 4m + 20 4m + 20 Draw a line underneath 8m2 - 2m + 4 ----------------------- m + 5) 8m3 + 38m2 - 6m + 20 8m3 + 40m2 ---------- -2m2 - 6m -2m2 - 10m ---------- 4m + 20 4m + 20 ------- Subtract vertically by mentally changing the sign of the 4m and the +20 and adding, getting 0 and 0. Write only the second 0. 8m2 - 2m + 4 ----------------------- m + 5) 8m3 + 38m2 - 6m + 20 8m3 + 40m2 ---------- -2m2 - 6m -2m2 - 10m ---------- 4m + 20 4m + 20 ------- 0 Since the remainder is 0, the quotient is 8m2 - 2m + 4 So the correct choice is (C) ---------------------------------------------- Multiply: (3y+11)(8y²-2y-9) (I have to show work.) (3y+11)(8y²-2y-9) Answers being: (A) 24y^3-6y^2-27y+11 (B) 112y^2-28y-126 (C) 24y^3+28y^2-49y-99 (D) 24y^3+94y^2+49y+99 Thank you. Write the second parenthetical expression (8y²-2y-9) as [(8y²-2y)-9] (3y+11)[(8y²-2y)-9] I colored the (8y²-2y) red so you can see that you are using "FOIL" Consider the parenthetical red factor as just ONE single term. FIRSTS + OUTERS + INNERS + LASTS | | | | 3y(8y²-2y) + (3y)(-9) + (11)(8y²-2y) + (11)(-9) (24y³-6y²) + (-27y) + (88y²-22y) + (-99) 24y³ - 6y² - 27y + 88y² - 22y - 99 Combining like terms and arranging terms in descending order: 24y³ + 82y² - 49y - 99 That answer is correct even though you don't have it listed. My guess is that it is (C) and you inadvertently reversed the digits of 82 as 28. Edwin AnlytcPhil@aol.com