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
