Question 147616: Please help me with these, I'm so confused!
Divide using polynomial long division.
a) (x^2 + 5x - 14) divided by (x-2)
b) (x^3 + x +30) divided by (x+3)
c) (8x^3 + 5x^2 - 12x + 10) divided by (x^2 - 3)
I have no idea how to even start them.
Found 2 solutions by scott8148, jim_thompson5910:
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! these work just like long division from grammar school
when you divide 12 into 156
__ 12 goes into 15, 1 times __ 1 times 12 is 12 __ 15 minus 12 is 3 __ bring down the 6 __ 12 goes into 36, 3 times
__ so; 12 goes into 156, 13 times
suppose we replace the 10 with x
__ so; 12 is x+2, and 156 is x^2+5x+6
__ x goes into x^2, x times __ x times x+2 is x^2+2x __ (x^2+5x)-(x^2+2x) is 3x __ bring down the 6 __ x goes into 3x, 3 times
__ 3 times x+2 is 3x+6 __ (3x+6)-(3x+6) is 0 (no remainder)
__ so; x+2 (12) goes into x^2+5x+6 (156), x+3 (13) times
a) x goes into x^2, x times __ x times x-2 is x^2-2x __ (x^2+5x)-(x^2-2x) is 7x __ bring down the -14
__ x goes into 7x, 7 times __ 7 times x-2 is 7x-14 __ (7x-14)-(7x-14) is 0
__ so; x-2 goes into x^2+5x-14, x+7 times
b) this works the same way, except that there is a "missing" term
__ if we replace 10 with x, then 103 becomes x^2+3 __ for dividing, it is easier if we write x^2+0x+3
__ so; x^3+x+30 can be x^3+0x^2+x+30 __ in decimal (regular) numbers, the zero acts as a "place holder"
__ x goes into x^3, x^2 times __ x+3 times x^2 is x^3+3x^2 __ (x^3+0x^2)-(x^3+3x^2) is -3x^2 __ bring down the x
__ x goes into -3x^2, -3x times __ -3x times x+3 is -3x^2-9x __ (-3x^2+x)-(-3x^2-9x) is 10x __ bring down the 30
__ x goes into 10x, 10 times __ 10 times x+3 is 10x+30 __ (10x+30)-(10x+30) is 0
__ so; x+3 goes into x^3+x+30, x^2-3x+10 times
try c) __ remember the "place holder" __ x^2+0x+3
Answer by jim_thompson5910(35256)  (Show Source):
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