Why isn’t there a commutative property for subtraction as well as for addition?
1 answer:
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Answer:
-1
Step-by-step explanation:
See the attachment for the polynomial long division. The constant in the quotient is -1.
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Here, there is a remainder of -x. If there were no remainder the constant in the quotient is the ratio of the constant in the dividend to the constant in the divisor: -2/2 = -1.
That could be a first guess in a "guess and check" solution approach.
<em>Guess</em>: first term of binomial quotient is (2x^3)/x^2 = 2x; last term of binomial quotient is -2/2 = -1. So, the quotient is guessed to be (2x -1).
<em>Check</em>: (2x -1)(x^2 -x +2) = 2x^3 -3x^2 +5x -2
Subtracting this from the actual dividend gives a remainder of -x. This has a lower degree than the divisor, so no further adjustment of the quotient is required.
