Question 94063
Descending order involves the exponents of x. In descending order the exponents of x decrease 
as you read from left to right.
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The following is an example of the descending order of x:
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{{{x^5}}} {{{x^4}}} {{{x^3}}} {{{x^2}}} {{{x^1}}} {{{x^0}}}
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Of course the exponents could be bigger than the 5 I started with.  And each of the terms 
in this descending order may have a multiplier associated with it ... for example, the series
of terms 
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{{{5x^5}}} {{{3x^4}}} {{{7x^3}}} {{{4x^2}}} {{{-2x^1}}} {{{14x^0}}}
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is arranged in descending powers of x.  Some of the multipliers could be zero, and that
would cause that power of x to disappear.
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One thing to note.  What is {{{x^0}}}? Remember the rule that says anything raised to
a zero power is equal to 1. So {{{x^0 = 1}}}. And also remember the rule that says 
anything raised to the first power is itself. So {{{x^1 = x}}}.
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Now back to your problem you are given {{{8 - x }}} and asked to arrange it in descending
powers of x.
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Note that {{{8 = 8 * 1 = 8*x^0}}}. And note that {{{-x = -1*x^1}}}
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Therefore, the answer to your problem is that in descending powers of x, the numbers 
are:
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{{{-x^1 + 8x^0 = -x + 8}}}
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and the minus x is to the first power, and the +8 is to the zero power. Therefore, 
the highest exponent involved is 1 and this makes it a first degree or first order binomial.
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Hope this is what you were looking for.