Andrew M. answered 07/20/15
Tutor
New to Wyzant
Mathematics - Algebra a Specialty / F.I.T. Grad - B.S. w/Honors
1. I think you left off the exponent on the squared term...
(25x3+25x2+5x-6 )/(x +4/5)
If x + 4/5 is a root of the polynomial then we should get no remainder if we set x = -4/5
Thus, in the synthetic division, we use -4/5 as the divisor
Also note that we must write the coefficients of all the terms in order of highest to lowest power, inputting zero for any missing exponents... Our coefficients are 25, 25, 5, -6
-4/5 / 25 25 5 -6
-20 -4 -4/5
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25 5 1 -6 4/5 The last term is the remainder R = -6 4/5
Bring down the 1st term.. 25... multiply it by the divisor .. -4/5... and take the product.. -20... up
under the 2nd term... Add down. Multiply that sum by the divisor and again put the product under
the next coefficient and again add down. Do this to the last term. The final number on the bottom
right is the remainder. The other terms are the coefficients in descending power order from left to right.
(25x3+25x2+5x-6 )/(x +4/5) = 25x2+5x + 1 + (-6 4/5)/(x+4/5)
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2. x+14/(x-9) = [x(x-9)]/(x-9) + 14/(x-9) = [x(x-9)+14]/(x-9) = (x2-9x+14)/(x-9)
= (x-7)(x-2)/(x-9)
There are many equivalent expressions. Do you have a multiple choice list from which you are
supposed to choose an answer?
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3.
(5/2)2(5/2)4 = (5/2)2+4=(5/2)6 = 56/26 = 15625/64 = 244.140625
The simplified expression they are probably looking for is 56/26
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4. x + y = 1 eqn 1
x + y = -7 eqn 2
I can already see that should be no solution to this because no two numbers will add
to both 1 and also add to -7. But, let's work it out.
From eqn 1 we can rearrange to solve for x in terms of y...
x = 1-y
Replace x with 1-y in eqn 2
1-y+y = -7
1=-7
Since we arrive at an equation which is patently untrue there is no solution.