Solution:
As per the problem
Maxine took out a loan for $3200 at 8% interest, compounded annually.
she makes yearly payments of $250.
she will be able to Payoff her loan iff yearly payments made is greater than yearly interest.
Interest payable in first year
Interest payable in first year
This payable interest will goes on increasing year on year.
Hence Maxine will never payoff the loan.
Answer:
The quotient when is divided by
is
with no remainder, so x+2 is a factor of p(x).
Step-by-step explanation:
As the question suggests, there are two ways to solve this problem: long division (which can always be used to divide polynomials), and synthetic division (which can only be used when you are dividing by something of the form . In this case, we may use synthetic division, since
is equal to
. I will use synthetic division here as it is slightly faster than long division.
The -2 on the left represents that we are dividing by x-(-2), and the rest of the numbers in the top row are the coefficients of p(x): 1, -4, -8, and 8.
The first step in synthetic division is to being down the leading coefficient of the polynomial: in this case, the 1 (as indicated in red). Now, we multiply the 1 by -2. We get -2, which we place directly below the -4.
Next, we add directly down the column, (-4 + (-2) is equal to -6), and this answer is placed in the box below. We can continue this process, getting the coefficients 1, -6, 4, and 0 in the bottom row.
This is the answer: the quotient is , and the remainder is zero (which indicates that x+2 is a factor of p(x)).
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Edit: long division
We may also solve this problem using long division (see the second image). The first step is to look at the leading coefficients: since is equal to
, the first term in the quotient will be
. Since
is equal to
, we must now subtract that from
.
We repeat the same process, as shown in the image. Since we eventually get to zero, the remainder is zero, and the polynomial at the top (x^2-6x+4) is the quotient.
<em>I know this might be difficult to follow, so please comment if you have any questions.</em>
