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Lerok [7]
3 years ago
10

the value of an investment decreased by 10%. The original amount of the investment was $75 what is the current value of the inve

stment?​
Mathematics
1 answer:
pishuonlain [190]3 years ago
6 0

Answer:

67.5

Step-by-step explanation:

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I don’t understand :(
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Answer:

The quotient when x^3-4x^2-8x+8 is divided by x+2 is x^2-6x+4 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 (x-a). In this case, we may use synthetic division, since x+2 is equal to x-(-2). 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 x^2-6x+4, and the remainder is zero (which indicates that x+2 is a factor of p(x)).

--

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 x \times x^2 is equal to x^3, the first term in the quotient will be x^2. Since x^2 \times (x+2) is equal to x^3+2x^2, we must now subtract that from x^3-4x^2-8x+8.

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>

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