3 thousands 2 hundreds 15 tens 2 ones

Mathematics

2 answers:

baherus [9]3 years ago

7 0

3,352 is the answear.

Leona [35]3 years ago

6 0

32,153
good luck hope this help need more help just send me a message

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I don’t understand :(

viktelen [127]

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.

I know this might be difficult to follow, so please comment if you have any questions.