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pickupchik [31]
3 years ago
14

Factor completely x3 + 4x2 + 8x + 32. (x + 4)(x2 + 8) (x − 4)(x2 − 8) (x − 4)(x2 + 8) (x + 4)(x2 − 8)

Mathematics
2 answers:
icang [17]3 years ago
7 0

\tt x^3+4x^2+8x+32\\\\=x^2(x+4)+8(x+4)\\\\=\boxed{\tt(x+4)(x^2+8)}

frutty [35]3 years ago
6 0

Answer:

(x+4)(x^2+8)

Step-by-step explanation:

x^3 + 4x^2 + 8x + 32

We will factor by grouping

Factor out an x^2 from the first two terms and an 8 from the last two terms

x^2 (x+4) + 8(x+4)

Now we can factor out an (x+4)

(x+4)(x^2+8)

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Six sophomores and 14 freshmen are competing for two alternate positions on the debate team. Which expression represents the pro
Elanso [62]

Answer:

<em>Choose the first alternative</em>

\displaystyle P=\frac{_{1}^{6}\textrm{C}\ _{1}^{5}\textrm{C}}{_{2}^{20}\textrm{C}}

Step-by-step explanation:

<u>Probabilities</u>

The requested probability can be computed as the ratio between the number of ways to choose two sophomores in alternate positions (N_s) and the total number of possible choices (N_t), i.e.

\displaystyle P=\frac{N_s}{N_t}

There are 6 sophomores and 14 freshmen to choose from each separate set. There are 20 students in total

We'll assume the positions of the selections are NOT significative, i.e. student A/student B is the same as student B/student A.

To choose 2 sophomores out of the 6 available, the first position has 6 elements to choose from, the second has now only 5

_{1}^{6}\textrm{C}\ _{1}^{5}\textrm{C} \text{ ways to do it}

The total number of possible choices is

_{2}^{20}\textrm{C} \text{ ways to do it}

The probability is then

\boxed{\displaystyle P=\frac{_{1}^{6}\textrm{C}\ _{1}^{5}\textrm{C}}{_{2}^{20}\textrm{C}}}

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