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sergiy2304 [10]
4 years ago
5

How to factor 3x^2-10x-8?

Mathematics
2 answers:
Soloha48 [4]4 years ago
4 0

<em>answer \\(x - 4)(3x + 2)  \\ solution \\  {3x}^{2}  - 10x - 8 \\  =  {3x}^{2}  - (12 - 2)x - 8 \\  =  {3x}^{2}  - 12x + 2x - 8 \\  = 3x(x - 4) + 2(x - 4) \\  = (x - 4)(3x  + 2) \\ hope \: it \: helps</em>

Elan Coil [88]4 years ago
3 0

Answer:

Step-by-step explanation:

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Answer:

The only possible number is 80.

Step-by-step explanation:

The number in question needs to be a multiple of all three of 2, 8, and 10. As a result, it must also be a multiple of the least common multiplier (lcm) of the three number.

Start by finding the least common multiplier of the three numbers.

Factor each number into its prime components:

  • 2 is a prime number itself.
  • 8 = 2^3.
  • 10 = 2 \times 5.

The only prime factors are 2 and 5.

  • The greatest power of 2 among the three numbers is 3.
  • The greatest power of 5 among the three numbers is 1.

Therefore, the least common multiplier of the three number should be the product of 2^3 and 5. That's equal to 2^3 \times 5 = 8 \times 5 = 40.

In other words, the number (or numbers) in question could be written in the form 40\, k, where k is an integer.

The question requires that this number be between 55 and 101. In other words,

55 \le 40\, k \le 101.

The goal is to find the possible values of k. Note that from integer division by 40,

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The inequality becomes:

1 \times 40 + 15 \le 40\, k \le 2 \times 40 + 21.

However,

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Hence,

1 \times 40 < 1\times 40 + 15 \le 40\, k \le 2 \times 40 + 21 < 3 \times 40.

1 \times 40 < 40\, k < 3 \times 40.

Divide by the positive number 40 to obtain:

1 < k < 3.

Since k is an integers, k = 2.

Indeed, 40 \, k = 80 is between 55 and 101.

Therefore, 80 is the number in question.

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