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STatiana [176]
1 year ago
11

Which polynomial is prime? x3 + 3x2 – 2x – 6 x3 – 2x2 + 3x – 6 4x4 + 4x3 – 2x – 2 2x4 + x3 – x + 2

Mathematics
2 answers:
zhuklara [117]1 year ago
5 0

A prime polynomial is a polynomial that can't be factored.

In order to check which polynomial is prime, we need to check which polynomials could be factored.

x^3 + 3x^2 – 2x – 6 can be factored as (x+3)(x^2-2)

x^3 – 2x^2 + 3x – 6 can be factored as (x-2)(x^2+3)

4x^4 + 4x^3 – 2x – 2 can be factored as 2(x+1)(2x^3-1)

2x^4 + x^3 – x + 2 can't be factored.

Therefore, 2x^4 + x^3 – x + 2 is a prime polynomial.


alexandr1967 [171]1 year ago
4 0
D. 2x4+x3–x+2 <span>is prime</span>
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Milton spilled some ink on his homework paper. He can't read the coefficient of $x$, but he knows that the equation has two dist
Lera25 [3.4K]

Answer:

Sum = -81

Step-by-step explanation:

See the comment for complete question.

Given

c = 36 ----- Constant

No coefficient of x^2

Required:

Determine the sum of all distinct positive integers of the coefficient of x

Reading through the complete question, we can see that the question has 3 terms which are:

x^2 ---- with no coefficient

x ---- with an unknown coefficient

36 ---- constant

So, the equation can be represented as:

x^2 + ax + 36 = 0

Where a is the unknown coefficient

From the question, we understand that the equation has two negative integer solution. This can be represented as:

x = -\alpha and x = -\beta

Using the above roots, the equation can be represented as:

(x + \alpha)(x + \beta) = 0

Open brackets

x^2 + (\alpha + \beta)x + \alpha \beta = 0

To compare the above equation to x^2 + ax + 36 = 0, we have:

a = \alpha + \beta

\alpha \beta = 36

Where: \alpha, \beta and \alpha \ne \beta

The values of \alpha and \beta that satisfy \alpha \beta = 36 are:

\alpha = -1 and \beta = -36

\alpha = -2 and \beta = -18

\alpha = -3 and \beta = -12

\alpha = -4 and \beta = -9

So, the possible values of a are:

a = \alpha + \beta

When \alpha = -1 and \beta = -36

a = -1 - 36 = -37

When \alpha = -2 and \beta = -18

a = -2 - 18 = -20

When \alpha = -3 and \beta = -12

a = -3 - 12 = -15

When \alpha = -4 and \beta = -9

a = -4 - 9 = -13

At this point, we have established that the possible values of a are: -37, -20, -15 and -9.

The required sum is:

Sum = -37 -20 -15 - 9

Sum = -81

7 0
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Flura [38]

Answer:

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