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2 years ago

Is X-1 a factor of x^5-1?

Mathematics

1 answer:

Musya8 [376]2 years ago

4 0

Answer: No

Explanation:

According to factor theorem, if f(x)=0 then x is a factor of the given function or equation.

As x-1 is a factor

We equate x-1=0

x=1

Substituting in x^5-1, we have 1^5-1 =1-1=0.

Hence, it's a factor.

When coming to x^5+1, it would become 1^5+1=1+1=2

So x-1 isn't a factor of x^5+1.

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Suppose the roots of the polynomial $x^2 - mx + n$ are positive prime integers (not necessarily distinct). Given that $m < 20

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<em>18</em> values for n are possible.

Step-by-step explanation:

Given the quadratic polynomial:

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To find:

How many possible values of $n$ are there ?

Solution:

First of all, let us have a look at the sum and product of a quadratic equation.

If the quadratic equation is:

$Ax^{2} +Bx+C$

and the roots are: $\alpha$ and $\beta$

Then sum of roots, $\alpha+\beta = -\frac{B}{A}$

Product of roots, $\alpha \beta = \frac{C}{A}$

Comparing the given equation with standard equation, we get:

A = 1, B = -m and C = n

Sum of roots, $\alpha+\beta = -\frac{-m}{1} = m$

Product of roots, $\alpha \beta = \frac{n}{1} = n$

We are given that $m$

$\alpha$ and $\beta$ are positive prime integers such that their sum is less than 20.

Let us have a look at some of the positive prime integers:

2, 3, 5, 7, 11, 13, 17, 23, 29, .....

Now, we have to choose two such prime integers from above list such that their sum is less than 20 and the roots can be repetitive as well.

So, possible combinations and possible value of $n (= \alpha \times \beta)$ are:

$1.\ 2, 2\Rightarrow n = 2\times 2 = 4\\2.\ 2, 3 \Rightarrow n = 6\\3.\ 2, 5 \Rightarrow n = 10\\4.\ 2, 7\Rightarrow n = 14\\5.\ 2, 11 \Rightarrow n = 22\\6.\ 2, 13 \Rightarrow n = 26\\7.\ 2, 17 \Rightarrow n = 34\\8.\ 3, 3\Rightarrow n = 3\times 3 = 9\\9.\ 3, 5 \Rightarrow n = 15\\10.\ 3, 7 \Rightarrow n = 21\\$

$11.\ 3, 11\Rightarrow n = 33\\12.\ 3, 13 \Rightarrow n = 39\\13.\ 5, 5 \Rightarrow n = 25\\14.\ 5, 7 \Rightarrow n = 35\\15.\ 5, 11 \Rightarrow n = 55\\16.\ 5, 13 \Rightarrow n = 65\\17.\ 7, 7 \Rightarrow n = 49\\18.\ 7, 11 \Rightarrow n = 77$

So,as shown above <em>18 values for n are possible.</em>

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