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guajiro [1.7K]
2 years ago
5

Find all zeros of x^4+x

Mathematics
1 answer:
Yuri [45]2 years ago
4 0

Answer:

-1 and 0

Step-by-step explanation:

X ^4+x = 0;

x*(x^3+1)=0;

x*[(x^3-x)+(x+1)]=0;

x*[x(x-1)(x+1)+(x+1)]=0;

x*(x+1)(x^2-x+1)=0;

Since x^2-x+1=0 has no intersection with the X-axis, there is no x value such that y=0

x=0 or -1

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Since we have a four digit number, there are four digit placements.

For the first digit, since there can either be a 5 or an 8, we have the arrangement as ²P₁ = 2 ways.

For the second digit, we have ten numbers to choose from, so we have ¹⁰P₁ = 10.

For the third digit, since it neither be a 5 or an 8, we have two less digit from the total of ten digits which is 10 - 2 = 8. So, the number of ways of arranging that is ⁸P₁ = 8.

For the last digit, we have ten numbers to choose from, so we have ¹⁰P₁ = 10.

So, the number of integers that can be formed are 2 × 10 × 8 × 10 = 20 × 80 = 1600 integers

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