prove that x^6-x^5-x^4-x^3-x^2-x+1>0 is true for any number
Please someone explain to me this, I couldn't understand the teacher when they talked about something similar to this and I didn't get it...
not even true,
e.g. let x = 1
x^6-x^5-x^4-x^3-x^2-x+1>0
1 - 1 - 1 - 1 - 1 - 1 - 1 + 1 > 0 is false
all I needed was ONE exception and the statement is false.
see:
http://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E6-x%5E5-x%5E4-x%5E3-x%5E2-x%2B1
To prove that the inequality x^6 - x^5 - x^4 - x^3 - x^2 - x + 1 > 0 is true for any number, we can use algebraic techniques.
First, let's factorize the expression:
x^6 - x^5 - x^4 - x^3 - x^2 - x + 1 = (x^6 + x) - (x^5 + x^4) - (x^3 + x^2 + x) + 1
Now, we can group the terms together:
x(x^5 + 1) - x^4(x + 1) - x(x^2 + 1) + 1
Next, let's focus on each group of terms individually:
1) x(x^5 + 1)
When x is positive, both x and (x^5 + 1) are positive, so their product is positive.
When x is negative, x is negative, but (x^5 + 1) is positive because raising a negative number to an odd power gives a negative result. Therefore, their product is negative.
2) -x^4(x + 1)
When x is positive, both x^4 and (x + 1) are positive, so their product is positive.
When x is negative, both x^4 and (x + 1) are negative, so their product is positive.
3) -x(x^2 + 1)
When x is positive, both x and (x^2 + 1) are positive, so their product is negative.
When x is negative, x is negative, but (x^2 + 1) is positive because squaring a negative number gives a positive result. Therefore, their product is positive.
Now, let's combine these results:
Positive term - Positive term + Negative term + Positive constant
Since the positive terms outweigh the negative terms and we have a positive constant, the entire expression is greater than zero for any value of x.
Therefore, the inequality is true for any number.