Descartes 4x^7+3x^6+x^5+2x^4-x^3+9x^2+x+1

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Using Descartes' Rule of Signs, determine the number of real solutions to 4x7 + 3x6 + x5 + 2x4 - x3 + 9x2 + x + 1

We first evaluate the possible positive roots using ƒ(x) = 4x7 + 3x6 + x5 + 2x4 - x3 + 9x2 + x + 1
There are 2 sign change(s) detailed below:
Sign Change 1) + to -
Sign Change 2) - to +

To find the remaining possible positive roots, we count down in pairs until we pass zero.
2 roots - 1 pair (2 roots) = 0
Therefore, we have a possible combination of (2 or 0) positive roots

Calculate possible negative roots:
Given ƒ(x) = 4x7 + 3x6 + x5 + 2x4 - x3 + 9x2 + x + 1, we first need to determine ƒ(-x)
ƒ(-x) = 4(-x)7 + 3(-x)6 + (-x)5 + 2(-x)4 - (-x)3 + 9(-x)2 + (-x) + 1

-x raised to an even power is positive. Odd exponents become negative:
4(-x)7 has a positive constant and odd exponent for a negative result of -4x7
3(-x)6 has a positive constant and even exponent for a positive result of + 3x6
(-x)5 has a positive constant and odd exponent for a negative result of - x5
2(-x)4 has a positive constant and even exponent for a positive result of + 2x4
-(-x)3 has a negative constant and odd exponent for a positive result of + x3
9(-x)2 has a positive constant and even exponent for a positive result of + 9x2
(-x) has a positive constant and odd exponent for a negative result of - x
1 has a positive constant and even exponent for a positive result of + 1
ƒ(-x) = -4x7 + 3x6 - x5 + 2x4 + x3 + 9x2 - x + 1

We first evaluate the possible negative roots using ƒ(x) = - 4x7 + 3x6 - x5 + 2x4 + x3 + 9x2 - x + 1
There are 5 sign change(s) detailed below:
Sign Change 1) - to +
Sign Change 2) + to -
Sign Change 3) - to +
Sign Change 4) + to -
Sign Change 5) - to +

To find the remaining possible negative roots, we count down in pairs until we pass zero.
5 roots - 1 pair (2 roots) = 3
3 roots - 1 pair (2 roots) = 1
Therefore, we have a possible combination of (5 or 3 or 1) negative roots