Question

given the polynomial
p(x)=x⁴ + 2x³ - 3x + 1
according to descartes rule of signs, find the number of possible negative real roots.
p(-x) has changes in sign, which means that the polynomial has.
three one two
one negative real root three negative real roots or one negative real root two negative real roots or zero negative real roots
two negative real roots or one negative real root

complete the sentence:
the following polynomial, p(x)=x⁴ - 6x³ + 18x² - 54x + 81, has roots
3 with a multiplicity of, and 3i and.
1 3 2

Explanation:

Part 1: Descartes' Rule of Signs for Negative Real Roots

Step1: Compute \(P(-x)\)

Substitute \(x\) with \(-x\) in \(P(x)=x^4+2x^3-3x+1\):
$$P(-x)=(-x)^4+2(-x)^3-3(-x)+1=x^4-2x^3+3x+1$$

Step2: Count sign changes in \(P(-x)\)

Coefficients of \(P(-x)\): \(+1, -2, +3, +1\). Sign changes: \(+1→-2\) (1st), \(-2→+3\) (2nd), \(+3→+1\) (no change). Total: 2.

Step3: Determine possible negative roots

By Descartes' Rule, number of negative real roots is equal to sign changes or less by even number: \(2\) or \(0\).

Step1: Factor \(P(x)=x^4-6x^3+18x^2-54x+81\)

Notice \(P(x)=(x^2-3x+9)^2\). Check if \(x=3\) is a root: \(3^2-3×3+9=9-9+9=9≠0\)? Wait, correct factorization: \(P(x)=(x-3)^2(x^2+0x+9)\)? No, direct division: Divide \(P(x)\) by \((x-3)\):
$$x^4-6x^3+18x^2-54x+81=(x-3)(x^3-3x^2+9x-27)=(x-3)^2(x^2+9)$$

Step2: Confirm multiplicity of 3

Factor form: \((x-3)^2(x^2+9)\). So root 3 has multiplicity 2.

Answer:

First blank: two
Second blank: two negative real roots or zero negative real roots

Part 2: Multiplicity of Root 3