Let the roots of polynomial P(x) = x^{3}+2x^{2}+3x+ 101 be r_1, r_2 and r_3, find the value of \\left|r_1^2+r_2^2+r_3^2\\right|.
", "hints_md": "Use [Vieta's Formulas](https://artofproblemsolving.com/wiki/index.php/Vieta%27s_formulas).", "hints_html": "Use Vieta’s Formulas.
", "editorial_md": "First, let's recall Vieta's formulas for a cubic polynomial $P(x)=ax^3+bx^2+cx+d$. \n\nThe roots $r_1$, $r_2$, and $r_3$\u200b of this polynomial satisfy the following relationships:\n\n1. $r_1 + r_2 + r_3 = -\\frac{b}{a}$\u200b\n2. $r_1r_2 + r_2r_3 + r_3r_1 = \\frac{c}{a}$\n3. $r_1r_2r_3 = -\\frac{d}{a}$\n\nFor our polynomial $P(x)=x^3+2x^2+3x+101$, the coefficients are:\n- $a=1$\n- $b=2$\n- $c=3$\n- $d=101$\n\nUsing Vieta's formulas, we can write:\n\n1. $r_1 + r_2 + r_3 = -\\frac{2}{1} = -2$\u200b\n2. $r_1r_2 + r_2r_3 + r_3r_1 = \\frac{3}{1} = 3$\n3. $r_1r_2r_3 = -\\frac{101}{1} = -101$\n\nTo find $r^2_1+r^2_2+r^2_3$\u200b, we must make an expression that relates the sum of the squares of the roots to the sums and products of the roots: \n\nTo do this, we start with the squared sum of roots expression:\n$$(r_1+r_2+r_3)^2 $$\nWhen expanded, this becomes:\n $$r^2_1+r^2_2+r^2_3 +2(r_1r_2 + r_2r_3 + r_3r_1) $$\nThus, by subtracting $2(r_1r_2 + r_2r_3 + r_3r_1)$ from this expanded form, we isolate the sum of the squares of the roots:\n $$(r_1+r_2+r_3)^2-2(r_1r_2 + r_2r_3 + r_3r_1) $$\n Substituting the values from Vieta's formulas into this expression then gives us our answer: \n $$(r_1+r_2+r_3)^2-2(r_1r_2 + r_2r_3 + r_3r_1) = (-2)^2-2(3) = 4-6 = -2 $$\n Thus, the value of $r^2_1+r^2_2+r^2_3$ is $-2$, so the answer is $2$.\n\n- botman", "editorial_html": "First, let’s recall Vieta’s formulas for a cubic polynomial P(x)=ax^3+bx^2+cx+d.
The roots r_1, r_2, and r_3 of this polynomial satisfy the following relationships:
For our polynomial P(x)=x^3+2x^2+3x+101, the coefficients are:
Using Vieta’s formulas, we can write:
To find r^2_1+r^2_2+r^2_3, we must make an expression that relates the sum of the squares of the roots to the sums and products of the roots:
To do this, we start with the squared sum of roots expression:
(r_1+r_2+r_3)^2
When expanded, this becomes:
r^2_1+r^2_2+r^2_3 +2(r_1r_2 + r_2r_3 + r_3r_1)
Thus, by subtracting 2(r_1r_2 + r_2r_3 + r_3r_1) from this expanded form, we isolate the sum of the squares of the roots:
(r_1+r_2+r_3)^2-2(r_1r_2 + r_2r_3 + r_3r_1)
Substituting the values from Vieta’s formulas into this expression then gives us our answer:
(r_1+r_2+r_3)^2-2(r_1r_2 + r_2r_3 + r_3r_1) = (-2)^2-2(3) = 4-6 = -2
Thus, the value of r^2_1+r^2_2+r^2_3 is -2, so the answer is 2.