Algebra Question #5

since r + s + t = -4 and rst = -2. And (1/st) + (1/rt) + (1/rs)= (r+s+t)/rst = -4/-2 = 2

Given the roots of $p(x) = x^3 + 4x^2 + 3x +2$ are r, s and t, find $\frac{1}{st} + \frac{1}{rt} + \frac{1}{rs}$ .

Note that the desired value involves the reciprocals of p(x)'s roots.

By reversing the order of p(x)'s coefficient $^1$ , we get a new polynomial, $q(x) = 2x^3 + 3x^2 + 4x +1$ with reciprocal roots, $\frac{1}{r} , \frac{1}{s} \text{ and } \frac{1}{t}$ .

Therefore, $\boxed{\frac{1}{st} + \frac{1}{rt} + \frac{1}{rs} = 4/2 = 2}$

$^1$ See Chapter 4, Transforming Polynomials, Example 4.1 of Khan, Adeel. "A Few Elementary Properties of Polynomials." (2006)

x^3 + 4x^2 +3x + 2 = 0

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Vieta's formulas

r + s + t = -b/a = -(4)/(1) = -4

rs + rt + st = c/a = (3)/(1) = 3

r s t = -d/a = -(2)/(1) = -2

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1/st + 1/rt + 1/rs = (r + s + t) / (r s t) = (-4) / (-2) = 2

$\color{#3D99F6}{\frac{1}{st}+\frac{1}{rt}+\frac{1}{rs}=\frac{r+s+t}{rst}}$ From Vieta,s formulas,we know that $r+s+t=-4.rst=-2$ so : $\color{#D61F06}{\frac{r+s+t}{rst}=\frac{-4}{-2}=\boxed{2}}$