1. ## finding zeros

Let r1, r2, and r3 be the zeros of f(x) = x^3 + 4x^2 - 3x - 7. Find the exact value of the expression (r1 + 2)(r2 + 2)(r3 + 2).

2. Originally Posted by Mr_Green
Let r1, r2, and r3 be the zeros of f(x) = x^3 + 4x^2 - 3x - 7. Find the exact value of the expression (r1 + 2)(r2 + 2)(r3 + 2).

If r1, r2, r3 are the roots of f(x) = x^3 + 4x^2 - 3x - 7, then:

f(x) = x^3 + 4x^2 - 3x - 7 = (x-r1)(x-r2)(x-r3)

so:

f(-2) = (-2-r1)(-2-r2)(-2-r3) = -(r1 + 2)(r2 + 2)(r3 + 2)

But f(-2) = -8+16+6-7 = +7, so (r1 + 2)(r2 + 2)(r3 + 2) = -7.

RonL

3. Hello, Mr_Green!!

Are you familiar with Vieta's Theorem?

Let $r_1,\:r_2,\:r_3$ be the zeros of $f(x) \:= \:x^3 + 4x^2 - 3x - 7$
Find the exact value of the expression: $(r_1 + 2)(r_2 + 2)(r_3 + 2).$

If $r_1,\:r_2,\:r_3$ are the zeros of $f(x) \:=\:x^3 + 4x^2 - 3x - 7$

then: . $\begin{array}{ccc}r_1+r_2+r_3 & =\;-4 \\
r_1r_2 + r_2r_3 + r_1r_3 & = \;-3 \\r_1r_2r_3 & = \;\;\;7 \end{array}
\begin{array}{ccc}(1)\\(2)\\(3)\end{array}$

We are given: . $R \;=\;(r_1+2)(r_2+2)(r_3+2)$

4. Hello, Mr_Green!!

Are you familiar with Vieta's Theorem?

Let $r_1,\:r_2,\:r_3$ be the zeros of $f(x) \:= \:x^3 + 4x^2 - 3x - 7$
Find the exact value of the expression: $(r_1 + 2)(r_2 + 2)(r_3 + 2).$
If $r_1,\:r_2,\:r_3$ are the zeros of $f(x) \:=\:x^3 + 4x^2 - 3x - 7$

then: . $\begin{array}{ccc}r_1+r_2+r_3 & =\;-4 \\
r_1r_2 + r_2r_3 + r_1r_3 & = \;-3 \\r_1r_2r_3 & = \;\;\;7 \end{array}
\begin{array}{ccc}(1)\\(2)\\(3)\end{array}$

We are given: . $R \;=\;(r_1+2)(r_2+2)(r_3+2)$

. . . . . . . . . . . . . $=\;r_2r_2r_3 + 2r_2r_2 + 2r_1r_3 + 4r_1 + 2r_2r_3 + 4r_2 + 4r_3 + 8$

. . . . . . . . . . . . . $= \;\underbrace{(r_1r_2r_3)} + 2\underbrace{(r_1r_2 + r_2r_3 + r_1r_3)} + 4\underbrace{(r_1 + r_2 + r_3)} + 8$

. . . . . . . . . . . . . $= \quad\;\:7\quad\;+\qquad\quad\; 2(-3)\qquad\quad + \qquad4(-4)\quad\;\; + 8$

. . Therefore: . $\boxed{R \;=\;-7}$