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Math Help - finding zeros

  1. #1
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    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).


    Thanks for your time.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Mr_Green View Post
    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).


    Thanks for your time.
    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
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  3. #3
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    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 \\<br />
r_1r_2 + r_2r_3 + r_1r_3 & = \;-3 \\r_1r_2r_3 & = \;\;\;7 \end{array}<br />
\begin{array}{ccc}(1)\\(2)\\(3)\end{array}


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








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  4. #4
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    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 \\<br />
r_1r_2 + r_2r_3 + r_1r_3 & = \;-3 \\r_1r_2r_3 & = \;\;\;7 \end{array}<br />
\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}

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