college algebra

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1. solve the following logarithmic equation log_8(x+8)+log_8(x+7)=2
what is the exact solution

2.se the rational theorem to find all the real zeros of the polynomial function. use the zeros to factor f over the real numbers.
f(x)=4x^4+5x^3+9x^2+10x+2
a.find the real zeros of f. x=?
b.then use the real zeros to factor f. f(x)=?

please show work.

  • college algebra -

    using rules of logs, your equation becomes
    log8 ((x+8)/(x+7)) = 2

    (x+8)(x+7) = 8^2 = 64
    x^2 + 15x + 56 - 64 = 0
    x^2 + 15x - 8 = 0
    x = (-15 ± √257)/2

    but x > -7 or else the log part would be undefined
    so x = (-15 + √257)/2

  • college algebra -

    2.
    if rational roots exist, they must be ±1, or ±2 or ±1/2, ±1/4
    we can rule out any positive x's since all the terms have positive and will never add up to zero

    f(-1) = 4 - 5 + 9 - 10 + 2 = 0 , yeahh
    by synthetic division
    4x^4+5x^3+9x^2+10x+2
    = (x+1)(4x^3 + x^2 + 8x + 2) , again all positives in the 2nd factor so try x = -1, -2, -1/2, -1/4
    x=-1
    f(-1) = -4+1 -8 + 2 ≠ 0
    f(-2) = -32 + ... ≠0
    ...
    f(-1/4) = ... 0
    so (4x+1) is a factor
    Using long algebraic division I ended up with
    4x^4+5x^3+9x^2+10x+2 = (x+1)(4x+1)(x^2 + 2)

    so there are 2 real roots or zeros:
    x = -1 and x = -1/4

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