Posted by **Ladybug-two questions- Reiny please help** on Saturday, December 1, 2012 at 11:08am.

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 -
**Reiny**, Saturday, December 1, 2012 at 11:45am
using rules of logs, your equation becomes

log_{8} ((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 -
**Reiny**, Saturday, December 1, 2012 at 3:38pm
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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