# college algebra

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)=?

1. 0
2. 1
1. 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

posted by Reiny
2. 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

posted by Reiny

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