What are the rational roots of the polynomial f(x)=20x4+x3+8x2+x-12

The rational root theorem tells you they will be of the form ...

... ±(divisor of 12)/(divisor of 20)

Hence any rational root will be one of ...

... {±1/20, ±1/10, ±3/20, ±1/5, ±1/4, ±3/10, ±2/5, ±1/2, ±3/5, ±3/4, ±4/5, ±1, ±6/5, ±3/2, ±2, ±12/5, ±3, ±4, ±6, ±12}

A graph shows them to be -4/5 and +3/4.

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Dividing (5x+4) and (4x-3) from f(x) gives a quotient of (x²+1). Since there are only two real zeros, we already know the remaining roots are complex. This factor tells us they are 1±i.

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752