Solve |3^4/4-2(3)^3+11(3)^2

$$| \frac { 3 ^ { 4 } } { 4 } - 2 ( 3 ) ^ { 3 } + 11 ( 3 ) ^ { 2 }$$

Evaluate

$$\frac{261}{4}=65.25$$

Solution Steps

Calculate 3 to the power of 4 and get 81.

$$|\frac{81}{4}-2\times 3^{3}+11\times 3^{2}|$$

Calculate 3 to the power of 3 and get 27.

$$|\frac{81}{4}-2\times 27+11\times 3^{2}|$$

Multiply 2 and 27 to get 54.

$$|\frac{81}{4}-54+11\times 3^{2}|$$

Convert 54 to fraction \frac{216}{4}.

$$|\frac{81}{4}-\frac{216}{4}+11\times 3^{2}|$$

Since \frac{81}{4} and \frac{216}{4} have the same denominator, subtract them by subtracting their numerators.

$$|\frac{81-216}{4}+11\times 3^{2}|$$

Subtract 216 from 81 to get -135.

$$|-\frac{135}{4}+11\times 3^{2}|$$

Calculate 3 to the power of 2 and get 9.

$$|-\frac{135}{4}+11\times 9|$$

Multiply 11 and 9 to get 99.

$$|-\frac{135}{4}+99|$$

Convert 99 to fraction \frac{396}{4}.

$$|-\frac{135}{4}+\frac{396}{4}|$$

Since -\frac{135}{4} and \frac{396}{4} have the same denominator, add them by adding their numerators.

$$|\frac{-135+396}{4}|$$

Add -135 and 396 to get 261.

$$|\frac{261}{4}|$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of \frac{261}{4} is \frac{261}{4}.

$$\frac{261}{4}$$

Factor

$$\frac{3 ^ {2} \cdot 29}{2 ^ {2}} = 65\frac{1}{4} = 65.25$$