\( \frac{6}{5} x^{8}+\frac{2}{5} x^{7}+\frac{1}{5} x^{5}+\frac{1}{5} x^{4} \) Factor out the greatest monomial factor so that the polynomial factor has all integer coefficient \( \frac{6}{5} x^{8}+\frac{2}{5} x^{7}+\frac{1}{5} x^{5}+\frac{1}{5} x^{4}=\square \) (Factor completely.)

To factor the expression \( \frac{6}{5} x^{8}+\frac{2}{5} x^{7}+\frac{1}{5} x^{5}+\frac{1}{5} x^{4} \), we first identify the greatest common factor (GCF) of all the terms. Each term contains a factor of \( \frac{1}{5} \) and the lowest power of \( x \) available is \( x^{4} \). Therefore, we can factor out \( \frac{1}{5} x^{4} \): \[ \frac{6}{5} x^{8}+\frac{2}{5} x^{7}+\frac{1}{5} x^{5}+\frac{1}{5} x^{4} = \frac{1}{5} x^{4} \left( 6 x^{4} + 2 x^{3} + x + 1 \right) \] Now, the polynomial inside the parentheses is \( 6 x^{4} + 2 x^{3} + x + 1 \). Next, let's look for integer coefficients by multiplying through by \( 5 \): \[ 5 \left( \frac{1}{5} x^{4} \left( 6 x^{4} + 2 x^{3} + x + 1 \right) \right) = x^{4} (6 x^{4} + 2 x^{3} + x + 1) \] And the complete factored form of the original expression is: \[ \frac{1}{5} x^{4} (6 x^{4} + 2 x^{3} + x + 1) \] Thus, the fully factored polynomial is: \[ \frac{1}{5} x^{4} (6 x^{4} + 2 x^{3} + x + 1) \]