Chapter 0 - Section 0.3 - Multiplying and Factoring Algebraic Expressions - Exercises - Page 21: 47

$x=\pm 2$ $or$ $x=\pm1$

$x^{4}-5x^{2}+4$ $Let$ $ x^{2}=y$ Then $x^{4}-5x^{2}+4=y^{2}-5y+4=y^{2}-4y-y+4=y(y-4)-1(y-4)=(y-4)(y-1)=(x^{2}-4)(x^{2}-1)$ $(x^{2}-4)(x^{2}-1)=0$ $x^{2}=4 $ $ or $ $x^{2}=1$ $x=\pm 2$ $or$ $x=\pm1$