## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Zeros: $-2,1$ $x$-intercepts: $-2,1$
To find the zeros of a function $f$, solve the equation $f(x)=0$ The zeros of the function are also the $x-$intercepts. Let $g(x)=0$: $$x^6+7x^3-8=0$$ Let $u=x^3$, the original equation becomes $$u^2+7u-8=0$$ By factoring $$(u+8)(u-1) = 0$$ Use the Zero-Product Property by equating each factor to zero, then solve each equation to obtain: \begin{align*} u+8 &=0 &\text{ or }& &u-1=0\\ u &= -8 &\text{ or }& &u=1\\ \end{align*} To solve for $x$, we use $u=x^3$ $$\because u = x^3$$ $$\therefore x = u^{\frac{1}{3}}$$ For $u=-8$ $$x=(-8)^{\frac{1}{3}}$$ $$x= -2$$ For $u=1$ $$x=(1)^{\frac{1}{3}}$$ $$x= 1$$ $\therefore x = -2,1$