How do you evaluate 2x^2 - 2z^4 + y^2 - x^2 + z^4 if x = -4, y = 3, and z = 2?

Jun 30, 2016

$9$

Explanation:

Given,

$2 {x}^{2} - 2 {z}^{4} + {y}^{2} - {x}^{2} + {z}^{4}$

Knowing that $x = - 4 , y = 3 , z = 2$, replace each variable with its appropriate number.

$= 2 {\left(- 4\right)}^{2} - 2 {\left(2\right)}^{4} + {\left(3\right)}^{2} - {\left(- 4\right)}^{2} + {\left(2\right)}^{4}$

Recall that a number raised to a power $n$ means to multiply the number by itself $n$ times. Hence,

$= 2 \left(\textcolor{red}{- 4} \cdot \textcolor{red}{- 4}\right) - 2 \left(\textcolor{b l u e}{2} \cdot \textcolor{b l u e}{2} \cdot \textcolor{b l u e}{2} \cdot \textcolor{b l u e}{2}\right) + \left(\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{3} \cdot \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{3}\right) - \left(\textcolor{p u r p \le}{- 4} \cdot \textcolor{p u r p \le}{- 4}\right) + \left(\textcolor{t e a l}{2} \cdot \textcolor{t e a l}{2} \cdot \textcolor{t e a l}{2} \cdot \textcolor{t e a l}{2}\right)$

Simplify.

$= 2 \left(\textcolor{red}{16}\right) - 2 \left(\textcolor{b l u e}{16}\right) + \left(\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{9}\right) - \left(\textcolor{p u r p \le}{16}\right) + \left(\textcolor{t e a l}{16}\right)$

$= 32 - 32 + 9 - 16 + 16$

$= \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{9} \textcolor{w h i t e}{\frac{a}{a}} |}}}$