# How do you solve this difficult algebra problem?

## F(x)= 2x^2-5x+3 and G(x)=1-4x-x^2 Calculate f(-2) + g(2)

Aug 12, 2018

$10$

#### Explanation:

$\text{substitute "x=-2" into" f(x) " and "x=2" into } g \left(x\right)$

$f \left(- 2\right) = 2 {\left(- 2\right)}^{2} - 5 \left(- 2\right) + 3$

$\textcolor{w h i t e}{f - \left(2\right)} = \left(2 \times 4\right) + \left(- 5 \times - 2\right) + 3$

$\textcolor{w h i t e}{f \left(- 2\right)} = 8 + 10 + 3 = 21$

$g \left(2\right) = 1 - \left(4 \times 2\right) - {\left(2\right)}^{2}$

$\textcolor{w h i t e}{g \left(2\right)} = 1 - 8 - 4 = - 11$

$f \left(- 2\right) + g \left(2\right) = 21 + \left(- 11\right) = 21 - 11 = 10$

Aug 13, 2018

$10$

#### Explanation:

The key realization is that $f \left(- 2\right)$ means that we will evaluate the function $f$ at $x = - 2$, and $g \left(2\right)$ means we evaluate $g$ at $x = 2$.

Plugging in $- 2$ for $f$, we get

$f \left(- 2\right) = 2 {\left(- 2\right)}^{2} - 5 \left(- 2\right) + 3 = 21$

Plugging in $2$ for $g$, we get

$g \left(2\right) = 1 - 4 \left(2\right) - {\left(2\right)}^{2} = 1 - 8 - 4 = - 11$

We have been asked to find

$f \left(- 2\right) + g \left(2\right)$, and since we have both values, we can sum them to get

$21 - 11 = 10$

Hope this helps!