Решите пожалуйста алгебру. Гдз платное. Желательно научите решать

To evaluate the expressions $x^2 - 3x + 5$ and $2x^3 + 3x^2 - 12x - 4$ at the given values of $x$, we can substitute each value into the expression and calculate the result. Let's do that:

For $x = 1$:
$1^2 - 3(1) + 5 = 1 - 3 + 5 = 3$

For $x = -1$:
$(-1)^2 - 3(-1) + 5 = 1 + 3 + 5 = 9$

For $x = 2$:
$2^2 - 3(2) + 5 = 4 - 6 + 5 = 3$

For $x = -2$:
$(-2)^2 - 3(-2) + 5 = 4 + 6 + 5 = 15$

For $x = 5$:
$5^2 - 3(5) + 5 = 25 - 15 + 5 = 15$

For $x = -5$:
$(-5)^2 - 3(-5) + 5 = 25 + 15 + 5 = 45$

So, the values of $x^2 - 3x + 5$ at $x = \pm 1, \pm 2, \pm 5$ are $3, 9, 3, 15, 15, 45$, respectively.

Now, let's evaluate the expression $2x^3 + 3x^2 - 12x - 4$ at the given values of $x":

For $x = 1$:
$2(1)^3 + 3(1)^2 - 12(1) - 4 = 2 + 3 - 12 - 4 = -11$

For $x = -1$:
$2(-1)^3 + 3(-1)^2 - 12(-1) - 4 = -2 + 3 + 12 - 4 = 9$

For $x = 2$:
$2(2)^3 + 3(2)^2 - 12(2) - 4 = 16 + 12 - 24 - 4 = 0$

For $x = -2$:
$2(-2)^3 + 3(-2)^2 - 12(-2) - 4 = -16 + 12 + 24 - 4 = 16$

For $x = 5$:
$2(5)^3 + 3(5)^2 - 12(5) - 4 = 250 + 75 - 60 - 4 = 261$

For $x = -5$:
$2(-5)^3 + 3(-5)^2 - 12(-5) - 4 = -250 + 75 + 60 - 4 = -119$

So, the values of $2x^3 + 3x^2 - 12x - 4$ at $x = \pm 1, \pm 2, \pm 5$ are $-11, 9, 0, 16, 261, -119$, respectively.