Polynomials and Factoring

1. Group like terms:

$(3x^3 + x^3) + (-2x^2 + 4x^2) + (5x - 3x) + (-1 + 7)$

$= 4x^3 + 2x^2 + 2x + 6$

This demonstrates that polynomials are closed under addition (A-APR.A.1): the sum of two polynomials is another polynomial.

2. Distribute the negative sign and combine like terms:

$(5x^2 + 3x - 8) - (2x^2 - 7x + 4)$

$= 5x^2 + 3x - 8 - 2x^2 + 7x - 4$

$= (5x^2 - 2x^2) + (3x + 7x) + (-8 - 4)$

$= 3x^2 + 10x - 12$

This demonstrates that polynomials are closed under subtraction (A-APR.A.1).

3. Distribute each term of $(2x + 3)$ across $(x^2 - 4x + 5)$:

$2x(x^2 - 4x + 5) + 3(x^2 - 4x + 5)$

$= 2x^3 - 8x^2 + 10x + 3x^2 - 12x + 15$

$= 2x^3 + (-8x^2 + 3x^2) + (10x - 12x) + 15$

$= 2x^3 - 5x^2 - 2x + 15$

This demonstrates that polynomials are closed under multiplication (A-APR.A.1).

4. Set each factor equal to zero:

$x - 2 = 0 \implies x = 2$

$x + 5 = 0 \implies x = -5$

$x - 1 = 0 \implies x = 1$

The zeros are $x = -5$, $x = 1$, and $x = 2$.

The $x$-intercepts of the graph are the points $(-5, 0)$, $(1, 0)$, and $(2, 0)$.

5. Zeros: Set each factor equal to zero:

$x + 3 = 0 \implies x = -3$

$x - 4 = 0 \implies x = 4$

The zeros are $x = -3$ and $x = 4$.

Direction: The leading coefficient is $-2$ (negative), so the parabola opens downward.

Additional point: The vertex lies at $x = \frac{-3 + 4}{2} = \frac{1}{2}$. Evaluate:

$g\!\left(\frac{1}{2}\right) = -2\!\left(\frac{1}{2} + 3\right)\!\left(\frac{1}{2} - 4\right) = -2\!\left(\frac{7}{2}\right)\!\left(-\frac{7}{2}\right) = -2 \cdot \left(-\frac{49}{4}\right) = \frac{49}{2} = 24.5$

The vertex is at $\left(\frac{1}{2},\; \frac{49}{2}\right)$. The rough graph passes through $(-3, 0)$ and $(4, 0)$, peaks at $\left(\frac{1}{2}, 24.5\right)$, and opens downward.