Factorize 2x2 + 5x + 2 and 10a2 - 12ab - 15ab2 + 18b3. Solve n2 = 2n + 8 and 2x2 − 10x − 3 = 0.

Question 1:

a) Factorize the following:

i. $2x^2 + 5x + 2$

Step 1: Find two numbers that multiply to $2 \times 2 = 4$ and add to 5. These numbers are 1 and 4.

Step 2: Rewrite the middle term using these numbers.

$$2x^2 + x + 4x + 2$$

Step 3: Group the terms and factor out common factors.

$$(2x^2 + x) + (4x + 2)$$ $$x(2x + 1) + 2(2x + 1)$$

Step 4: Factor out the common binomial.

$$\boxed{(2x + 1)(x + 2)}$$

ii. $10a^2 - 12ab - 15ab^2 + 18b^3$

Step 1: Group the terms.

$$(10a^2 - 12ab) - (15ab^2 - 18b^3)$$

Step 2: Factor out the greatest common factor from each group.

$$2a(5a - 6b) - 3b^2(5a - 6b)$$

Step 3: Factor out the common binomial.

$$\boxed{(5a - 6b)(2a - 3b^2)}$$

b) Solve the following:

i. $n^2 = 2n + 8$ using the quadratic formula.

Step 1: Rewrite the equation in standard form $an^2 + bn + c = 0$.

$$n^2 - 2n - 8 = 0$$

Here, $a=1$, $b=-2$, $c=-8$.

Step 2: Apply the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

$$n = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-8)}}{2(1)}$$ $$n = \frac{2 \pm \sqrt{4 + 32}}{2}$$ $$n = \frac{2 \pm \sqrt{36}}{2}$$ $$n = \frac{2 \pm 6}{2}$$

Step 3: Calculate the two possible values for $n$.

$$n_1 = \frac{2 + 6}{2} = \frac{8}{2} = 4$$ $$n_2 = \frac{2 - 6}{2} = \frac{-4}{2} = -2$$

The solutions are $\boxed{n=4, n=-2}$.

ii. $2x^2 - 10x - 3 = 0$ using the completing the square method.

Step 1: Move the constant term to the right side.

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

Step 2: Divide the entire equation by the coefficient of $x^2$ (which is 2).

$$x^2 - 5x = \frac{3}{2}$$

Step 3: Take half of the coefficient of $x$ (which is $-5$), square it $\left(-\frac{5}{2}\right)^2 = \frac{25}{4}$, and add it to both sides.

$$x^2 - 5x + \frac{25}{4} = \frac{3}{2} + \frac{25}{4}$$

Step 4: Factor the left side as a perfect square and simplify the right side.

$$\left(x - \frac{5}{2}\right)^2 = \frac{6}{4} + \frac{25}{4}$$ $$$$