91Ó°ÊÓ

To find the roots of the quadratic equation $$8x^2 - 2x - 15$$, we can use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$, with $$a=8, b=-2, c=-15$$. Plugging these values into the formula, we get:$$ x=\frac{2 \pm \sqrt{(-2)^{2} - 4 \cdot 8 \cdot (-15)}}{2 \cdot 8} $$

Before solving for x, calculate the discriminant:$$ \Delta = b^2 - 4ac = (-2)^2 - 4(8)(-15) = 4 + 480 = 484 $$The discriminant is $$\Delta = 484$$, which is a positive perfect square, meaning we will have two real and rational roots.

Now, we can solve for x with the discriminant:$$ x = \frac{2 \pm \sqrt{484}}{16} $$Simplifying, we get:$$ x = \frac{2\pm22}{16} $$Thus, we have two roots:$$ x_1 = \frac{(2+22)}{16}=\frac{24}{16}= \frac{3}{2}\\ x_2 = \frac{(2-22)}{16}=\frac{-20}{16}=-\frac{5}{4} $$

Now that we have both roots $$x_1 = \frac{3}{2}$$ and $$x_2 = -\frac{5}{4}$$, we can rewrite the quadratic equation in the factored form:$$ y = a(x-x_1)(x-x_2)\\ y = a(x-\frac{3}{2})(x+\frac{5}{4}) $$To find the value of a, we can use an original equation and plug one of the roots with its corresponding y-coordinate, for example, use the root $$x_1 = \frac{3}{2}$$ :$$ y = 8(\frac{3}{2})^2 - 2(\frac{3}{2}) - 15 $$Solving, we get:$$ y = 8 \cdot \frac{9}{4} - 3 - 15 = 18 - 3 - 15 = 0 $$Since we are trying to find the original equation, the y value will be 0, so:$$ 0 = a(x_1-\frac{3}{2})(x_1+\frac{5}{4}) $$With $$x_1 = \frac{3}{2}$$, we have:$$ 0 = a(\frac{3}{2} - \frac{3}{2})(\frac{3}{2} + \frac{5}{4}) $$Since the left side is 0, the right side of the equation must also be 0. Given the factor of $$x_1 - \frac{3}{2}$$ will be equal to 0, the value of "a" can be anything. We use the coefficient of $$x^2$$ term in the original equation:$$ a=8 $$

So, finally, the quadratic equation in its factored form is:$$ y = 8(x-\frac{3}{2})(x+\frac{5}{4}) $$The constants in the factored form are $$a=8$$, $$r=\frac{3}{2}$$, and $$s=-\frac{5}{4}$$.