How can you use factoring to solve quadratic equations in standard form?

Step-by-step explanation:

The roots or the zeroes of a quadratic equation are its x-intercepts, which are the points on the graph where it crosses the x-axis. When a quadratic equation is in standard form, ax² + bx + c, the usual goal is to find the binomial factors that could help us determine the roots or solutions to the given equation.  Once you find the factors, then you will use the Zero-Product Property to solve for the solutions.  

For example, if you were given the following quadratic equation in standard form, x² + 6x + 8 = 0, and were tasked to find its factors.

x² + 6x + 8 = 0

where a = 1, b = 6, and c = 8:

In order to find the factors of the given quadratic equation, we must find the  factors with a product ac, and sum b.  

Using 2 and 4 as possible factors produce a product that is equal to

product of ac:   2 × 4 = 8

sum of b:           2 + 4 = 6.

Therefore, the factors of the given quadratic equation, x² + 6x + 8 = 0 are:

(x + 2) (x + 4).

Using the Zero-Product Property, where it states that for any real numbers, u and v, if uv = 0, then u = 0 or v = 0.  

Thus, setting the binomial factors of the given quadratic equation to zero will provide the solutions.

(x + 2) (x + 4)

x + 2 = 0

x + 2 - 2 = 0 - 2

x = -2

x + 4 = 0

x + 4 - 4 = 0 - 4

x = -4

Therefore, the solutions to the quadratic equation, x² + 6x + 8 = 0 are x = -2, and x = -4.

135 is your answer

step-by-step explanation:

try 180 times three then divide that number by 4 (sides) to get your answer. im glad to : )

8/k

step-by-step explanation: