Y=x^3+kx^2-9x-18

SAT

1 answer:

deff fn [24]2 years ago

6 0

Answer:

Explanation:

A solution to a function is when the graph of the function crosses the x-axis. In other words, it's a point where y = 0. So, we can set x^3 + kx^2 - 9x -18 equal to 0. Note that there is an unknown coefficient in this equation (k) that we have to solve for. We are given that one solution to this equation is -2. This means that we can plus x = -2 into the equation and solve for the unknown k:

$(-2)^3+k(-2)^2-9(-2)-18=0\\-8+4k+18-18=0\\4k=8\\k=2$

Therefore, the complete equation is the following:

$y=x^3+2x^2-9x-18$

As it turns out, we can use grouping to factor this polynomial. Let's start by re-ordering the terms to visualize it better:

$x^3-9x+2x^2-18=0$

The first two terms have a GCF of x, and the last two terms have a GCF of 2:

$x(x^2-9)+2(x^2-9)$

Now, there is a common term that we can bring to the front to complete our factoring process:

$(x^2-9)(x+2)$

First of all, note that x+2 is one of the factors, and if this was set to 0, we would get -2, which is consistent with the problem. Also, note that our other term can be factored more because it is a difference of two squares:

$(x-3)(x+3)(x+2)=0$