f(x)=x^3-4x^2-x+4

How do I find the x-intercept for this?

This is from my book:
step 1: 0=x^3-4x^2-x+4
step 2: 0=x^2(x-4)+(-1)(x-4)
step 3: 0=(x^2-1)(x-4)

I know how to solve once I get to step 3, but in step 2, I don't know what they've done or why. Where does that (-1) come from? Why is it (x-4)?

Please help explain this to me. I'd be really thankful.

step 1: 0=x^3-4x^2-x+4

we would like to factor that.
rewrite as
x^3-4x^2 - (x-4)
now you can see that I can factor an (x-4) out of the first two terms
x^2 (x-4) - (x-4)
which is
x^2 (x-4) -1 (x-4)

step 2: 0=x^2(x-4)+(-1)(x-4)
step 3: 0=(x^2-1)(x-4)

To find the x-intercepts of the function f(x)=x^3-4x^2-x+4, you can follow the steps you mentioned:

Step 1: Set f(x) equal to zero: 0=x^3-4x^2-x+4

Step 2: You factor the equation using the grouping method. By grouping, you find common factors that can be factored separately. In this case, we group the first two terms and the last two terms:

0 = (x^3-4x^2) + (-x+4)

Now, let's factor out the common factor from each group:

0 = x^2(x-4) + (-1)(x-4)

The common factor in the first group is x^2, and in the second group, it is -1. Now, notice that (x-4) is a common factor between the two terms. To factor it out, we write it as a separate term for both groups, and we get (x-4)(x^2-1).

Step 3: Now you have the factored form: 0 = (x-4)(x^2-1).

To solve for x, you set each factor equal to zero:

x-4 = 0 and x^2-1 = 0

From x-4 = 0, you get x = 4. This is one x-intercept of the function.

From x^2-1 = 0, you can solve for x by factoring or using the difference of squares formula:

x^2-1 = (x-1)(x+1)

Now, set each factor equal to zero:

x-1 = 0 gives x = 1

x+1 = 0 gives x = -1

So, the x-intercepts of the function f(x) = x^3-4x^2-x+4 are x = 4, x = 1, and x = -1.

To find the x-intercepts of a function, you need to solve the equation f(x) = 0. In this case, your function is f(x) = x^3 - 4x^2 - x + 4.

The steps provided in your book are a way to factorize the equation so that it is easier to solve. Let's break down each step and understand why it is done:

Step 1: 0 = x^3 - 4x^2 - x + 4
Here, we set the equation equal to zero to find the x-intercepts.

Step 2: 0 = x^2(x - 4) + (-1)(x - 4)
In this step, the goal is to factorize the four-term polynomial into two binomial factors. The two terms on the left side, x^2(x - 4) and (-1)(x - 4), share a common factor of (x - 4).

So, the first term, x^2(x - 4), can be rewritten using the distributive property as x^2 * x - x^2 * 4 = x^3 - 4x^2.
Similarly, the second term, (-1)(x - 4), can be written as -x +4.

Combining these results, we get 0 = x^3 - 4x^2 - x + 4, which is the same equation we started with in step 1.

Step 3: 0 = (x^2 - 1)(x - 4)
In this step, the equation has been factored further. By using the distributive property, we expand the equation derived in step 2: x^2(x - 4) + (-1)(x - 4) = (x^2 - 1)(x - 4).

At this point, we now have a factored form that expresses the equation as a product of two binomial factors: (x^2 - 1) and (x - 4).

Now, to find the x-intercepts, we set each factor equal to zero:

For (x^2 - 1) = 0, we solve for x^2 = 1, which gives us two solutions: x = 1 and x = -1.

For (x - 4) = 0, we solve for x = 4 as our third solution.

Thus, the x-intercepts of the function f(x) = x^3 - 4x^2 - x + 4 are x = 1, x = -1, and x = 4.

It's important to note that these steps are specific to this particular equation, and may not be applicable to all equations. Factoring can be a useful technique for solving polynomial equations and finding the x-intercepts, especially when the equation can be factorized.