Polynomials finding zeros

Question

what are the zeros of this function f(x)= 3(x²-25) (4x²+4x+1)

Answer 1

So in order to find the zeros of this function we have to determine where the function crosses the x axis.

First, we can disregard the coefficient of 3, as it does not effect the zeros, just the function as a whole.

Now we can begin with the first part of the function. (x^2-25) is an example of a perfect square (because square root of 25 is 5). That factors into (x+5)(x-5).

With the second part we would have to factor out the binomial. Since the coefficient of the x^2 value is 4 and the constant is 1, we need to find two numbers that can multiply to equal 4 and be added together to equal 4. In this case those two factors are both +2.

We rewrite the equation to include +2: 4x^2 +2x +2x +1 ,

Then factor it out: 2x(2x+1) + 1(2x+1) which gives us (2x+1) and (2x+1), (We don't have to count this one twice, unless you are accounting for multiplicity).

Our 3 factors from both parts are (2x+1), (x+5), and (x-5). This information gives us information that the zeros are located at x= -5, -1/2, and at 5 (remember that (x+1) would give us the zero at -1). Hope this helps!

Answer 2

Find zeros of f(x)= 3(x²-25) (4x²+4x+1)

So whats a zero? It's when f(x)=0, so we need to find all the values of x that cause f(x) to equal 0.

We can ignore the 3 because since 3 * 0 = 0, it isn't consequential to this problem

Lets factor the parenthesis separately. We'll start with (x²-25).

This factors to (x-5)(x+5).

As a side note: You can check it by just multiplying this out by using the acronym FOIL. It stands for First (multiply x * x), Outer (multiply x * 5), Inner (multiply -5 * x), Last (multiply -5 * 5). Once you've multiplied them all, sum them up and see that whats left is x² + 5x - 5x - 25 = x² - 25.

Next, we'll factor the second group: 4x²+4x+1.

This one is a little harder to factor but it factors to (2x+1)(2x+1) or (2x+1)². Again, you can check using the same method as before. 2x*2x + 2x*1 + 2x*1 +1*1 =4x² + 4x +1.

So, why did we find these factors? Because what we really want is to find out how to make f(x)=0. Factoring is kind of like unmultiplying a function. To get the original function, you have to multiply all the factors together back together; and as we know, anything multiplied by 0 is 0. So if we can make any of the factors equal to zero, the whole function will equal 0. But how do we know what will make the factors equal 0? We just say they do and then solve for x! It's like this

Our first one is (x+5): So we'll say x+5=0 -> x=-5. This means that when x=-5, that factor will equal zero, causing f(x) to equal 0.

Similarly, for (x-5): x-5=0 -> x=5. Again, when x=5, that factor will equal zero, causing f(x) to equal 0.

and lastly, (2x+1)²: (2x+1)²=0 (since 0^1/2=0) -> 2x+1=0 -> 2x=-1 -> x=-1/2. Since technically there are two of this factor, we say it has a multiplicity of 2.

So, the zeros are 5 (multiplicity of 1), -5 (multiplicity of 1), and -1/2 (multiplicity of 2).