Question Video: Finding the Set of Zeros of a Polynomial Function | Nagwa Question Video: Finding the Set of Zeros of a Polynomial Function | Nagwa

Question Video: Finding the Set of Zeros of a Polynomial Function Mathematics • Third Year of Preparatory School

Join Nagwa Classes

Attend live Mathematics sessions on Nagwa Classes to learn more about this topic from an expert teacher!

Find the set of zeros of the function 𝑓(π‘₯) =βˆ’9π‘₯⁴ + 225π‘₯Β².

03:40

Video Transcript

Find the set of zeros of the function 𝑓 of π‘₯ is equal to negative nine π‘₯ to the fourth power plus 225π‘₯ squared.

In this question, we’re given a function 𝑓 of π‘₯ and we’re asked to determine the set of zeroes of this function. We can start by recalling the set of zeroes of a function 𝑓 of π‘₯ is the set of all values of π‘₯ such that 𝑓 evaluated at π‘₯ is equal to zero. Therefore, we can determine the set of zeros of the given function by solving the equation 𝑓 of π‘₯ is equal to zero. This gives us the equation negative nine π‘₯ to the fourth power plus 225π‘₯ squared is equal to zero.

And there are a few different methods we can use to solve this equation. One way is to notice the first term on the left-hand side of our equation and the second term on the left-hand side of this equation share a factor of π‘₯ squared. If we take out the shared factor of π‘₯ squared, we have the equation π‘₯ squared multiplied by negative nine π‘₯ squared plus 225 is equal to zero.

And now we have a product which is equal to zero. For the product of two numbers to be equal to zero, one of the two factors must be equal to zero. Therefore, either π‘₯ squared is equal to zero or negative nine π‘₯ squared plus 225 is equal to zero. And we can solve each of these equations separately. First, if π‘₯ squared is equal to zero, then we know π‘₯ itself must be equal to zero. There’s many different ways of seeing this. For example, we could sketch the graph 𝑦 is equal to π‘₯ squared and note that the only π‘₯-intercept is at zero. Or we could take the square root of both sides of the equation. Normally, we would get a positive and a negative square root. However, the square root of zero is just zero. So the only root would be π‘₯ is equal to zero.

Let’s now move on to solving the second factor equal to zero. There’s several different ways we could do this. Since our coefficient of π‘₯ squared is negative, let’s multiply the entire equation through by negative one. This gives us the equation nine π‘₯ squared minus 225 is equal to zero. And now there’s several different methods we could use to evaluate this equation. For example, we could notice that this is a difference between two squares. This would allow us to factor the equation and solve for π‘₯. However, it’s easier just to solve this equation directly.

We’ll start by adding 225 to both sides of the equation. This gives us that nine π‘₯ squared is equal to 225. Next, we can divide both sides of the equation through by nine. This gives us that π‘₯ squared is equal to 225 divided by nine. Finally, we need to take the square root of both sides of the equation. Remember, we’ll get a positive and a negative solution. This then gives us that π‘₯ is equal to positive or negative the square root of 225 divided by nine.

Finally, we can simplify this equation by using the laws of exponents. We recall the square root of π‘Ž divided by 𝑏 is equal to the square root of π‘Ž divided by the square root of 𝑏. This gives us that π‘₯ is equal to positive or negative root 225 divided by root nine. And now we can evaluate this expression. The square root of 225 is 15, and the square root of nine is three. Therefore, π‘₯ is equal to positive or negative 15 divided by three. And we can cancel the shared factor of three in the numerator and denominator of this expression, giving us that π‘₯ is equal to positive or negative five. And this means we found all of the zeroes of our function. Either π‘₯ is negative five, zero, or five.

But remember, the question wants us to write this as the set of zeroes. So we need to write these three elements as a set. Therefore, we were able to show the set of zeroes of the function 𝑓 of π‘₯ is equal to negative nine π‘₯ to the fourth power plus 225π‘₯ squared is the set containing negative five, zero, and five.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy