Video: Finding the Set of Zeros of a Cubic Function by Factorisation

Find the set of zeroes of the function 𝑓(π‘₯) = π‘₯(π‘₯Β² βˆ’ 81) βˆ’ 2(π‘₯Β² βˆ’ 81).

03:22

Video Transcript

Find the set of zeroes of the function 𝑓 of π‘₯ equals π‘₯ times π‘₯ squared minus 81 minus two times π‘₯ squared minus 81.

We might begin by recalling that the zeros of a polynomial function are the values of π‘₯ that make it equal to zero. And so we need to set the expression π‘₯ times π‘₯ squared minus 81 minus two times π‘₯ squared minus 81 equal to zero and solve for π‘₯.

Now, if we were to distribute our parentheses, we’d probably spot that we have a cubic. That’s a polynomial whose order is three. And so we might be thinking that we need to distribute the parentheses and go from there. However, if we look really carefully, we see that the two terms share a common factor. They share a factor of π‘₯ squared minus 81. And so to solve this equation, we’re going to factor by removing that common factor of π‘₯ squared minus 81.

If we divide the first term by π‘₯ squared minus 81, that leaves us simply with π‘₯. And then if we divide the second term by π‘₯ squared minus 81, we get negative two. And so we can see that the expression π‘₯ times π‘₯ squared minus 81 minus two times π‘₯ squared minus 81 is equal to π‘₯ squared minus 81 times π‘₯ minus two. And so if we can fully factor an expression when finding the zeros, it’s really helpful because we now need to ask ourselves, which values of π‘₯ make this expression equal to zero?

And of course, since we’re multiplying π‘₯ squared minus 81 by π‘₯ minus two, and that gives us zero, we can say that either π‘₯ squared minus 81 must be equal to zero or π‘₯ minus two must be equal to zero. We’ll now solve each of these equations in turn. We’ll solve our first equation by adding 81 to both sides. And that tells us that π‘₯ squared is equal to 81. We’re now going to take the square root of both sides. But we do need to be a little bit careful. We’ll need to take both the positive and negative square root of 81.

And so the solutions to the equation π‘₯ squared minus 81 equals zero are π‘₯ equals nine and π‘₯ equals negative nine. The other equation is a little bit more straightforward. We’re just going to add two to both sides. And so we find another solution to the equation 𝑓 of π‘₯ equals zero and, therefore, a zero of our function to be two. And we found the solutions to the equation 𝑓 of π‘₯ equals zero, but the question asks us to find the set of zeros of the function. And so we use these squiggly brackets to represent the set containing the elements negative nine, two, and nine.

And that’s the answer to our question. The set of zeros of the function 𝑓 of π‘₯ equals π‘₯ times π‘₯ squared minus 81 minus two times π‘₯ squared minus 81 contains the elements negative nine, two, and nine. Note, of course, that we could go back to our original function to check whether these answers are correct. We would need to substitute each one in turn and double-check that we do indeed get an answer of zero.

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