Question Video: Finding Two Roots of a Cubic Equation given the Value of the Third Root | Nagwa Question Video: Finding Two Roots of a Cubic Equation given the Value of the Third Root | Nagwa

Question Video: Finding Two Roots of a Cubic Equation given the Value of the Third Root Mathematics

Given that 𝑓(π‘₯) = π‘₯Β³ + 3π‘₯Β² βˆ’ 13π‘₯ βˆ’ 15 and 𝑓(βˆ’1) = 0, find the other roots of 𝑓(π‘₯).

03:30

Video Transcript

Given that 𝑓 of π‘₯ equals π‘₯ cubed plus three π‘₯ squared minus 13π‘₯ minus 15 and 𝑓 of negative one is zero, find the other roots of 𝑓 of π‘₯.

Here, we have a cubic equation. And we know that when π‘₯ is equal to negative one, the function itself is equal to zero. We’re going to begin by fully factoring our expression for 𝑓 of π‘₯. And to do that, we recall the factor theorem. If 𝑓 of π‘Ž is equal to zero, then π‘₯ minus π‘Ž must be a factor of 𝑓 of π‘₯. This tells us that π‘₯ plus one must be a factor of our function 𝑓 of π‘₯. So how do we work out the other factors? Well, we could use polynomial long division.

We could divide π‘₯ cubed plus three π‘₯ squared minus 13 π‘₯ minus 15 by π‘₯ plus one. Alternatively, we can say that π‘₯ cubed plus three π‘₯ squared minus 13π‘₯ minus 15 is π‘₯ plus one times some quadratic: π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐. We then distribute the parentheses, making sure we multiply each of the terms π‘₯ and one by π‘Žπ‘₯ squared, 𝑏π‘₯, and 𝑐 individually. When we do, we find that π‘₯ cubed plus three π‘₯ squared minus 13π‘₯ minus 15 is equal to π‘Žπ‘₯ cubed plus π‘Žπ‘₯ squared plus 𝑏π‘₯ squared plus 𝑏π‘₯ plus 𝑐π‘₯ plus 𝑐.

We’re now going to equate coefficients. We’ll begin by looking at the π‘₯ cubed terms. On the left-hand side, the coefficient of π‘₯ cubed is one. And on the right-hand side, we can see it’s equal to π‘Ž. So we found the value of π‘Ž. It’s equal to one. Next, we equate the π‘₯ squared terms. On the left-hand side, the coefficient of π‘₯ squared is three. And on the right, it’s π‘Ž plus 𝑏. We just worked out, though, that π‘Ž itself is equal to one. So we can say that three is equal to one plus 𝑏, which means that 𝑏 must be equal to two.

We’ll repeat this process with the coefficient of π‘₯ to the power of one. On the left-hand side, that’s negative 13. And on the right, that’s 𝑏 plus 𝑐. We just calculated 𝑏 to be equal to two. So we can say that negative 13 must be equal to two plus 𝑐. By solving for 𝑐 and subtracting two from both sides, we see that 𝑐 is equal to negative 15. And we could actually check this by equating the coefficient of π‘₯ to the power of zero. That’s the constants. And we once again see that negative 15 is indeed equal to 𝑐. We can now say that π‘₯ cubed plus three π‘₯ squared minus 13π‘₯ minus 15 must be equal to π‘₯ plus one times π‘₯ squared plus two π‘₯ minus 15.

The roots of our function are found by setting this entire equation equal to zero and solving for π‘₯. We know that for this statement to be true, either π‘₯ plus one must be equal to zero or π‘₯ squared plus two π‘₯ minus 15 is equal to zero. We do already know that negative one is a root. So we go on to solving the equation π‘₯ squared plus two π‘₯ minus 15 equals zero. It’s a quadratic equation whose coefficient of π‘₯ squared is one. So we’re going to have an π‘₯ and an π‘₯.

We need two numbers whose product is negative 15 and whose sum is two. That’s five and negative three. Now for the product of these two expressions to be equal to zero, we can say that either π‘₯ plus five must be equal to zero or π‘₯ minus three must be equal to zero. Then, by subtracting five from both sides in our first equation, we find π‘₯ to be equal to negative five. And adding three in our second, we find π‘₯ is equal to three. And these are the other roots of 𝑓 of π‘₯.

It’s useful to know that we can check these solutions. If we substitute π‘₯ equals negative five and π‘₯ equals three into our original function, we should get zero. And you might wish to check this for yourself. But we do indeed get a value of zero. So π‘₯ equals negative five and π‘₯ equals three are the roots of 𝑓 of π‘₯.

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