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

Given that π(π₯) = π₯Β³ + 3π₯Β² β 13π₯ β 15 and π(β1) = 0, find the other roots of π(π₯).

03:30

### Video Transcript

Given that π of π₯ equals π₯ cubed plus three π₯ squared minus 13π₯ minus 15 and π of negative one is zero, find the other roots of π 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. Weβre going to begin by fully factoring our expression for π of π₯. And to do that, we recall the factor theorem. If π of π is equal to zero, then π₯ minus π must be a factor of π of π₯. This tells us that π₯ plus one must be a factor of our function π 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π₯ squared plus ππ₯ plus π. We then distribute the parentheses, making sure we multiply each of the terms π₯ and one by ππ₯ squared, ππ₯, and π individually. When we do, we find that π₯ cubed plus three π₯ squared minus 13π₯ minus 15 is equal to ππ₯ cubed plus ππ₯ squared plus ππ₯ squared plus ππ₯ plus ππ₯ plus π.

Weβre now going to equate coefficients. Weβ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 itβs equal to π. So we found the value of π. Itβ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, itβs π plus π. We just worked out, though, that π itself is equal to one. So we can say that three is equal to one plus π, which means that π must be equal to two.

Weβll repeat this process with the coefficient of π₯ to the power of one. On the left-hand side, thatβs negative 13. And on the right, thatβs π plus π. We just calculated π to be equal to two. So we can say that negative 13 must be equal to two plus π. By solving for π and subtracting two from both sides, we see that π is equal to negative 15. And we could actually check this by equating the coefficient of π₯ to the power of zero. Thatβs the constants. And we once again see that negative 15 is indeed equal to π. 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. Itβs a quadratic equation whose coefficient of π₯ squared is one. So weβre going to have an π₯ and an π₯.

We need two numbers whose product is negative 15 and whose sum is two. Thatβ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 π of π₯.

Itβ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 π of π₯.