Video: Factorizing Third-Degree Polonomials

Find the remainder when 3π‘₯Β³ βˆ’ 2π‘₯Β² + 4π‘₯ + 5 is divided by 3π‘₯ + 4.

02:27

Video Transcript

Find the remainder when three π‘₯ cubed minus two π‘₯ squared plus four π‘₯ plus five is divided by three π‘₯ plus four.

The question is asking us to find the remainder term when a cubic polynomial is divided by a linear polynomial. One way of doing this is by using polynomial long division. However, we know this is a long process. Instead, let’s notice we’re dividing by a linear polynomial. And we’re only asked to find the remainder term. This should remind us of the remainder theorem. We recall the remainder theorems tells us when 𝑝 of π‘₯ is divided by a linear polynomial π‘₯ minus π‘Ž, then the remainder is constant and equal to 𝑝 evaluated at π‘Ž.

We do have to be a little careful with how we use the remainder theorem in this case. We are dividing by a linear polynomial three π‘₯ plus four, but it’s not in the form π‘₯ minus π‘Ž. So instead of actually doing our long division, let’s call our quotient polynomial π‘ž of π‘₯ and our remainder polynomial π‘Ÿ of π‘₯. This means we’ll have three π‘₯ cubed minus two π‘₯ squared plus four π‘₯ minus five is equal to three π‘₯ plus four times π‘ž of π‘₯ plus π‘Ÿ of π‘₯ for some polynomials π‘ž of π‘₯ and π‘Ÿ of π‘₯. We can see our divisor is a linear polynomial; it has degree one. Our remainder must have a lower degree than our divisor. This means it must have degree zero. In other words, it’s a constant. We’ll call this constant π‘Ÿ.

And at this point, there’s two similar ways of solving this equation. If we solve our linear factor is equal to zero, this gives us π‘₯ is equal to negative four over three. One way to find the value of π‘Ÿ is to substitute this value directly into this expression. Doing this, we get our cubic polynomial evaluated at negative four over three is equal to zero times π‘ž evaluated at negative four over three plus π‘Ÿ. But this simplifies to just give us π‘Ÿ. And this is a perfectly valid way of solving this equation. However, we’re going to do this by taking a factor of three outside of our divisor. Doing this, we can rewrite this as three times π‘₯ plus four over three.

And now, we’re starting to see something interesting. Let’s consider three as part of π‘ž of π‘₯. So what do we now have? We have our cubic polynomial is equal to π‘₯ plus four over three times some polynomial plus a constant. In fact, what we’ve done here is found an expression for our quotient polynomial when we divide our cubic by π‘₯ plus four over three. In other words, the remainder term when we divide by π‘₯ plus four over three or when we divide by three π‘₯ plus four are the same. This means we can just use the remainder theorem to find our value of π‘Ÿ. And we’ll do this by evaluating our cubic polynomial at π‘₯ is equal to negative four over three. This gives us the following expression. And then, by evaluating this expression, we were able to show that our remainder term must be equal to negative 11.

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