Applying the rational root theorem
The final factored form is given as (x + 1)(x - 1/3)(x + 3), which drops the leading coefficient 3; g(x) = 3(x + 1)(x - 1/3)(x + 3). A correcting note and an after-question were added.
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The video finds all zeros of g(x) = 3x^3 + 11x^2 + 5x - 3 using the rational root theorem. Testing candidates p/q shows x = -1 is a root, so (x + 1) is a factor; long division leaves 3x^2 + 8x - 3, and the quadratic formula gives the other roots 1/3 and -3.
You'll be able to
- list the possible rational roots of a polynomial using factors of the constant term (p) and the leading coefficient (q).
- test a candidate root by evaluating the polynomial and use a root to find a factor.
- divide out a known factor and solve the remaining quadratic to find all zeros.
Key words
- Rational root theorem
- Any rational root of a polynomial with integer coefficients has the form p/q, where p divides the constant term and q divides the leading coefficient.
- Root (zero)
- An x-value that makes the polynomial equal to zero.
- Algebraic long division
- Dividing one polynomial by another, term by term, like ordinary long division.
- Quadratic formula
- x = (-b ± sqrt(b^2 - 4ac)) / (2a), which solves ax^2 + bx + c = 0.
Transcript
- [Instructor] All right, we're told to identify all of the real and imaginary roots or zeros of the function g(x). So, pause this video, have a go at this and I'll give you a hint. The rational root theorem will be useful. Okay, now, let's think about what the rational root theorem tells us.
It tells us if this polynomial here does have rational roots, it can be made up based on the factors of the constant term and we could call the constant term right over here, p, and the factors of the coefficient on the highest-degree term, which is also a three, let's call that q.
So, let's think about the factors of p. So, I'm gonna list p factors right over here. And we want the plus or minus version of it. Well, luckily, three is a prime number, so, it only has two factors. So, we could have a positive one, we could have a negative one, we could have a positive three, or we could have a negative three.
And similarly, q is also a three. So, its factors are the same thing, 1, -1, 3 and -3. Now, we wanna test not these numbers, but we want to test numbers that are made up of the possible p over q's and both the positives and the negatives of them. So, for example, we could first try to take one over one right over here, which is, of course, going to be equal to one.
Now, how can we test whether that is actually a root of g? Well, we can just try to evaluate what g(1) is going to be equal to, g(1) is going to be equal to three times one of the third, which is just three, plus 11 times one-squared, which is just 11, plus five times one, which is just five, minus three.
So, if I just eyeball it here, these two things cancel out and I am left with 11 plus five, which is equal to 16. So, one is definitely not a zero. If it was, then we would've gotten a zero here, not a 16. So, we can rule that one out. So, this is not a root. Now, let's try negative one.
And there's a couple of ways we can get negative one. We could have a -1p over a 1q, you could have a 1p over a -1q, or you could even have a -3p over a 3q, or you could have a 3p over a -3q. So, those are all ways of getting to negative one. And so, let's see if that's a root.
So, g(-1) is going to be equal to three, negative one of the third power is negative one, so, this is gonna be negative three, plus negative one-squared is one still, so, this is still gonna be a positive 11 and then plus five times negative one, which is negative five and then we have minus three.
So, let's see what happens here. We end up with negative three plus 11 is eight, minus five is three, minus three is equal to zero. So, we have found one of our roots, x equals negative one is one of our roots. Now, how can we find the other roots? There could be, especially, if we're looking at imaginary roots here, there should be three of them.
So, what we can do if we know negative one is a root, then that also tells us that x plus one must be a factor. 'Cause if you said x plus one equals zero, then you'd get x equals negative one. And so, let's divide x plus one into the rest of this polynomial and start to factor it.
(Instructor sneezes) So, if I take x plus one and I'm gonna do algebraic long division, you could also do synthetic division, but I prefer algebraic long division, because it feels a little less mysterious. So, x plus one goes into 3x to the third, plus 11x-squared plus 5x minus three.
So, you just look at the highest-degree terms, x goes into 3x to the third, it goes 3x-squared times, so, I'll write here in the second-degree column, 3x-squared times and all I did is I took 3x to the third divided by x to get that. 3x-squared times x is 3x to the third.
3x-squared times one is plus 3x-squared. And this is just like regular long division. It has a lot of similarities. Now, we subtract this, that cancels with that. And 11x-squared minus 3x-squared is 8x-squared. Bring down the first-degree term, plus 5x, x goes into 8x-squared, 8x times.
If you just take 8x-squared divided by x, you get 8x. So, 8x times x is 8x-squared, 8x times one is, so we have plus 8x here, we subtract again and we get 5x minus 8x is negative 3x. And then, bring down this constant term, minus three, x goes into negative 3x, negative three times x plus one is negative 3x and then minus three.
And of course, now, if you were to subtract this, we get no remainder. That makes sense because we all already shoot. So, we already saw that negative one is a root, or that x plus one is a factor. So, now, we could actually rewrite g(x), g(x) is equal to x plus one times this business, 3x-squared plus 8x minus three.
So, now, this is a standard quadratic. We can try to factor this, or we could use a quadratic formula. I'm gonna use the quadratic formula here. So, we remember negative b plus or minus, well, let me just write it out here. So, this is our a, let me write this in another color.
This is our a, this is our b and that is our c. So, quadratic formula, negative b, so that's negative eight, plus or minus the square-root of b-squared, so, that's eight-squared, so, it's 64, minus four times a, which is three, times c, which is negative three.
So, I could put a three here and then make this positive. And then, all of that is going to be over 2a. So, 2a is two times three is six. So, this is going to be the same thing as negative eight plus or minus, four times three times three, this is all 36. So, under the radical, you have 64 plus 36, which is a hundred.
Square-root of a hundred is 10. So, negative eight plus or minus 10 over six. And so, this gives us x could be equal to, if we take negative eight plus 10 over six, negative eight plus 10 is two, over six is positive one-third, or if we take negative eight minus 10, that's negative 18 over six, which is negative three.
So, there you have it. We have figured out what our roots are. Our roots are negative one, that's where we started and also, one-third and negative three. If someone asks you to factor this whole thing, we could now write, maybe I do this in a new color, we could write, I'm running out of space here, but let me just write in this white color, g(x) is equal to x plus one times x minus one-third, times x plus three.
That would be it in a completely factored form. And we all were able to get there by using that rational root theorem.
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