Finding zeros of polynomials (example 2) | Mathematics III | High School Math | Khan Academy

• So I have the polynomial ( p(x) ) here, and ( p(x) ) is being expressed as a fourth degree polynomial times ( (3x - 8)^2 ). So this would actually give you some, this would give you ( 9x^2 ) and a bunch of other stuff, and then you multiply that times this. It would actually give you a sixth degree polynomial all in all, but our goal is to find the ( x ) values where that makes ( p(x) = 0 ), or another way to find the roots or the zeros of this polynomial, and in particular, we're going to focus on the real zeros, the real roots of this polynomial. And like always, I encourage you to give a go at it, and then we'll do it together. Alright, let's tackle this.

So, the way I want to solve ( p(x) = 0 ). I want to solve ( p(x) = 0 ), and figure out what, and when I say solve it, I want to say, well what ( x ) values will make the polynomial equal to zero. So I just need to set this right-hand side to be equal to zero and then solve for ( x ). The best way that I can think of doing that is by factoring this out as much as I can, and if I can rewrite it as a product of a bunch of expressions equaling zero, well, a product of a bunch of things equaling zero, you can make it equal zero by any one of them equaling zero. And so let's do that.

So ( (3x - 8)^2 ), this is already factored quite nicely. Let's see if we can factor all of this business in white, and the way I will tackle it is to see if I can factor by grouping. So let me group together these first two terms, and then let me group together these second two terms. Essentially, factoring by grouping is doing the distributive property in reverse twice.

So from these first two terms, I could factor out, let's see, what could I factor out? I could factor out a—let me see—I'll just factor out ( x^3 ). So I get ( x^3(3x - 8) ). Interesting, we have a ( 3x - 8 ) over there as well. Now these second two terms, I could factor out a ( 5 ), so this is going to be ( +5(x(3x - 8)) ). Very interesting, and of course I have these parentheses around all of that, and then I have ( (3x - 8)^2 ).

This ( 3x - 8 ) is showing up a lot, and so, and of course this is going to be equal to zero. So we're gonna be equal to zero, and now I can factor out a ( 3x - 8 ) over here. I could factor that ( 3x - 8 ) out, and I'm going to get ( (3x - 8)(x^3 + 5) ) times ( (3x - 8)^2 ) is all going to be equal to zero. It's all equal to zero.

Now if what I just did looks a little like voodoo, just realize I have two terms, both of them are multiples of ( 3x - 8 ). I just factored out, I just factored out the ( 3x - 8 ). I did distributive property in reverse, so I factored it out, and what you're left with this term you just look for the next of third, and in this term you're just left with a ( +5 ).

Now ( (3x - 8)(x^3 + 5)(3x - 8)^2 ). Well, I could just rewrite this as ( (3x - 8)^3(x^3 + 5) = 0 ). So let me do that. I can just rewrite this as ( (3x - 8)^3 ), that's that times that, and then times—let's do this in a nicer color—times ( (x^3 + 5) = 0 ).

Now, in order to get this to be equal zero, either ( (3x - 8)^3 ) is going to be equal to zero, or ( (x^3 + 5) ) is going to be equal to zero. So let's first think about ( (3x - 8)^3 = 0 ).

So I can write this as ( 3x - 8 = 0 ) or ( x^3 + 5 = 0 ). So to make ( (3x - 8)^3 ) equal zero, well that means ( 3x - 8 = 0 ) or ( 3x = 8 ). Divide both sides by ( 3 ), ( x = \frac{8}{3} ). So that's one way to make this polynomial equal zero, ( x = \frac{8}{3} ). In fact, just this right over there will become zero, zero times anything is zero.

So this is a zero of our polynomial. And let's see, so for ( x^3 ), we could say, if we subtract ( 5 ) from both sides, we have ( x^3 = -5 ), and so if we take both to the one-third power, we could say ( x = \sqrt[3]{-5} ).

Now at first, you might say, "Wait, can I take the square root of a negative number?" and I would say, "Of course you can!" The cube root of (-1) is (-1). The cube root of (-8) is (-2). In fact, you could, even if we're dealing with reals. This is going to be a negative number. This is not going to be an imaginary number right over here, and so these are, these are the two zeros of the polynomial. There's gonna be negative, I think, negative ( 1 ) point something. I'm sure we could figure it, figure out it exactly.

So let's raise—so let's raise ( 5^{(1/3)} ) is equal to—so that's ( -5^{(1/3)} ), so negative ( 5^{(1/3)} ) power is going to be approximately equal to ( -1.71 ).

So we have two real roots, two real roots to this polynomial, or two zeros, two real zeros for this polynomial. And so those are going to be the two places where we intercept the ( x )-axis. The two ( x ) values for which where the two places where we intercept the ( x )-axis is the easiest way to say it.