- Zeros of polynomials introduction
- Zeros of polynomials: plotting zeros
- Zeros of polynomials: matching equation to zeros
- Zeros of polynomials: matching equation to graph
- Zeros of polynomials (factored form)
- Zeros of polynomials (with factoring): grouping
- Zeros of polynomials (with factoring): common factor
- Zeros of polynomials (with factoring)
When a polynomial is given in factored form, we can quickly find its zeros. When it's given in expanded form, we can factor it, and then find the zeros! Here is an example of a 3rd degree polynomial we can factor by first taking a common factor and then using the sum-product pattern.
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- How did we get (x+3)(x-2) from (x^2+x-6)? I can see where the +3 and -2 came from, but what's going on with the x^2+x part? And, how would I apply this to an equation such as (x^2+7x-6)?(5 votes)
- There are numerous ways to factor, this video covers getting a common factor. This will not work for x^2 + 7x - 6. Ic an tell you a way that works for it though, in fact my prefered way works for all quadratics, and that i why it is my preferred way.
Anyway, with x^2 + x - 6 Sal used this method.
if you have a quadratic ax^2 + bx + c where a, b and c are some numbers, you want to find two more numbers d and e. you want d + e = b and d * e = a * c, so it's kinda a guess and check to find them. Anyway, you then change the original from ax^2 + bx + c to ax^2 + dx + ex + c, so you replace bx with dx + ex
In x^2 + x - 6 a = 1, b = 1 and c = -6. With this we wand d + e = 1 and d * e = 1 * -6 = -6. You usually want to start with the multiplication one, so what are all the numbers that multiply to equal equal -6?
1 * -6
-1 * 6
2 * -3
-2 * 3
3 * -2
-3 * 2
Now of these, do any work out to add together to be 1? Yes, -2 and 3 do. So you change the equation.
x^2 + x - 6
x^2 - 2x + 3x - 6
You could make -2x + 3x into 3x - 2x if you want, it works either way. I encourage you to try it the other way after I show you this way
Anyway, you ant to put parenthesis around the left two numbers and the right two numbers.
x^2 - 2x + 3x - 6
(x^2 - 2x) + (3x - 6)
Keep in mind x^2 + 3 - 2x - 6 would become (x^2 + 3) + (- 2x - 6) i mention it because the - could throw you off. Anyway, continuing
(x^2 - 2x) + (3x - 6)
From (x^2 - 2x) you can pull out a x to get x(x - 2) and a 3 from the other to get 3(x - 2) so now you have
x(x-2) + 3(x - 2)
Notice how both have x-2? That was what our d and e did, so if you can find a d and e this will always work out. Sometimes you won't be able to find d and e though. After this you can change it to
x(x - 2) + 3(x - 2)
(x + 3)(x - 2)
If you don't understand how to do that I can explain, as well as anything else you don't understand. Anyway, that's how to get from x^2 + x - 6 to (x+3)(x-2)(13 votes)
- What if you have a function that = x^3 + 8 when finding the zeros? Would you just cube root?(3 votes)
- You probably could if you had a function that was a cube root. Otherwise, if the function can't be factored, you write the answer out in cube root form(2 votes)
- Could you also factor 5x(x^2 + x - 6) as 5x(x+2)(x-3) = 0 to get x=0, x= -2, and x=3 instead of factoring it as 5x(x+3)(x-2)=0 to get x=0, x= -3, and x=2?(3 votes)
- No because -3 and 2 adds up to -1 instead of 1. They have to add up as the coefficient of the second term.(2 votes)
- how did you get -6 out of -30(1 vote)
- I have almost this same problem but it is 5x² -5x -30. What should I do there? Should I group them together?(2 votes)
- When you are factoring a number, the first step tends to be to factor out any common factors, if possible. In this problem that common factor is 5, so we can factor it out to get 5(x² - x - 6). Then we can factor again to get 5((x - 3)(x + 2)). This isn't the only way to do this, but it is the first one that came to mind.(2 votes)
- We want to find the zeros of this polynomial: p(x)=2x3+5x2−2x−5 Plot all the zeros (x-intercepts) of the polynomial in the interactive graph.(1 vote)
- The first way to approach this is to see if you can factor out something in first two terms and second two terms and get another common factor. So p(x)= x^2 (2x + 5) - 1 (2x+5) works well, then factoring out common factor and setting p(x)=0 gives (x^2-1)(2x+5)=0. From there, note first is difference of perfect squares and can be factored, then you use zero product rule to find the three x intercepts.(2 votes)
- how to find more values of x(1 vote)
- In this example, he used p(x)=(5x^3+5x^2-30x)=0. Using that equation will show us all the places that touches the x-axis when y=0. But if we want to find all the x-value for when y=4 or other real numbers we could use p(x)=(5x^3+5x^2-30x)=4. Simply replace the f(x)=0 with f(x)= ANY REAL NUMBER. I hope this helps.(2 votes)
is there another way to factor polynomials with 4 terms
without seperating into 2 group , first group : first term, second term ,
second group : third term and fourth term.(1 vote)
- the only factors of the function f are (x+3)^2 and (3x-2) what is the product of its three zeroes?(1 vote)
- (𝑥 + 3)² and (3𝑥 − 2) are the only factors of 𝑓.
Thus, 𝑓(𝑥) = (𝑥 + 3)²⋅(3𝑥 − 2)
In practice, 𝑓 only has two zeros.
𝑓(𝑥) = 0 ⇒ 𝑥 = −3 or 𝑥 = 2∕3
But, 𝑥 = −3 is what we call a double zero (because the factor (𝑥 + 3) occurs twice in the factorization of 𝑓).
So technically, 𝑓 has the three zeros
𝑥₁ = −3
𝑥₂ = −3
𝑥₃ = 2∕3
And their product is (−3)⋅(−3)⋅2∕3 = 6(1 vote)
- So we're given a p of x, it's a third degree polynomial, and they say, plot all the zeroes or the x-intercepts of the polynomial in the interactive graph. And the reason why they say interactive graph, this is a screen shot from the exercise on Kahn Academy, where you could click and place the zeroes. But the key here is, lets figure out what x values make p of x equal to zero, those are the zeroes. And then we can plot them. So pause this video, and see if you can figure that out. So the key here is to try to factor this expression right over here, this third degree expression, because really we're trying to solve the X's for which five x to third plus five x squared minus 30 x is equal to zero. And the way we do that is by factoring this left-hand expression. So the first thing I always look for is a common factor across all of the terms. It looks like all of the terms are divisible by five x. So let's factor out a five x. So this is going to be five x times, if we take a five x out of five x to the third, we're left with an x squared. If we take out a five x out of five x squared, we're left with an x, so plus x. And if we take out a five x of negative 30 x, we're left with a negative six is equal to zero. And now, we have five x times this second degree, the second degree expression and to factor that, let's see, what two numbers add up to one? You could use as a one x here. And their product is equal to negative six. And let's see, positive three and negative two would do the trick. So I can rewrite this as five x times, so x plus three, x plus three, times x minus two, and if what I did looks unfamiliar, I encourage you to review factoring quadratics on Kahn Academy, and that is all going to be equal to zero. And so if I try to figure out what x values are going to make this whole expression zero, it could be the x values or the x value that makes five x equal zero. Because if five x zero, zero times anything else is going to be zero. So what makes five x equal zero? Well if we divide five, if you divide both sides by five, you're going to get x is equal to zero. And it is the case. If x equals zero, this becomes zero, and then doesn't matter what these are, zero times anything is zero. The other possible x value that would make everything zero is the x value that makes x plus three equal to zero. Subtract three from both sides you get x is equal to negative three. And then the other x value is the x value that makes x minus two equal to zero. Add two to both sides, that's gonna be x equals two. So there you have it. We have identified three x values that make our polynomial equal to zero and those are going to be the zeros and the x intercepts. So we have one at x equals zero. We have one at x equals negative three. We have one at x equals, at x equals two. And the reason why it's, we're done now with this exercise, if you're doing this on Kahn Academy or just clicked in these three places, but the reason why folks find this to be useful is it helps us start to think about what the graph could be. Because the graph has to intercept the x axis at these points. So the graph might look something like that, it might look something like that. And to figure out what it actually does look like we'd probably want to try out a few more x values in between these x intercepts to get the general sense of the graph.