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## Integrated math 3

### Course: Integrated math 3 > Unit 4

Lesson 1: Zeros of polynomials- 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)

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# Zeros of polynomials: matching equation to zeros

When we are given a list of the zeros of a polynomial, we can conclude the polynomial must have certain factors, which gives us information about the equation of the polynomial.

## Want to join the conversation?

- Why not just substitute each value into each equation using your calculator instead of having to go through that long, drawn out process?(1 vote)
- who would need a calculator? Even though Sal goes into a long explanation, if you understand what he is saying, it would take about 2 seconds to recognize the right answer. What if you were given the problem without answer choices? It is simple to learn what to do and more broadly applicable.(8 votes)

- Why not just substitute?(2 votes)
- Because you can be given an question without multiple choices so you will need to figure out the answer, this way you can see the logic behind the problem and make it more easily to you to understand what is going on(2 votes)

- if you were applying this to the set up of f(x)=ax^2+bx+c and given two zeros how would you find c?(1 vote)
- From the video we know that if a polynomial has the solutions

𝑥 = 𝑥₁ and 𝑥 = 𝑥₂,

then the polynomial could be

𝑓(𝑥) = (𝑥 − 𝑥₁)(𝑥 − 𝑥₂)

Given that we are dealing with a 2nd-degree polynomial the two zeros will have multiplicity 1, so the factors (𝑥 − 𝑥₁) and (𝑥 − 𝑥₂) won't have any exponents.

However, there could be some coefficient 𝑎, so in general we will have

𝑓(𝑥) = 𝑎(𝑥 − 𝑥₁)(𝑥 − 𝑥₂), which expands to

𝑓(𝑥) = 𝑎𝑥² − 𝑎(𝑥₁ + 𝑥₂)𝑥 + 𝑎𝑥₁𝑥₂

With 𝑏 = −𝑎(𝑥₁ + 𝑥₂) and 𝑐 = 𝑎𝑥₁𝑥₂, we get

𝑓(𝑥) = 𝑎𝑥² + 𝑏𝑥 + 𝑐

– – –

As an example, let's say the solutions are 𝑥 = 5 and 𝑥 = −2.

Then we have 𝑏 = −𝑎(5 + (−2)) = −3𝑎

and 𝑐 = 𝑎⋅5⋅(−2) = −10𝑎

So, 𝑓(𝑥) = 𝑎𝑥² − 3𝑎𝑥 − 10𝑎

– – –

As you can see, 𝑏 and 𝑐 both depend on 𝑎, and just given the two zeros we can't determine the coefficients 𝑎, 𝑏, and 𝑐.

However, if we know one of the coefficients, then we can solve for the other two.

If in the above example we knew that 𝑏 = −6,

then we would know that −3𝑎 = −6 ⇒ 𝑎 = −6∕(−3) = 2

and 𝑐 = −10𝑎 = −10⋅2 = −20,

and thus 𝑓(𝑥) = 2𝑥² − 6𝑥 − 20

– – –

Another way to determine 𝑓(𝑥) would be if we, in addition to the two zeros, knew a third point on the parabola.

In our example we found that 𝑓(𝑥) = 𝑎𝑥² − 3𝑎𝑥 − 10𝑎.

Let's say we knew that 𝑓(6) = 16

Plugging 𝑥 = 6 into our formula, we get

𝑓(6) = 𝑎⋅6² − 3𝑎⋅6 − 10𝑎

⇒ 16 = 36𝑎 − 18𝑎 − 10𝑎

⇒ 16 = 8𝑎

⇒ 𝑎 = 16∕8 = 2

𝑏 = −3𝑎 = −3⋅2 = −6

𝑐 = −10𝑎 = −10⋅2 = −20

Again, we get 𝑓(𝑥) = 2𝑥² − 6𝑥 − 20

– – –

Conclusion:

If 𝑓(𝑥) = 𝑎𝑥² + 𝑏𝑥 + 𝑐 has the solutions 𝑥 = 𝑥₁ and 𝑥 = 𝑥₂,

then 𝑏 = −𝑎(𝑥₁ + 𝑥₂)

and 𝑐 = 𝑎𝑥₁𝑥₂

If we know one of the coefficients 𝑎, 𝑏, and 𝑐, we can use these formulas to determine the other two coefficients.

Also, we can write

𝑓(𝑥) = 𝑎𝑥² − 𝑎(𝑥₁ + 𝑥₂)𝑥 + 𝑎𝑥₁𝑥₂,

which we can use to determine 𝑎 if we know a third point on the parabola.(3 votes)

- At3:33, how can Sal multiply the last binomial by eight without doing it to the other side?(1 vote)
- The problem here is not to manipulate a given equation, like when we're told to solve x²+3x=4, it's to
*find*an equation given some properties.

If we're manipulating a given equation, we avoid multiplying only one side because that changes the nature of the equation; we're now solving a different problem than the one given to us.

But when we're trying to construct an equation, we can change whatever we like, because we're not trying to preserve what was given to us. When Sal multiplied only one side by 8, he switched to working on a slightly different equation. That was the point. The old equation didn't match any of the answers, and the new one did.

Specifically, by multiplying the expression by a constant, Sal changed the equation*without*changing its roots/zeros, because we already had the roots how we wanted them. We just needed to tweak it to match one of the answers provided.(1 vote)

- I'm having issues understanding solving to find zeros of a polynomial expression. is there a video for that?(1 vote)
- The first video of this topic could perhaps help(1 vote)

## Video transcript

- A polynomial P has zeros when
X is equal to negative four, X is equal to three, and X is equal to one-eighth. What could be the equation of P? So pause this video and
think about it on your own before we work through it together. All right. So the fact that we have
zeros at these values, that means that P of X, when X is equal to one of
these values is equal to zero. So P of negative four is equal to zero, P of three is equal to zero, and P of one-eighth is equal to zero. And before I even look at these choices, I could think about
constructing a polynomial for which that is true. That's going to be true if I
can express this polynomial as the product of expressions
where each of these would make each of those
expressions equal to zero. So what's an expression that would be zero when X is equal to negative four? Well the expression X plus four, this is equal to zero when
X is equal to negative four, so I like that. What would be an expression
that would be equal to zero when X is equal to three? Well what about the
expression X minus three? If X is equal to three, then this is going to
equal to be equal zero. Zero times anything is
going to be equal to zero. So P of three would be zero in this case. And then what is an expression
that would be equal to zero when X is equal to one-eighth? Well that would be X minus one-eighth. Now tho-- these aren't the only expressions. You could multiply them
by constants and still the principles that I just
talked about would be true. But our polynomial would
look something like this. You could try it out. If X is equal to negative fou-- (chuckles) if X is equal to negative four, well then this first expression is zero. Zero times something
times something is zero. Same thing for X equals three. If this right over-- If X equals three, then X minus three is equal to zero, and then zero times something
times something is zero. And then if X is equal to one-eighth, this expression's going
to be equal to zero. Zero times something times something is going to be equal to zero. So which of these choices look like that? So let's see. X plus four, I actually see that in choices B, and I see that in choices D. Choice C has X minus four there. So that would have a
zero at X equals four. If X equals four, this first-- this first expression-- this first part of the expression would be equal to zero. But we care about that happening when X is equal to negative four. So I would actually rule out C, and then for the same
reason I would rule out A. So we're between B and D, and now let's see. Which of these have an
X minus three in them. Well I see an X minus three here. I see an X minus three there. So I like the-- I still like B and D. I'll put another check
mark right over there. And then last but not least, which of these would be equal to zero when X is equal to one-eighth? Well, let's see. If I do one-eighth times one-eighth here I'm gonna get one-sixty-fourth for this part of the expression. And so that's not going
to be equal to zero. And these other two
things aren't going to be equal to zero when X
is equal to one-eighth, so this one is not looking so good, but let's verify this one. This has-- If X is equal to one-eighth, we have eight times
one-eighth which is one, minus one. That is going to be equal to zero, so this one checks out. And you might be thinking hey! This last polynomial looks
a little bit different than this polynomial that I wrote up here when I just tried to come up
with a polynomial for which this would be true. And as I mentioned, you could take this and
multiply it by constants and it would still be true.
So if you just take this, and if we were to multiply it by eight, you would get P of X down here, because if we were to
multiply this times eight, which wouldn't change the zeros, well then if you distribute this eight, this last expression would
become eight X minus one. Which is exactly what we have down here.