Question

Please help!!!!!!

What polynomial has roots of -6, 1, and 4?

x3 - 9x2 - 22x + 24
x3 - x2 - 26x - 24
x3 + x2 - 26x + 24
x3 + 9x2 + 14x - 24

Answer 1

A polynomial with roots \(r_1, r_2,r_3,...r_n\) can be written in the form\[P(x) = a(x-r_1)(x-r_2)...(x-r_n)\]where \(a\) is a scaling constant to make the curve pass through a specific point, or 1 if that isn't required.

Answer 2

\(a\) is also the coefficient of the highest power term in the polynomial.

So you can write out the polynomial in factored form directly from the list of roots, then multiply it out to find the expanded form.

Answer 3

Another approach would be to simply try the roots in each prospective polynomial, and discard any polynomial which does not evaluate to 0 at all of the roots.

Answer 4

If you do that, take a second to see if one of the roots might be particularly convenient arithmetically and use that for the first pass to discard as many candidates as easily as possible.

Answer 5

still not grasping this
how do i set this up?
P(x)=a(x-6)(x-1)(x-4) ?
?

Answer 6

So, an example:

a polynomial with roots 3 and 4 can be written in factored form

\[P(x) = a(x-3)(x-4)\]
clear so far?

Answer 7

yeah i guess

Answer 8

wherever we have a root, there will be a zero in that list of products terms, and that will make the value of the polynomial at the root be 0, just as we want. I hope that you agree that there won't be any other places where we get 0 for the value of the polynomial except those that we specifically selected.

So if we had to choose from a list of polynomials:

\[x^2-2x-3\]\[x^2-6x+8\]\[x^2-7x+12\]
we know that \(a=1\) because the coefficient of the highest power term is 1.

That means \[P(x) = 1(x-3)(x-4) = x^2-4x-3x+12=x^2-7x+12\]

Answer 9

And if you plug the values in,

\[P(3) = (3)^2-7(3)+12 = 9-21+12=0\]\[P(4) = (4)^2-7(4)+12 = 16-28+12=0\]

and with one of the polynomials you get at least one non-zero result:
\[(3)^2-2(3)-3 = 9-6-3=0\][(4)^2-2(4)-3=16-8-3=5\]

So that second polynomial I tried cannot be the one with our desired roots.

Answer 10

sorry, typing too fast on a lagging system:
\[(4)^2-2(4)-3=16-8-3=5\]

Answer 11

So you take your 3 roots \(r_1,r_2,r_3\) and construct the factored polynomial:
\[P(x) = a(x-r_1)(x-r_2)(x-r_3)=(x-r_1)(x-r_2)(x-r_3)\](because we know that \(a=1\)) and then you multiply that sucker out and compare it with the answer choices.

Answer 12

A bit of cleverness will allow you to strike half of the candidates from consideration by just multiplying the three roots together. The constant term in a polynomial that can be written in factored form like this is simply the product of all of the roots. In my example, the roots were \(3,4\) so their product is \(3*4=12\) and any polynomial with a constant term that is not \(12\) is not the right polynomial.

\[(x+a)(x+b) = x*x + x*a + x*b + a*b = x^2 + (a+b)x + ab\]

Answer 13

i appreciate your help but i dont understand anything you really just said so im just gonna close the question.

Answer 14

no, let's try again

Answer 15

you understand what a root is, right?

Answer 16

it's a place where the curve of the polynomial intersects the x-axis, or \[P(x)=0\]

Answer 17

yes okay

Answer 18

the only way we can construct a polynomial with multiple roots is by multiplying stuff together, and that stuff is simple terms of the form \((x-r)\) where \(r\) is a root. Any time \(x\) is the same value as a root, the term that contains the root will equal 0, and the product will be 0, which makes that value of \(x\) a root.

Answer 19

That means there is a one-to-one correspondence between a set of roots and a polynomial, as long as we know that scaling factor \(a\), which we can get in a number of different ways.

Do you agree that there are infinitely many polynomials that have the same set of roots?

For example, look at this graph of 3 different equations. They all have the same roots, right? (assuming all roots are in view)

Answer 20

The only thing that is different about them is the scaling factor.

One is \[P(x) = 1(x-1)(x+1) = x^2-1\]another is\[P(x)=2(x-1)(x+1) = 2x^2-2\]and the last is \[P(x)=4(x-1)(x+1) = 4x^2-4\]

Answer 21

at \(x=1\) or \(x=-1\), they all are equal to \(0\). But at other values, such as \(x=0\), they are different.

But all polynomials with only those two roots will have a graph of the same shape, though it may be more or less tall, and may be flipped upside, depending on the scaling factor.

Answer 22

Does that make sense to you?

Answer 23

I'll take that as a yes :-)

Our roots are \(-6, 1, 4\) so our polynomial must have factors \((x-(-6)) = (x+6)\), \((x-1)\) and \((x-4)\)

We know the scaling factor is \(1\) because the coefficient of the \(x^3\) term in all of the answer choices is \(1\). Therefore, our polynomial with those roots is \[P(x)=(x+6)(x-1)(x+4) = \](whatever you get when you multiply that out)

Answer 24

I also pointed out that an alternative method to choose between a set of polynomials that includes the right answer (a luxury you won't always have) is to simply plug in the values of the roots as \(x\) and discard any polynomial that doesn't give you a result of 0 for all of the roots.