Make A Polynomial Function

[mathdad]

Use f(x) = a(x - r_1)(x - r_2)(x - r_3), where r represents the zeros, to find a polynomial function.

Zeros: a, b, c; degree 3

x - r_1 = x - a

x - r_2 = x - b

x - r_3 = x - c

Let a = 1

f(x) = (x - a)(x - b)(x - c)

Can I leave f(x) like this? Must I expand?

 

[JeffM]

Use f(x) = a(x - r_1)(x - r_2)(x - r_3), where r represents the zeros, to find a polynomial function.

Zeros: a, b, c; degree 3

x - r_1 = x - a

x - r_2 = x - b

x - r_3 = x - c

Let a = 1

f(x) = (x - a)(x - b)(x - c)

Can I leave f(x) like this? Must I expand?

What you have specified is a generic cubic, which is indeed a polynomial. So your answer is formally correct.

I suspect, however, that the question is expecting you to: (1) substitute numeric values rather than new variables, and (2)
expand the product so that it is obvious how the coefficients of the polynomial relate to the roots.

 

[mathdad]

What you have specified is a generic cubic, which is indeed a polynomial. So your answer is formally correct.

I suspect, however, that the question is expecting you to: (1) substitute numeric values rather than new variables, and (2)
expand the product so that it is obvious how the coefficients of the polynomial relate to the roots.

Sounds good to me. I will expand on paper.

 

[Steven G]

Use f(x) = a(x - r_1)(x - r_2)(x - r_3), where r represents the zeros, to find a polynomial function.

Zeros: a, b, c; degree 3

x - r_1 = x - a

x - r_2 = x - b

x - r_3 = x - c

Let a = 1

f(x) = (x - a)(x - b)(x - c)

Can I leave f(x) like this? Must I expand?

I am sorry but I do not believe that the problem was written as you showed. I just can't see that the coefficient a in front is also the value of one of the roots. Sure this could happen but as I said I do doubt it.

Also if you choose a to be 1, then why do you have (x-a) instead of (a-1)???? Shouldn't the answer be f(x) = (x - 1)(x - b)(x - c)

 

[mathdad]

I am sorry but I do not believe that the problem was written as you showed. I just can't see that the coefficient a in front is also the value of one of the roots. Sure this could happen but as I said I do doubt it.

Also if you choose a to be 1, then why do you have (x-a) instead of (a-1)???? Shouldn't the answer be f(x) = (x - 1)(x - b)(x - c)

Yes, you got the right polynomial.

 

[Otis]

... I do not believe that the problem was written as you showed ...

I think mathdad makes up his own exercises, from time to time.

\(\;\)

 

[mathdad]

I think mathdad makes up his own exercises, from time to time.

\(\;\)

I never make up my own questions. I'm not that smart.

 

[mathdad]

I am sorry but I do not believe that the problem was written as you showed. I just can't see that the coefficient a in front is also the value of one of the roots. Sure this could happen but as I said I do doubt it.

Also if you choose a to be 1, then why do you have (x-a) instead of (a-1)???? Shouldn't the answer be f(x) = (x - 1)(x - b)(x - c)

It is the right function according to the answer section.

 

[Dr.Peterson]

Use f(x) = a(x - r_1)(x - r_2)(x - r_3), where r represents the zeros, to find a polynomial function.

Zeros: a, b, c; degree 3

Let a = 1

f(x) = (x - a)(x - b)(x - c)

It is the right function according to the answer section.

Yes, your work is correct. And the problem, while not ideal, is also valid.

Note that there is an "a" in the generic formula they say to use, and there is a different "a" in the problem. When you said, "let a = 1", you were clearly referring to the former, not the latter. The only quibble I might give (and in fact I wouldn't, because you are not the teacher) is that it would be clearer to distinguish the two "a"s when you talk about them, so no one would confuse them. The textbook, not you, caused this little difficulty!

This is not uncommon; we might, for example, have the general rule that a^m * a^n = a^{m+n}, and then apply it to simplify (a+1)^2 * (a+1)^3. In explaining it, I might say things like "the a in the formula represents (a+1) in this problem"; or I might just rewrite the formula as "b^m * b^n = b^{m+n} to avoid confusion. Students need this sort of help; but teachers should have no trouble understanding what you are saying, because we're used to juggling multiple levels of symbols.

 

[mathdad]

Yes, your work is correct. And the problem, while not ideal, is also valid.

Note that there is an "a" in the generic formula they say to use, and there is a different "a" in the problem. When you said, "let a = 1", you were clearly referring to the former, not the latter. The only quibble I might give (and in fact I wouldn't, because you are not the teacher) is that it would be clearer to distinguish the two "a"s when you talk about them, so no one would confuse them. The textbook, not you, caused this little difficulty!

This is not uncommon; we might, for example, have the general rule that a^m * a^n = a^{m+n}, and then apply it to simplify (a+1)^2 * (a+1)^3. In explaining it, I might say things like "the a in the formula represents (a+1) in this problem"; or I might just rewrite the formula as "b^m * b^n = b^{m+n} to avoid confusion. Students need this sort of help; but teachers should have no trouble understanding what you are saying, because we're used to juggling multiple levels of symbols.

Thank you very much.