Found this in my year 10 textbook on polynomials and doesn't know the meaning

[sbaamu1910]

Hello, I am just looking for an explanation on subscripts and I need help with this question:

Let P(x) = a_*x^n + a_n-1*x^n-1 + ... + a_1 +a_0 be a polynomial where the coefficients are integers. Also let P(w) = 0 where w is an integer. Show that w is a factor of a_0.

I am not sure if I denoted that problem correctly,

 

[stapel]

Hello, I am just looking for an explanation on subscripts and I need help with this question:

Let P(x) = a_*x^n + a_n-1*x^n-1 + ... + a_1 +a_0 be a polynomial where the coefficients are integers. Also let P(w) = 0 where w is an integer. Show that w is a factor of a_0.

I am not sure if I denoted that problem correctly,

The subscripts are just a way of naming the different numerical coefficients, and pairing them with their terms (being the variable having the same superscript/power as the coefficient has subscript/name).

 

[lev888]

I think it should be "a_1*x", not "a_1". 1 is the power of x.
You are told that if x is w, the value of the polynomial is 0. Given this, you are asked to prove that w is a factor of a_0.
I would start by plugging in w and writing down the resulting equality. Then "solve it" for a_0. What do you get?

 

[Dr.Peterson]

Hello, I am just looking for an explanation on subscripts and I need help with this question:

Let P(x) = a_*x^n + a_n-1*x^n-1 + ... + a_1 +a_0 be a polynomial where the coefficients are integers. Also let P(w) = 0 where w is an integer. Show that w is a factor of a_0.

I am not sure if I denoted that problem correctly,

Here is the problem formatted more nicely and slightly corrected:

Let [imath]P(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x +a_0[/imath] be a polynomial where the coefficients are integers. Also let [imath]P(w) = 0[/imath] where [imath]w[/imath] is an integer. Show that [imath]w[/imath] is a factor of [imath]a_0[/imath].​

A simple example would be [imath]P(x)=2x^2-8x+6[/imath]; the coefficients, [imath]a_2=2[/imath], [imath]a_1=-[/imath]8, and [imath]a_0=6[/imath] are integers; taking [imath]w=3[/imath], we find that [imath]P(3)=2(3)^2-8(3)+6=0[/imath], and the claim is that therefore 3 is a factor of 6, as it is.

Now please tell us (a) what it is about subscripts that you don't understand, and (b) what you have learned about polynomials that might be relevant. Do you know things like the factor theorem, or the rational root theorem?

 

[Steven G]

I may be mistaken but the rational root theorem is taught in precalculus. Then again, we don't have a precalculus section.

Continuing with Dr Peterson's post we can solve for 6. We get that 6=8(3) - 2(3)². Clearly 3 is a factor of the left side. Then 3 is a factor of the right hand side which is 6.