Idon’t understand this topic. someone ! according to the rational roots theorem what are all the possible roots of f(x)=15x^11-6x^8+x^3-; x+3

Let's take a simple example, a quadratic. If "a" and "b" are roots of the quadratic, then it can be factored as

... (x -a)(x -b) = 0

When this is multiplied out, you get

... x² -(a+b)x +ab

That is, the constant term of the quadratic is the product of the roots of the quadratic. (Note that the coefficient of x² is 1.) If we multiply by more binomial terms, the result is still the same: the constant term is the product of the roots (when the leading coefficient is 1).

This says that if we divide by the leading coefficient, so we get

... f(x)/15 = x¹¹ -(6/15)x⁸ +(1/15)x³ -(1/15)x +3/15

the product of the roots of f(x) will be 3/15. That is, any rational roots will be factors of 3/15, so will be of the form (factor of 3)/(factor of 15).

Factors of 3 include ±1, ±3. Factors of 15 include ±1, ±3, ±5, ±15. To find the possible rational roots, we combine these in all the possible ways, then list the unique values.

(±1)/(±1), (±1)/(±3), (±1)/(±5), (±1)/(±15), (±3)/(±1), (±3)/(±3), (±3)/(±5), (±3)/(±15) . . . the complete list

Sorting out unique values, we have ±1/15, ±1/5, ±1/3, ±3/5, ±1, ±3.

yes

step-by-step explanation: