According to the fundamental theorem of algebra, which polynomial function has exactly 11 roots

f(x) = (x+2)³(x²−7x+3)⁴

Step-by-step explanation:

The fundamental theorem of algebra (in its simplest definition), tells us that a polynomial with a degree of n will have n number of roots.

recall that the degree of a polynomial is the highest power that exists in any variable.

i.e.

polynomial, p(x) = axⁿ + bxⁿ⁻¹ + cxⁿ⁻² + + k

has the degree (i.e highest power on a variable x) of n and hence has n-roots

In our case, if we expand all the polynomial choices presented, if we consider the 2nd choice:

f(x) = (x+2)³(x²−7x+3)⁴   (if we expand and simplify this, we end up with)

f(x) = x¹¹−22x¹⁰+150x⁹−116x⁸−2077x⁷+3402x⁶+11158x⁵−8944x⁴−10383x³+13446x²−5076x+648

we notice that the term with the highest power is x¹¹

hence the polynomial has a degree of 11 and hence we expect it to have exactly 11 roots

8

step-by-step explanation:

this is a subtraction problem.

10 - 2 = 8

$8

step-by-step explanation:

it is pretty much backwards.

you take the 4x2 multiply that, and that is = $8