According to the fundamental theorem of algebra, which polynomial function has exactly 11 roots
![According to the fundamental theorem of algebra which polynomial function has exactly 11 roots class=](https://us-static.z-dn.net/files/d7d/78e79b7b78eeed161117cb615c54ae91.png)
Answer:
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