Find a degree 6 polynomial with - 8 as a root, no other roots, and in which the coefficient of x^6 is 16. Assume that all (non-constant) factors of the polynomial correspond to real zeroes.
To find a degree 6 polynomial with -8 as a root and no other roots, we need to use the factor theorem. According to the factor theorem, if a polynomial has a root "r", then (x - r) is a factor of the polynomial. In this case, (x - (-8)) or (x + 8) is a factor of the polynomial.
Now, since we want the coefficient of x^6 to be 16, we can start building our polynomial by multiplying it by (x + 8):
p(x) = 16(x + 8) = 16x + 128
However, this is only a linear polynomial, and we need it to be degree 6. To achieve this, we can multiply it by additional (x - r) terms, where "r" represents any other root except -8.
So, let's say we choose to multiply p(x) by (x - a_1), where "a_1" is any real number other than -8. This would give us:
p(x) = (x - a_1)(16x + 128)
Now, since we want no other roots, we need to ensure that (x - a_1) is not a factor of our polynomial. To do this, we want to ensure that the term (x - a_1) doesn't divide evenly into (16x + 128), thus making the polynomial equal to zero when "x" is equal to "a_1". One way to guarantee this is to set the coefficient of "x" in (16x + 128) to a non-zero value.
Let's choose a value, say 2:
p(x) = (x - a_1)(16x + 128)
= (x - a_1)(2(8x + 64))
= 2(x - a_1)(8x + 64)
Now, it's important to note that we can choose any value for "a_1", as long as it's different from -8.
By repeating this process five more times, choosing different values for "a_2", "a_3", ..., "a_6", we can create a degree 6 polynomial with -8 as a root, no other roots, and a coefficient of x^6 equal to 16.