Question

A polynomial function has a root of –6 with multiplicity 3 and a root of 2 with multiplicity 4. If the function has a negative leading coefficient and is of odd degree, which could be the graph of the function

Answer 1

When you set the factors of a polynomial equal to zero, you can find the roots. This is because the roots are the values of x that make the function equal zero. Multiplicity is the number of times the roots repeat.

Root of -6 with multiplicity 1 is the factor (x + 6).
Root of -2 with multiplicity 3 is the factor (x + 2)3.
Root of 0 with multiplicity 2 is the factor x2.
Root of 4 with multiplicity 3 is the factor (x - 4)3.


The function is

f(x) = x2(x + 6)(x + 2)3(x - 4)3

f(x) = x*x(x + 6)(x + 2)(x + 2)(x + 2)(x - 4)(x - 4)(x - 4)


If we expand this polynomial, the leading term of the polynomial will be x9. The leading term was obtained by multiplying out the first terms of each binomial together. The last term of the polynomial will be -3072. This was obtained by multiplying out the last terms of each binomial together.

Answer 2

Solution:

When zero is set to the polynomial factor, we can easily find the root. This is due to the fact that root are the value of x that makes to zero. The root repeats based on the number of multiplicity.

The root of "- 6" with multiplicity of 1 then factor is  (x + 6).

The root of "- 2" with multiplicity of 3 then factor is $(x + 2)^{3}$.

The root of "0" with multiplicity of 2 then factor is $x^{2}$.

The root of "4" with multiplicity of 3 then factor is $(x - 4)^{3}$.

The function is

$f (x) = x^{2} \times (x + 6) \times (x + 2)^{3} \times (x - 4)^{3}$

$f (x) = x \times x \times (x + 6) \times (x + 2) \times (x + 2) \times (x + 2) \times (x - 4) \times (x - 4) \times (x - 4)$