Question

For the Polynomial function p(x) = `2x^3 + 4x^2` + 8x + 16, Determine the maximum number of real zeroes that the polynomial function may have ? Determine the right and left behavior of the graph of the polynomial function.?

Answer 1

Possible rational zeroes:
`1, -1, 2, -2, 4, -4, 8, -8, 16, -16, 1/2, -1/2.`
Real zeroes: `x=-2` Factoring over the complex number system:
`p(x)=2(x+2i)(x-2i)(x+2)` Complex zeroes: `x=2i, x=-2i` e. `2x^3` f. Right end behavior is heading upwards (approaching infinity), left end behavior is heading downwards (approaching minus infinity).

Explanation:

a. We determine the potential zeroes by rational root theorem , which tells us for a polynomial `p(x)=ax^n+bx^(n-1)+...+cx+d`, the possible roots are given by the positive and negative factors of `d` divided by the positive and negative factors of `a,` or the positive and negative factors of the constant term divided by the positive and negative factors of the coefficient of the term of highest degree.

Here, `d=16, a=2,` factors of `16` include `(+-1, +-2, +-4, +-8, +-16)`, factors of `2` include `+-1, +-2`, so possible zeroes are given by:

`1, -1, 2, -2, 4, -4, 8, -8, 16, -16, 1/2, -1/2.`

b. For the real zeroes, we'll need to factor `p(x)=2x^3+4x^2+8x+16` and solve by setting `p(x)=0`. Applying the rational root theorem and using it to find the actual roots is tedious. Fortunately, this can be factored by grouping:

`2x^3+4x^2+8x+16=0`

Group together the relevant terms:

`(2x^3+4x^2)+(8x+16)=0`

Factor:

`2x^2(x+2)+8(x+2)=0`

`(2x^2+8)(x+2)=0`

For `2x^2+8=0:`

`2(x^2+4)=0, x^2+4=0, x^2=-4, x=+-sqrt(-4)=+-2i` -- these are not real zeroes, so they're not relevant to the problem so far.

For `x+2=0: x=-2`

c. To factor over the complex number system, let's get back to our factored form (which is not factored over complex numbers yet):

`p(x)=(2x^2+8)(x+2)`

The term involving complex solutions is `2x^2+8`. Recall we solved this already, seeing that `x=2i, -2i`. So, the factored form of `2x^2+8` is `2(x+2i)(x-2i),` and the polynomial factored over the complex number system is thus

`p(x)=2(x+2i)(x-2i)(x+2)`

d. `x=2i, x=-2i`, these were solved for earlier on.

e. For large values of `x,` the graph of the function simply resembles the graph of the term of highest degree (with the highest exponent), or `2x^3.`

graph{2x^3+4x^2+8x+16 [-10, 10, -5, 5]}

f. In general, for polynomials of third degree (the highest exponent is three) and positive coefficients on the cubed term, the right behavior is heading upwards (approaching infinity) and the left behavior is heading downards (approaching minus infinity).