## Polynomial Functions Questions

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.

### Polynomial Functions Questions

Hello,

Thanks for taking your time reading. I have encountered 2 problems as I was going through my textbook, so may I please kindly ask of some assistance on this.

"Find the polynomial f such that f(2x-1) = 16x^4 - 32x^3 +12x^2"

And

"When polynomial f is divided by (x-3) the remainder is 2, and when divided by (x+1) the remainder is -4. Find the remainder when polynomial f is divided by (x^2 -2x - 3)"

Thank you very much.
cruxxfay

Posts: 13
Joined: Tue Aug 14, 2012 10:56 am

### Re: Polynomial Functions Questions

Under normal conditions, a reply would not be so complete as is the one that follows. However, Purplemath does not, to my knowledge, currently offer tutorials on these particular topics. Therefore, I will provide a more extensive response.

cruxxfay wrote:"Find the polynomial f such that f(2x-1) = 16x^4 - 32x^3 +12x^2"

The result polynomial is the composition of the unknown f(x) with g(x) = 2x - 1. By nature of composition of functions, we know that (f o g o g-1)(x) = f(x). Since g(x) = 2x - 1, then g-1(x) = (x + 1)/2. We are given that (f o g)(x) = 16x4 - 32x3 + 12x2. Combining these results, we find:

. . . . .$\left(f \circ g \circ g^{-1}\right)(x)\, =\, \left(f \circ g\right)\left(\frac{x\, +\, 1}{2}\right)$

. . . . . . .$=\, 16\left(\frac{x\, +\, 1}{2}\right)^4\, -\, 32\left(\frac{x\, +\, 1}{2}\right)^3\, +\, 12\left(\frac{x\, +\, 1}{2}\right)^2$

Simplify to obtain the required polynomial answer.

cruxxfay wrote:"When polynomial f is divided by (x-3) the remainder is 2, and when divided by (x+1) the remainder is -4. Find the remainder when polynomial f is divided by (x^2 -2x - 3)"

The Factor Theorem states that any polynomial f(x) may be restated as p(x)q(x) + r(x), where p(x) is the (polynomial) divisor, q(x) is the result (that is, the quotient), and r(x) is the remainder. The Remainder Theorem, related to the Factor Theorem, states that the value of a polynomial f(x) at x = a is the same as the value of the remainder when f(x) is divided by x - a. We use these as follows:

We are given that f(x) must be divisible by the given quadratic. Therefore:

. . . . .$f(x)\, =\, (x^2\, -\, 2x\, -\, 3)q(x)\, +\, r(x)\, =\, (x\, -\, 3)(x\, +\, 1)q(x)\, +\, r(x)$

Immediately, we note that the specified linear divisors are factors of the quadratic divisor. Continuing directly and applying the Remainder Theorem, we find:

. . . . .$f(3)\, =\, (3\, -\, 3)(3\, +\, 1)q(3)\, +\, r(3)\, =\, (0)(4)q(x)\, +\, r(3)\, =\, 2$

So r(3) = 2.

. . . . .$f(-1)\, =\, (-1\, -\, 3)(-1\, +\, 1)q(-1)\, +\, r(-1)\, =\, (-4)(0)q(-1)\, +\, r(-1)\, =\, -4$

So r(-1) = -4. Since we are dividing by a quadratic, then logically the remainder can have a degree of 0 or 1. Since r(x) is clearly not constant, then r(x) must be linear. Then r(x) = ax + b for some values a and b. One may then find the equation of the remainder as being r(x) = (3/2)x - (5/2).

Since f(x) = (x2 - 2x - 2)p(x) + r(x), then the remainder, upon division by (x2 - 2x - 2), must be r(x).

Kindly please reply with queries, should any of the above leave questions in your mind. Thank you.
nona.m.nona

Posts: 256
Joined: Sun Dec 14, 2008 11:07 pm

### Re: Polynomial Functions Questions

So r(-1) = -4. Since we are dividing by a quadratic, then logically the remainder can have a degree of 0 or 1. Since r(x) is clearly not constant, then r(x) must be linear. Then r(x) = ax + b for some values a and b. One may then find the equation of the remainder as being r(x) = (3/2)x - (5/2).

I know how it is in the form y=ax+b because it is a quadratic, but how did you get to (3/2) x - (5/2)?
cruxxfay

Posts: 13
Joined: Tue Aug 14, 2012 10:56 am

### Re: Polynomial Functions Questions

cruxxfay wrote:I know how it is in the form y=ax+b because it is a quadratic, but how did you get to (3/2) x - (5/2)?

Given the two points on the line, one may use algebraic methods to find the equation of that line: Find the slope, and then utilize that slope value and either of the original points, together with one of the forms of the generic line equation, to find the equation of this particular line.
nona.m.nona

Posts: 256
Joined: Sun Dec 14, 2008 11:07 pm