Question

What is the quotient and remainder, written as partial fractions, of startfraction 2 x cubed minus 25 x 40 over x squared minus 2 x minus 8 endfraction? 2 x minus 4 minus startfraction 2 over x 4 endfraction startfraction 1 over x minus 2 endfraction 2 x minus 4 startfraction 2 over x 4 endfraction startfraction 1 over x minus 2 endfraction 2 x minus 4 minus startfraction 2 over x 4 endfraction minus startfraction 1 over x minus 2 endfraction 2 x minus 4 startfraction 2 over x 4 endfraction minus startfraction 1 over x minus 2 endfraction

Answer 1

The question involves performing polynomial long division to divide a polynomial by a quadratic polynomial and then expressing the remainder as partial fractions. Through long division, we would obtain a quotient and remainder which can then be broken down into partial fractions.

Step-by-step explanation:

The question appears to be asking for the division of the polynomial 2x³ - 25x + 40 by the quadratic polynomial x² - 2x - 8, and expressing the result as a quotient with a remainder, followed by breaking the remainder down into partial fractions. To find the quotient and remainder, we perform polynomial long division.

First, we factor the denominator: x² - 2x - 8 can be factored as (x - 4)(x + 2). Polynomial long division would show that the quotient is 2x + 3 and the remainder is 16x - 4. We then express the remainder over the original denominator as a partial fraction: \(rac{16x - 4}{(x - 4)(x + 2)}\).

To further break this down into partial fractions, we write it as \(A/(x - 4) + B/(x + 2)\), and solve for A and B, which gives us the complete expression in terms of partial fractions.

Answer 2

The correct choice is D.

To find the partial fraction decomposition of the given rational function
`(2 x^3-25 x+40)/(x^2+2 x-8)`, we first factor the denominator.

The denominator factors as follows:

`x^2+2 x-8=(x-2)(x+4)`

Now, we can express the given rational function as a sum of partial fractions with undetermined coefficients:

`(2 x^3-25 x+40)/(x^2+2 x-8)=(A)/(x-2)+(B)/(x+4)`

where A and B are constants.

Next, we clear the fractions by multiplying both sides by the common denominator (x−2)(x+4):

`2 x^3-25 x+40=A(x+4)+B(x-2)`

Now, we can determine the values of A and B by comparing coefficients. We can do this by substituting appropriate values for x that eliminate one of the terms, making it easier to solve for the unknowns.

Let's start with x=2:

`2(2)^3-25(2)+40=A(2+4)`

Simplifying the left side:

`$16-50+40=6 A$`

which gives A=−1.

Now, let's try x=−4:

`2(-4)^3-25(-4)+40=B(-4-2)`

Simplifying the left side:

`-128+100+40=-6 B`

which gives B=−12.

Now we can write the original expression as partial fractions:

`$(2 x^3-25 x+40)/(x^2+2 x-8)=(-1)/(x-2)+(-12)/(x+4)$`

To write this in the form given in the choices, factor out the common factor from the numerators:

`(2 x^3-25 x+40)/(x^2+2 x-8)=-(1)/(x-2)-(12)/(x+4)=2-(2)/(x+4)-(1)/(x-2)`

Now, compare this to the provided choices:

`2 x-4+(2)/(x+4)-(1)/(x-2)`

Complete Question:
What is the quotient and remainder, written as partial fractions, of
`(2 x^3-25 x+40)/(x^2+2 x-8)` ?

a.
`& 2 x-4-(2)/(x+4)+(1)/(x-2) \\`

b.
`& 2 x-4+(2)/(x+4)+(1)/(x-2) \\`

c.
`2 x-4-(2)/(x+4)-(1)/(x-2)`

d.
`2 x-4+(2)/(x+4)-(1)/(x-2)`