- When the function f(x) is divided by x 2, the quotient is x² 10x 8 and the
remainder is – 8. Find the function f(x) and write the result in standard form.

3.2 The Factor Theorem and The Remainder Theorem 257

3.2 The Factor Theorem and The Remainder Theorem

Suppose we wish to find the zeros of f(x) = x

3 + 4x

2 − 5x − 14. Setting f(x) = 0 results in the

polynomial equation x

3 + 4x

2 − 5x − 14 = 0. Despite all of the factoring techniques we learned1

in Intermediate Algebra, this equation foils2 us at every turn. If we graph f using the graphing

calculator, we get

The graph suggests that the function has three zeros, one of which is x = 2. It’s easy to show

that f(2) = 0, but the other two zeros seem to be less friendly. Even though we could use the

‘Zero’ command to find decimal approximations for these, we seek a method to find the remaining

zeros exactly. Based on our experience, if x = 2 is a zero, it seems that there should be a factor

of (x − 2) lurking around in the factorization of f(x). In other words, we should expect that

x

3 + 4x

2 − 5x − 14 = (x − 2) q(x), where q(x) is some other polynomial. How could we find such

a q(x), if it even exists? The answer comes from our old friend, polynomial division. Dividing

x

3 + 4x

2 − 5x − 14 by x − 2 gives

x

2 + 6x + 7

x−2 x

3 + 4x

2 − 5x − 14

−

x

3 −2x

2

6x

2 − 5x

−

6x

2 −12x)

7x − 14

− (7x −14)

0

As you may recall, this means x

3 + 4x

2 − 5x − 14 = (x − 2)

x

2 + 6x + 7

, so to find the zeros of f,

we now solve (x − 2)

x

2 + 6x + 7

= 0. We get x − 2 = 0 (which gives us our known zero, x = 2)

as well as x

2 + 6x + 7 = 0. The latter doesn’t factor nicely, so we apply the Quadratic Formula to

get x = −3 ±

√

2. The point of this section is to generalize the technique applied here. First up is

a friendly reminder of what we can expect when we divide polynomi

Step-by-step explanation:

last option

step-by-step explanation:

given that δdef is right and isoceles,

assuming that side df is the hypotenuse means that de an ef are the other 2 sides that intersect at 90° and are of equal length.

because they are of equal length, we can say that they are congruent

because they intersect at 90°, their slopes must be opposite reciprocals.

only the last option satisfies these conditions.

answer:

your answer is: c. 196 7/8 in^3.