Question

Can someone please help me with a few "Solving Polynomial Equations" problems? Any help will be greatly appreciated! :D

Answer 1

Here's the first one: Determine the zeros of f(x) = x3 – 3x2 – 16x + 48.

Answer 2

@amistre64 @phi

Answer 3

did you try plotting it in geogebra ?

Answer 4

we have to show our work... :(

Answer 5

What approach do they expect you to use to solve this ?

Answer 6

The zeros of the function represent the values of x that make the function equal to zero. So you may replace the function notation with 0.

f(x) = x2 + 7x + 6
0 = x2 + 7x + 6

Since the depressed equation is a quadratic equation, you may algebraically solve for x by factoring, using the quadratic formula, or by completing the square. Remember, looking at the discriminant is the best way to find out which method should be used.

0 = x2 + 7x + 6
a = 1, b = 7, c = 6

b2 – 4ac
(7)2 – 4(1)(6)
49 – 4(1)(6)
49 – 24
25

The value of the discriminant is a perfect square, which means this quadratic may be solved
by factoring.

Review

Show additional review on the discriminant

f(x) = (x + 1)(x + 6)

Set both factors equal to 0 and solve for x.

0=x + 1 0=x + 6
–1 –1 –6 –6
–1 = x –6 = x

The two remaining zeros of f(x) = x3 + 3x2 – 22x – 24 are –1 and –6. These zeros may be tested using synthetic division.

synthetic division with the factor, negative 1, in the top left corner, the coefficients 1, 3, negative 22, and negative 24 to the right, and 1, 2, negative 24, and 0 in the last rowsynthetic division with the factor, negative 6, in the top left corner, the coefficients 1, 3, negative 22, and negative 24 to the right, and 1, negative 3, negative 4, and 0 in the last row

Both values of –1 and –6 have a remainder of 0. Therefore, they are both zeros of the function!

Answer 7

That's what it says in the lesson

Answer 8

yes, but you need at least one root (then you use synthetic division to reduce the problem to a quadratic)

do they expect you to know the rational root test ?

Answer 9

The rational root theorem?

Answer 10

First, use the Rational Root Theorem to find the factors of p over q. The p value is the constant term –24. This is divisible by ±1, ±2, ±3, ±4, ±6, ±8, ±12, and ±24.

The q value is the leading coefficient of 1. One is divisible only by itself and –1.

Take each of the factors from p and write them as a fraction with the factors of q as the denominator. Make sure you account for all combinations.

Factors of p: ±1, ±2, ±3, ±4, ±6, ±8, ±12, and ±24 Factors of q: ±1

Factors of p over q: ±1 over 1, ±2 over 1, ±3 over 1, ±4 over 1, ±6 over 1, ±8 over 1, ±12 over 1, ±24 over 1

These fractions may all be simplified so the factors of p over q are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24. This means there are sixteen rational values that could possibly be zeros of the function. That's a lot
of possibilities!

Answer 11

This is the problem they're using f(x) = x3 + 3x2 – 22x – 24

Answer 12

yes, that is it

for your problem
f(x) = x^3 – 3x^2 – 16x + 48

you need the factors of 48 divided by the factors of 1 (1 is the number in front of the x^3)

the factors of 1 are just 1, and dividing by 1 does not change the answer.

so your *possible factors* are the factors of 48

can you list the factors of 48 ?

Answer 13

can you list the factors of 48 ?

Answer 14

1 and 48 are the first set, so you put ±1, ±48 on the list

can you find more factors ?

Answer 15

does 2 divide into 48 ?

Answer 16

+/-(1,2,3,4,6,8,12,24,48)

Answer 17

yes, a very long list. now the idea is try each of those numbers in the equation

f(x) = x^3 – 3x^2 – 16x + 48
replace x with a candidate number, do the arithmetic, and see if you get 0
once you find a number that gives 0, stop and we go on to the next step.
there are 18 numbers (counting positive and negative), which is a boatload of work

can you do it ?

Answer 18

Since the degree of the polynomial is 3, doesn't that mean that there will be 3 zeros?

Answer 19

yes, but once we find the first one, we can use synthetic division to make the problem a quadratic (degree 2), and then factor it. that will be less work

Answer 20

okay. So I just plug in the factors until I find one that works?

Answer 21

yes, it is not my idea of fun. start with ±1,±2±3
(assuming the answer will be a small number )

Answer 22

3 works

Answer 23

yes, so (x-3) is one of the factors and x=3 is one of the roots (makes the equation 0)

Answer 24

okay, now what?

Answer 25

do you know how to do synthetic division

Answer 26

yeah

Answer 27

the setup is