Calculus Help with Synthetic Division?

[Nathalie]

The instruction is "Use synthetic division to find the required function values."

The problem is
g(x) = x^3 - x^2 + 25x - 25
(a) g(5)
(b) g(1/5)
(c) g(-1.5)
(d) g(-1)

So, I understand how to use Synthetic Division when I am giving a dividend and a divisor such as (x - 5) or things like that. But... I don't get this questions because if g(5) doesn't that mean all the x is 5? So why don't I just simply replace all the x with the 5 or whatever given value?

Please help clarify. Thanks in advance.

 

[mmm4444bot]

The instruction is "Use synthetic division … [to find g(5)] …

… why don't I just simply replace all the x with the 5 … The reason why is because you're required to follow the given instruction.

Do you know that the remainder IS g(5), after dividing g(x) by x - 5 ?

In other words, g(5) = 200 because 200 is the last number we get from synthetic division (i.e., the remainder).

     5     |     1     -1     25     -25
           |            5     20     225
           |----------------------------
           |     1      4     45     200

 

[kristopher0123]

Whoa that helped me so much lol, I just understood what the remainder means and what 'getting the zeros' does hahaha. And I thought I knew it all =P

 

[Nathalie]

I am a bit lost now

So whatever value that is inside the () after g is the value that I am supposed to divide the dividend with?

As in...

g(x) = x^3 - x^2 + 25x - 25
(a) g(5)
(b) g(1/5)
(c) g(-1.5)
(d) g(-1)

For example, for (b) what I will be doing is...

     1/5   |     1     -1     25     -25
           |            
           |----------------------------
           |

Right?
What do I do with negative? When I put it on the left side of the Synthetic Division bar, I use positive value right?

Please correct me if I have misunderstood... I am not good with Math :/
Thanks a lot.

p.s. that x-5 is just a random term that I pulled off... not given for this problem.

 

[mmm4444bot]

I am a bit lost now

So whatever value that is inside the () after g is the value that I am supposed to [use in the synthetic-division process]? YES!

As in...

g(x) = x^3 - x^2 + 25x - 25
(a) g(5)
(b) g(1/5)
(c) g(-1.5)
(d) g(-1)

For example, for (b) what I will be doing is...

     1/5   |     1     -1     25     -25
           |            
           |----------------------------
           |

Right? CORRECT !

What do I do with negative? When I put it on the left side of the Synthetic Division bar, I use positive value right? NO.

With -1.5, it means we're dividing g(x) by (x + 1.5), so go through the synthetic-division process in the usual way, multiplying by -1.5.

 

[kristopher0123]

Idk if this will be helpful or not but heres how i learned it. you drop the first number to the bottom of the box (1/5), then multiply it by the number in the box and put it in the space under the second number, then add those two numbers together, put it below the line under the second column, then multiply it by the number in the box, then put that number in third space, add those together, and put it below the numbers in the third column, etc. until you're done. sorry i can't show you an example, i have no idea how to make the equations and what not like you guys do haha...

 

[mmm4444bot]

Idk if this will be helpful or not …

Kristopher, the original poster already stated understanding the synthetic-division algorithm; their questions lie elsewhere.

 

[Nathalie]

Ah. I get it now. Thank you very much =]!

Though I am confused with other questions, which I will post soon if I can't figure out how to do from the textbooks... Sorry for stupid question like this one. I did not take Algebra courses so I am confused with a lot of stuff. I am trying to self study for the upcoming test.. but yeah... having a lot of questions. Anyways! Thank you very much for the help and also with the correction of the wording

 

[kristopher0123]

Re:

Idk if this will be helpful or not …

Kristopher, the original poster already stated understanding the synthetic-division algorithm; their questions lie elsewhere.

Well haha, like I said 'Idk if this will be helpful or not' =P