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Sunny1982

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Hi folks
I was wondering if any folks can help me with this revision question, f(x)=x3+x2-4x-4 hence I need to solve this equation x3+x2-4x-4=0. I just can't get my head around it what could it be. I'm guessing this bit x3+x2-4x-4=0 is the last bit of the equation.
 
Sunny1982 said:
Hi folks
I was wondering if any folks can help me with this revision question, f(x)=x3+x2-4x-4 hence I need to solve this equation x3+x2-4x-4=0. I just can't get my head around it what could it be. I'm guessing this bit x3+x2-4x-4=0 is the last bit of the equation.
Click to expand...

\(\displaystyle x^3+x^2-4x-4=x^2(x+1)-4(x+1)=(x+1)(x^2-4)=(x+1)(x+2)(x-2)\)
 
is this correct then?


1. Usethe factor theorem to factorise completely f(x)= x³+x²-4x-4.
Hence solve the equation x³+x²-4x-4=0

x³+x²-4x-4=0
x² (x + 1) - 4(x + 1) = 0
(x + 2)(x + 1)(x - 2) = 0
x = -2 , -1 , 2
 
Sunny1982 said:
is this correct then?
1. Usethe factor theorem to factorise completely f(x)= x³+x²-4x-4.
Hence solve the equation x³+x²-4x-4=0
x³+x²-4x-4=0
x² (x + 1) - 4(x + 1) = 0
(x + 2)(x + 1)(x - 2) = 0
x = -2 , -1 , 2
Click to expand...
Yes, it is correct.
 
Sunny1982 said:
I'm not being rude how come you wrote it wrong to the way I wrote it? I just want to know and learn different ways of interpreting answers.
Click to expand...

I have absolutely no idea what you mean by that.
What I posted is exactly the same as what you posted, the only difference is the order of the factors.
But multiplication is commutative. So order makes no difference whatsoever;
 
I was pretty confident this question was correct but when I handed it to my tutor he wrote the following notes back and told me to correct it, I don't understand where I have gone wrong:(
1. Usethe factor theorem to factorise completely f(x)= x³+x²-4x-4.
Hence solve the equation x³+x²-4x-4=0

x³+x²-4x-4=0
x² (x + 1) - 4(x + 1) = 0
(x + 2)(x + 1)(x - 2) = 0
x = -2 , -1 , 2

Initially you had f(x) =x³+x²-4x-4.

Then checking you find f(-1) = 0.

Hence f(x)=(x+1)g(x) and work out what g is.

And repeat for g(x).

It's not the easiest way to do it. And your method would probably be the best way to go about it if they hadn't said "Use the factor theorem".
 
Sunny1982 said:
I was pretty confident this question was correct but when I handed it to my tutor he wrote the following notes back and told me to correct it, I don't understand where I have gone wrong:(
1. Usethe factor theorem to factorise completely f(x)= x³+x²-4x-4.
Hence solve the equation x³+x²-4x-4=0

x³+x²-4x-4=0
x² (x + 1) - 4(x + 1) = 0
(x + 2)(x + 1)(x - 2) = 0
x = -2 , -1 , 2

Initially you had f(x) =x³+x²-4x-4.

Then checking you find f(-1) = 0.

Hence f(x)=(x+1)g(x) and work out what g is.

And repeat for g(x).

It's not the easiest way to do it. And your method would probably be the best way to go about it if they hadn't said "Use the factor theorem".
Click to expand...
It seems to me that what your tutor is saying is very simple. You were told to solve the problem using the Factor Theorem, not by factoring. You solved the problem by factoring so you did not do what the problem asked. As you have specified the problem, however, the Factor Theorem does not seem to me to lead directly to a solution. Moreover your tutor's notes seem to imply that there was part of the problem statement that you did not give to us, something about f(- 1). Did you give an exact and complete problem statement?
 
To use the Factor Theorem, you need to start with the solutions,
Hence solve the equation x³+x²-4x-4=0, (by inspection is probably ok -- but NOT by factoring)

As Jeff M pointed out, it is probably the form of your answer that is troubling your tutor.
Use these steps..
1) Find roots of x³+x²-4x-4=0, (use newton, guess, cubic formula -- but NOT by factoring)
2) Try each root in f(x) and demonstrate that f(x)=0 for each of the three x-values
3) Apply (state the) Factor Theorem
4) Write f(x) in factored form.
 
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