MHB Determine the values of f(6) and g(6)

[laprec]

Kindly help with this problem solving:

f(x) is divisible by x^2-5x-6 and g(x) is divisible by x^2-2x-3
f(x)=g(x)+2x+2
a) Determine the values of f(6) and g(6)
b) State common factor for the two functions

Attempts:
For f(x):factors are x^2-5x-6=(x-6)(x+1)

For g(x):Factors are x^2-2x-3=(x-3)(x+1)

∴for question b: the common factor will be (x+1)for the two functions

Kindly assist with question (a)

 

[Opalg]

Kindly help with this problem solving:

f(x) is divisible by x^2-5x-6 and g(x) is divisible by x^2-2x-3
f(x)=g(x)+2x+2
a) Determine the values of f(6) and g(6)
b) State common factor for the two functions

Attempts:
For f(x):factors are x^2-5x-6=(x-6)(x+1)

For g(x):Factors are x^2-2x-3=(x-3)(x+1)

∴for question b: the common factor will be (x+1)for the two functions

Kindly assist with question (a)

What you have done so far is correct. You have shown that $f(x)$ has factors $x-6$ and $x+1$ (but it may also have other factors). From that, you can say that $f(x) = (x-6)(x+1)p(x)$, for some unknown function $p(x)$. What happens if you put $x=6$ in that equation?

You are told that $f(x) = g(x) + 2x + 2$. Once you know $f(6)$, you can put $x=6$ in that equation to get $g(6)$.

You have shown that $g(x)$ has factors $x-3$ and $x+1$ (but it may also have other factors). From that, you can say that $g(x) = (x-3)(x+1)q(x)$, for some unknown function $q(x)$. So you know that $f(x)$ and $g(x)$ have a common factor $x+1$. But could $f(x)$ and $g(x)$ have any other common factors? In other words, could those functions $p(x)$ and $q(x)$ have common factors? To rule out that possibility, you need to go back to the equation $f(x)=g(x)+2x+2.$

 

[laprec]

What you have done so far is correct. You have shown that $f(x)$ has factors $x-6$ and $x+1$ (but it may also have other factors). From that, you can say that $f(x) = (x-6)(x+1)p(x)$, for some unknown function $p(x)$. What happens if you put $x=6$ in that equation?

You are told that $f(x) = g(x) + 2x + 2$. Once you know $f(6)$, you can put $x=6$ in that equation to get $g(6)$.

You have shown that $g(x)$ has factors $x-3$ and $x+1$ (but it may also have other factors). From that, you can say that $g(x) = (x-3)(x+1)q(x)$, for some unknown function $q(x)$. So you know that $f(x)$ and $g(x)$ have a common factor $x+1$. But could $f(x)$ and $g(x)$ have any other common factors? In other words, could those functions $p(x)$ and $q(x)$ have common factors? To rule out that possibility, you need to go back to the equation $f(x)=g(x)+2x+2.$

Thanks a lot!
when I substituted x=6 into (x-6)(x+1)(px), f(x)=0. This make g(x)=-2x-2. I'm sure this is wrong.

- - - Updated - - -

What you have done so far is correct. You have shown that $f(x)$ has factors $x-6$ and $x+1$ (but it may also have other factors). From that, you can say that $f(x) = (x-6)(x+1)p(x)$, for some unknown function $p(x)$. What happens if you put $x=6$ in that equation?

You are told that $f(x) = g(x) + 2x + 2$. Once you know $f(6)$, you can put $x=6$ in that equation to get $g(6)$.

You have shown that $g(x)$ has factors $x-3$ and $x+1$ (but it may also have other factors). From that, you can say that $g(x) = (x-3)(x+1)q(x)$, for some unknown function $q(x)$. So you know that $f(x)$ and $g(x)$ have a common factor $x+1$. But could $f(x)$ and $g(x)$ have any other common factors? In other words, could those functions $p(x)$ and $q(x)$ have common factors? To rule out that possibility, you need to go back to the equation $f(x)=g(x)+2x+2.$

Thanks a lot!
when I substituted x=6 into (x-6)(x+1)(px), f(x)=0. This make g(x)=-2x-2. I'm sure this is wrong.

I meant my substitution is wrong!

 

[Opalg]

when I substituted x=6 into (x-6)(x+1)(px), f(x)=0.

That should be $f(6)=0.$

This make g(x)=-2x-2.

No, it makes $g(6) = -2\times6 - 2.$

 

[laprec]

That should be $f(6)=0.$No, it makes $g(6) = -2\times6 - 2.$

Thanks a lot! I am grateful for your help. How do I determine if functions p(x) and q(x) have a common factor using f(x)=g(x)+2x+2?

 

[I like Serena]

Thanks a lot! I am grateful for your help. How do I determine if functions p(x) and q(x) have a common factor using f(x)=g(x)+2x+2?

Hi laprec,

Suppose they have a common factor (x-a) for some constant a.
Then what is f(a) respectively g(a)?
Is that possible given that f(x)=g(x)+2x+2?

 

[laprec]

Hi laprec,

Suppose they have a common factor (x-a) for some constant a.
Then what is f(a) respectively g(a)?
Is that possible given that f(x)=g(x)+2x+2?

Thanks I like Serena,
So since x+1 is a common factor, therefore g(-1)= (-1-3)(-1+1)q(x)=0
g(-1)=0
Therefore f(-1)= 0+2(-1)+2=0
Thus x+1 is a common factor of both functions since the remainder equal zero.
I will appreciate if you can confirm that this is order.
Thank you.

 

[I like Serena]

Thanks I like Serena,
So since x+1 is a common factor, therefore g(-1)= (-1-3)(-1+1)q(x)=0
g(-1)=0
Therefore f(-1)= 0+2(-1)+2=0
Thus x+1 is a common factor of both functions since the remainder equal zero.
I will appreciate if you can confirm that this is order.
Thank you.

Yep. It is.
But then, we already knew that (x+1) was a common factor.
So we can divide left and right by (x+1) leaving p(x) = q(x) + 2, can't we?
Could there be another common factor (x-a)?

 

[laprec]

Yep. It is.
But then, we already knew that (x+1) was a common factor.
So we can divide left and right by (x+1) leaving p(x) = q(x) + 2, can't we?
Could there be another common factor (x-a)?

If we divide (x-6)(x+1)p(x)=(x-3)(x+1)q(x)+2(x+1) by (x+1)

We will get (x-6)p(x)=(x-3)q(x)+2

p(x)=q(x)+2 ??

I am also struggling to see any common factor in the equation
Thanks!

 

[I like Serena]

If we divide (x-6)(x+1)p(x)=(x-3)(x+1)q(x)+2(x+1) by (x+1)

We will get (x-6)p(x)=(x-3)q(x)+2

p(x)=q(x)+2 ??

I am also struggling to see any common factor in the equation
Thanks!

Nope. We can't just leave out (x-6) or (x-3).
However, suppose that p(x) and q(x) both contain a common factor (x-a).
Then what do we get if we fill in x=a in the equation?

 

[laprec]

Nope. We can't just leave out (x-6) or (x-3).
However, suppose that p(x) and q(x) both contain a common factor (x-a).
Then what do we get if we fill in x=a in the equation?

(a-6)p(a)=(a-3)q(a)-2
Thanks so much but I couldn't make headway with the substitution.

 

[I like Serena]

(a-6)p(a)=(a-3)q(a)-2
Thanks so much but I couldn't make headway with the substitution.

If (x-a) is a factor of p(x), then we can write p(x)=(x-a)r(x) for some unknown polynomial r(x), can't?
Now fill in x=a again...

 

[laprec]

If (x-a) is a factor of p(x), then we can write p(x)=(x-a)r(x) for some unknown polynomial r(x), can't?
Now fill in x=a again...

Therefore p(a)= (x-a)r(x)=0

 

[I like Serena]

Therefore p(a)= (x-a)r(x)=0

Yep! So?

 

[laprec]

Yep! So?

So (x-a) is a factor!

 

[I like Serena]

So (x-a) is a factor!

Not so fast!
If (x-a) is a factor of both p(x) and q(x).
Then p(a)=0 and q(a)=0, yes?
What does that mean for (a-6)p(a)=(a-3)q(a)+2?

 

[laprec]

Not so fast!
If (x-a) is a factor of both p(x) and q(x).
Then p(a)=0 and q(a)=0, yes?
What does that mean for (a-6)p(a)=(a-3)q(a)+2?

Substituting p(a)=0 and q(a)=0 in the equation will give 0=2 (this cannot be true)

 

[I like Serena]

Substituting p(a)=0 and q(a)=0 in the equation will give 0=2 (this cannot be true)

Indeed.
Therefore our assumption that (x-a) is a common factor of both p(x) and q(x) must be false.
This is called a proof by contradiction.

 

[laprec]

Indeed.
Therefore our assumption that (x-a) is a common factor of both p(x) and q(x) must be false.
This is called a proof by contradiction.

So, we can conclude that (x+1) is the only common factor.
Wow! You are the best. Thanks for taking me through this process. I surely learned a lot