What Are the Values of A, B, and C for the Given Expression?

[chwala]

Homework Statement

Find the values of $A,B$ and $C$ if the expression

$Ax(x-2)(x+3)+Bx(x-2)+Cx(x+3)+ (x-2)(x+3)$

has a constant value for all values of $x$.

Relevant Equations

Partial fractions

My working follows here,

Let
$(x-2)[Ax(x+3)+Bx] + (x+3)[Cx +(x-2)]=0$

$(x-2)(x+3) \left[Ax+\dfrac{Bx}{(x+3)}\right]+(x+3)(x-2)\left[\dfrac{Cx}{(x-2)} +1\right]$=0

$⇒Ax+\dfrac{Bx}{(x+3)} = -\left[\dfrac{Cx}{(x-2)} +1\right]$

$Ax(x+3)(x-2)+Bx(x-2) = -[Cx(x+3)+(x-2)(x+3)]$

On comparing coefficients we end up with;

$A=0$

...and the simultaneous equation;

$B+C=-1$
$-2B+3C=-1$

Giving me

$B=-\dfrac{2}{5}$ and $C=-\dfrac{3}{5}$

There could be a better approach...And also your insight on my steps are welcome. Cheers!

 

[Delta2]

I think you should 've stated the problem a bit better:
"Find the values A,B,C for which the following expression equals zero for every x"

Other than that I can almost immediately see that A=0 because it would be the only coefficient of $x^3$. Hold on while I check about B and C.

One could work in a brute force manner, that is expand all the products and sort the powers $x,x^2,x^3$ and sum their coefficients

And now I see that your work seems like going from Athens to Paris , then from Paris back to Athens and then from Athens to Moscow which is what we should have done at first place.

 

[chwala]

I think you should 've stated the problem a bit better:
"Find the values A,B,C for which the following expression equals zero for every x"

Other than that I can almost immediately see that A=0 because it would be the only coefficient of $x^3$. Hold on while I check about B and C.

One could work in a brute force manner, that is expand all the products and sort the powers $x,x^2,x^3$ and sum their coefficients

And now I see that your work seems like going from Athens to Paris , then from Paris back to Athens and then from Athens to Moscow which is what we should have done at first place.

The original text book problem had no $0$. I came up with the $0$ on the rhs...The problem is posted as it is on the textbook.

 

[Delta2]

According to Wolfram expansion, either you forgot a +6 at the end of the expression or we just want this expression to equal -6 (for any x).

Values of A,B,C seem correct

Oh I see now, probably the textbook means find the values of A,B,C such that the given expression is the same for any x.

 

[fresh_42]

Homework Statement: Find the values of $A,B$ and $C$ if the expression

$Ax(x-2)(x+3)+Bx(x-2)+Cx(x+3)+ (x-2)(x+3)$

has a constant value for all values of $x$.

This means that $Ax(x-2)(x+3)+Bx(x-2)+Cx(x+3)+ (x-2)(x+3)=D$ for all $x.$

Why do you assume $D=0$ right from the start?

Let
$(x-2)[Ax(x+3)+Bx] + (x+3)[Cx +(x-2)]=0$

Differentiation could be another idea to approach the problem.

 

[WWGD]

If it's a constant, D, it must be 0, as f(0)=0.

 

[fresh_42]

If it's a constant, D, it must be 0, as f(0)=0.

$f(0)=-6$

 

[WWGD]

$f(0)=-6$

Guess my phone couldn't read that far right. Last term I see is Cx.

 

[SammyS]

Homework Statement: Find the values of $A,B$ and $C$ if the expression

$Ax(x-2)(x+3)+Bx(x-2)+Cx(x+3)+ (x-2)(x+3)$

has a constant value for all values of $x$.
Relevant Equations: Partial fractions

My working follows here,

There could be a better approach...And also your insight on my steps are welcome. Cheers!

You should pay attention to Posts #5 by @fresh_42 .

This means that $Ax(x-2)(x+3)+Bx(x-2)+Cx(x+3)+ (x-2)(x+3)=D$ for all $x.$

Why do you assume $D=0$ right from the start?Differentiation could be another idea to approach the problem.

@chwala,

It appears that you tried to use partial fractions, but you messed that up in addition to setting the expression to zero.

Set the expression equal to a constant, $D$, as suggested above.

Following that you could expand the left hand side followed by collecting terms. The equate corresponding coefficients.

Alternatively, take the expression given by fresh and substitute well chosen values for $x$.

 

[chwala]

This means that $Ax(x-2)(x+3)+Bx(x-2)+Cx(x+3)+ (x-2)(x+3)=D$ for all $x.$

Why do you assume $D=0$ right from the start?Differentiation could be another idea to approach the problem.

Noted @fresh_42

Amending my post we shall have,$Ax(x-2)(x+3)+Bx(x-2)+Cx(x+3)+ (x-2)(x+3)=D$

Where, $f(D)=0$

$(x-2)(x+3)[Ax(x+3)(x-2)+Bx(x-2)] = D-[(x+3)(x-2)[Cx(x+3)+(x-2)(x+3)]$

Solving will lead us to the solutions in post $1$, the only extra equation is

$D+6=0, D=-6$.

 

[SammyS]

$Ax(x-2)(x+3)+Bx(x-2)+Cx(x+3)+ (x-2)(x+3)=D$

Where, $f(D)=0$

No way !

Assuming that you have defined $f(x)$ as $\displaystyle f(x)=Ax(x-2)(x+3)+Bx(x-2)+Cx(x+3)+ (x-2)(x+3)$

Then , $f(0)=D$. In fact, $f(x)=D$ for all $x$.

 

[fresh_42]

Noted @fresh_42

Amending my post we shall have,$Ax(x-2)(x+3)+Bx(x-2)+Cx(x+3)+ (x-2)(x+3)=D$

Where, $f(D)=0$

$(x-2)(x+3)[Ax(x+3)(x-2)+Bx(x-2)] = D-[(x+3)(x-2)[Cx(x+3)+(x-2)(x+3)]$

Solving will lead us to the solutions in post $1$, the only extra equation is

$D+6=0, D=-6$.

What do you mean by $f$?

@WWGD used it as $f(x):=Ax(x-2)(x+3)+Bx(x-2)+Cx(x+3)+ (x-2)(x+3).$ Then we would have $f(0)=D$
and $f(0)=-6$ so $D=-6$ and $g(x):=Ax(x-2)(x+3)+Bx(x-2)+Cx(x+3)+ (x-2)(x+3) +6 =0.$

From here on you have several possible paths:

1. Insert values like $x\in \{0,2,-3,\pm 1\}$ and see which equations you get.
2. Multiply the entire polynomial and compare $a_0x^3+a_1x^2+a_2x+a_3 = 0$ to get the equations.
3. Differentiate the polynomial and solve for $g'(x)=0.$ This is quadratic and is easier to solve.

 

[chwala]

Error on my part guys! hey! ... its a straightforward fact that using Remainder Theorem $f(x)=D$ where $D$ is the Remainder. In this problem we know that $f(0)=-6$; $⇒D=-6$.

Let me explore on the differentiation approach...

 

[chwala]

What do you mean by $f$?

@WWGD used it as $f(x):=Ax(x-2)(x+3)+Bx(x-2)+Cx(x+3)+ (x-2)(x+3).$ Then we would have $f(0)=D$
and $f(0)=-6$ so $D=-6$ and $g(x):=Ax(x-2)(x+3)+Bx(x-2)+Cx(x+3)+ (x-2)(x+3) +6 =0.$

From here on you have several possible paths:

1. Insert values like $x\in \{0,2,-3,\pm 1\}$ and see which equations you get.
2. Multiply the entire polynomial and compare $a_0x^3+a_1x^2+a_2x+a_3 = 0$ to get the equations.
3. Differentiate the polynomial and solve for $g'(x)=0.$ This is quadratic and is easier to solve.

I think my amended version is clear @fresh_42. The only missing term was the constant term $D$ which i have since inserted.

Let

$f(x)= Ax(x-2)(x+3)+Bx(x-2)+Cx(x+3)+ (x-2)(x+3)$

Using Remainder theorem; i shall have $f(x)=D$ where $D$ is the remainder.

thus

$Ax(x-2)(x+3)+Bx(x-2)+Cx(x+3)+ (x-2)(x+3)=D$

$(x-2)(x+3)[Ax(x+3)(x-2)+Bx(x-2)] = D-[(x+3)(x-2)[Cx(x+3)+(x-2)(x+3)]$

Solving for the constants $A,B,C$ and $D$ will lead us to the values found in post $1$, the only extra constant to be determined is

$D+6=0, D=-6$.

 

[chwala]

What do you mean by $f$?

@WWGD used it as $f(x):=Ax(x-2)(x+3)+Bx(x-2)+Cx(x+3)+ (x-2)(x+3).$ Then we would have $f(0)=D$
and $f(0)=-6$ so $D=-6$ and $g(x):=Ax(x-2)(x+3)+Bx(x-2)+Cx(x+3)+ (x-2)(x+3) +6 =0.$

From here on you have several possible paths:

1. Insert values like $x\in \{0,2,-3,\pm 1\}$ and see which equations you get.
2. Multiply the entire polynomial and compare $a_0x^3+a_1x^2+a_2x+a_3 = 0$ to get the equations.
3. Differentiate the polynomial and solve for $g'(x)=0.$ This is quadratic and is easier to solve.

Your approach works but i think it may be longer than my approach. In your way one has to come up with a system of equations presumably $4$ equations to solve for the unknown constants.

 

[fresh_42]

Your approach works but i think it may be longer than my approach. In your way one has to come up with a system of equations presumably $4$ equations to solve for the unknown constants.

I would follow @SammyS suggestion and insert those values for $x.$

$f(0)=-6=D$ already gave one unknown.
$f(2)$ gives a second unknown.
$f(-3)$ gives a third unknown,
and $f(1)$ should do the rest.

 

[Steve4Physics]

@chwala, the approach from @SammyS and @fresh_42 is really simple/quick. Here it is if required...

Spoiler

$f(x) = Ax(x-2)(x+3)+Bx(x-2)+Cx(x+3)+ (x-2)(x+3)$
Let $D$ be the fixed value of $f(x)$, i.e. $f(x) = D$ for all $x$.

By inspection we can see evaluating $f(x)$ for $x=0, 2$ and $– 3$ will make terms vanish.

Find $D$ by evaluating $f(0)$:
$D = f(0) = A.0 + B.0+ C.0 + (0-2)(0+3) = -6$

Find $C$ by using $f(2)=D = -6$:
$f(2) = A.0 + B.0 + C.2(2+3) + 0.(0+3) = -6$
$10C = -6 \implies C = -3/5$

Find $B$ by using $f(-3)= D =-6$:
$f(-3) = A.0+B(-3)(-3-2)+C.0+ 0 = -6$
$15B = -6 \implies B = -2/5$

Find $A$ using $f(\text {any convenient value of x}) =D = -6$ and the known values of $B$ and $C$:
$f(1) = A.1.(-1)(4) + (-2/5).1(-1) +(-3/5).1.4 + (-1)(4) = -6$
$-4A + 2/5 - 12/5 -4 = -6 \implies A = 0$

If you remove all my commentary, it’s just a few lines of elementary algebra/arithmetic.

 

[WWGD]

Homework Statement: Find the values of $A,B$ and $C$ if the expression

$Ax(x-2)(x+3)+Bx(x-2)+Cx(x+3)+ (x-2)(x+3)$

has a constant value for all values of $x$.
Relevant Equations: Partial fractions

My working follows here,

Let
$(x-2)[Ax(x+3)+Bx] + (x+3)[Cx +(x-2)]=0$

$(x-2)(x+3) \left[Ax+\dfrac{Bx}{(x+3)}\right]+(x+3)(x-2)\left[\dfrac{Cx}{(x-2)} +1\right]$=0

$⇒Ax+\dfrac{Bx}{(x+3)} = -\left[\dfrac{Cx}{(x-2)} +1\right]$

$Ax(x+3)(x-2)+Bx(x-2) = -[Cx(x+3)+(x-2)(x+3)]$

On comparing coefficients we end up with;

$A=0$

...and the simultaneous equation;

$B+C=-1$
$-2B+3C=-1$

Giving me

$B=-\dfrac{2}{5}$ and $C=-\dfrac{3}{5}$

There could be a better approach...And also your insight on my steps are welcome. Cheers!

You already have D=-6. Now compute f(2) gives you C=-6/15. Now try f(-3), to get C=3/5.
Now you have just one variable left, so you're done. Factor out the (x-2) to simplify.

 

[SammyS]

Error on my part guys! hey! ... its a straightforward fact that using Remainder Theorem $f(x)=D$ where $D$ is the Remainder. In this problem we know that $f(0)=-6$; $⇒D=-6$.

Let me explore on the differentiation approach...

Can you state the Remainder Theorem?

In the case of your given exercise, what polynomial are you dividing by what monomial ? ... and why is this useful.?

 

[chwala]

Can you state the Remainder Theorem?

In the case of your given exercise, what polynomial are you dividing by what monomial ? ... and why is this useful.?

I will refer you to the link below:

https://www.cuemath.com/algebra/remainder-theorem/

Specifically look at the sub titles;

1. Remainder theorem statement and proof.
2. Proof of Remainder theorem.
3. Remainder theorem for polynomials.

In our case;

$p(x) = (x-0)⋅ q(x) + r$

Substitute

$x=0$

$p(0)=r$

Useful because we do not have $0$ on the rhs of the equation but a constant $D$.
Cheers.