Remainder theorem question - combine divisor

[tony24810]

Homework Statement

when f(x) is divided by (x+1), remainder is -9; when f(x) is divided by (x-3), remainder is -1; what is the remainder if f(x) is divided by (x+1)(x-3)?

Homework Equations

f(x) = divider * q(x) + remainder

The Attempt at a Solution

f(x) = (x+1) * a(x) -9
f(x) = (x-3) * b(x) -1

unable to perform further useful calculation

 

[eumyang]

When you divide f(x) by (x+1)(x-3), a quadratic, the highest possible degree of the remainder polynomial is 1.

So you have f(x) = (x+1)(x-3)*q(x) + (ax + b).

By the Remainder Theorem...
"when f(x) is divided by (x+1), remainder is -9; when f(x) is divided by (x-3), remainder is -1
... what does the above tell us? You'll then need to plug some things into x and f(x) above to come up with two equations involving a and b. Solve for a and b.

 

[Harrisonized]

Let's assume that a(x)=ax+b and b(x)=cx+d. This is a valid assumption, but it's not the only one that works (for example, you could have a higher order polynomial for a(x) and b(x)).

Let a(x)=ax+b
Let b(x)=cx+d

Then
f(x)=(x+1)(ax+b)-9 = ax²+(a+b)x+b-9
f(x)=(x-3)(cx+d)-1 = cx²+(d-3c)x-3d-1

We want to determine the coefficients a, b, c, and d that will make this work. Let's subtract the two equations.

0=[ax²+(a+b)x+b-9] - [cx²+(d-3c)x-3d-1]
0=(a-c)x²+(a+b+3c-d)x+(b+3d-8)

Hence,

(a-c)=0
(a+b+3c-d)=0
(b+3d-8)=0

Three equations, four unknowns. Observe that there is not one unique solution, but a set of valid solutions.

 

[tony24810]

thanks eumyang

ok so i give it another go

i have

f(-1) = -9
f( 3) = -1

thus

f(x) = (x+1)(x-3)*q(x) + (ax + b)

f(-1) = [(-1)+1][(-1)-3]*q(x) + [a(-1) + b]
f(3) = [(3)+1][(3)-3]*q(x) + [a(3) + b]

-9 = -a + b
-1 = 3a + b

a=2
b=-7

solution = 2x - 7

would this be correct?

 

[tony24810]

Hence,

(a-c)=0
(a+b+3c-d)=0
(b+3d-8)=0

Three equations, four unknowns. Observe that there is not one unique solution, but a set of valid solutions.

thanks for help, but i cannot deduce a unique equation from this set of equations nor a set of rules

this question came from an exam paper so there should be some sort of reasonable answer, I'm not sure if my solution of 2x - 7 is correct, now I'm trying to plug in something to verify

 

[tony24810]

the solution (2x-7) seems to be wrong

for f(x) = (x+1)(x-3)*q(x) + (2x -7)

i tried to plug in q(x) = (x+2)

thus f(x) = x^3 + x + 1

when f(x) is divided by (x+1) and (x-3), i don't get -9 and -1please help

 

[Dick]

the solution (2x-7) seems to be wrong

for f(x) = (x+1)(x-3)*q(x) + (2x -7)

i tried to plug in q(x) = (x+2)

thus f(x) = x^3 + x + 1

when f(x) is divided by (x+1) and (x-3), i don't get -9 and -1please help

I get (x+1)(x-3)(x+2)+(2x -7)=x^3-5x-13. I think your example does work.

 

[tony24810]

thanks dick

yes i made a mistake, when x-2 is plug in i got

(x+1)(x-3)(x+2)+(2x -7)=x^3-5x-13

but still it does not work?

can someone tell me if 2x-7 is the correct answer

and

is there a unique answer to this question?

 

[Dick]

thanks dick

yes i made a mistake, when x-2 is plug in i got

(x+1)(x-3)(x+2)+(2x -7)=x^3-5x-13

but still it does not work?

can someone tell me if 2x-7 is the correct answer

and

is there a unique answer to this question?

Why doesn't it work? If I divide that polynomial by (x+1) I get remainder -9, if I divide by (x-3) I get remainder -1. And I don't see anything wrong with eumyang's original hint. Reread it and see if you disagree. 2x-7 is the unique correct answer. The x+2 you plugged in for q(x) to check isn't unique. You could have plugged anything in for that and it will still work.

 

[tony24810]

thanks Dick! i made some more arithmetic errors, i guess I'm too tired. now i got it.

but in the 3rd post, Harrisonized mentioned there is a set a solution. i couldn't see a flaw in his work out. is it invalid to assume

Let a(x)=ax+b
Let b(x)=cx+d

 

[eumyang]

but in the 3rd post, Harrisonized mentioned there is a set a solution. i couldn't see a flaw in his work out. is it invalid to assume

Let a(x)=ax+b
Let b(x)=cx+d

Well, Harrisonized did say that

Let's assume that a(x)=ax+b and b(x)=cx+d. This is a valid assumption, but it's not the only one that works (for example, you could have a higher order polynomial for a(x) and b(x)).

But note that Harrisonized is trying to find the coefficients for the two quotient polynomials a(x) and b(x). (To be honest, I'm not sure why he did that, since you're being asked for the remainder polynomial.)

 

[Harrisonized]

But note that Harrisonized is trying to find the coefficients for the two quotient polynomials a(x) and b(x). (To be honest, I'm not sure why he did that, since you're being asked for the remainder polynomial.)

Above, I found that

(a-c)=0
(a+b+3c-d)=0
(b+3d-8)=0

From the first equation, we have a=c. Since we have three equations and four unknowns, we have one degree of freedom, so we can rewrite all four coefficients a, b, c, and d in terms of a single variable. Let's choose that single variable to be a. (This is an arbitrary choice.) Solving the remaining two equations:

(4a+b-d)=0 → b=d-4a
(b+3d-8)=0 → d-4a+3d=8 → d=a+2
b=d-4a → b=-3a+2

Therefore,

f(x)=(x+1)(ax+b)-9=(x+1)(ax-3a+2)-9=ax²+(-2a+2)x-3a-7
f(x)=(x-3)(cx+d)-1=(x-3)(ax+a+2)-1=ax²+(-2a+2)x-3a-7

I didn't have to plug into both to find f(x), but since both forms result in the same f(x), I confirm that the solutions found are correct. Since we have determined it up to a constant, a, we have found all quadratic solutions of what you described in your problem.

Now you want the remainder, right? Divide through by (x+1)(x-3)=x²-2x-3.

[ax²+(-2a+2)x-3a-7]/[x²-2x-3]
= [ax²-2ax-3a]/[x²-2x-3] + [2x-7]/[x²-2x-3]
= a + [2x-7]/[x²-2x-3]

2x-7 is your unique remainder.

 

[Dick]

Above, I found that

(a-c)=0
(a+b+3c-d)=0
(b+3d-8)=0

From the first equation, we have a=c. Since we have three equations and four unknowns, we have one degree of freedom, so we can rewrite all four coefficients a, b, c, and d in terms of a single variable. Let's choose that single variable to be a. (This is an arbitrary choice.) Solving the remaining two equations:

(4a+b-d)=0 → b=d-4a
(b+3d-8)=0 → d-4a+3d=8 → d=a+2
b=d-4a → b=-3a+2

Therefore,

f(x)=(x+1)(ax+b)-9=(x+1)(ax-3a+2)-9=ax²+(-2a+2)x-3a-7
f(x)=(x-3)(cx+d)-1=(x-3)(ax+a+2)-1=ax²+(-2a+2)x-3a-7

I didn't have to plug into both to find f(x), but since both forms result in the same f(x), I confirm that the solutions found are correct. Since we have determined it up to a constant, a, we have found all quadratic solutions of what you described in your problem.

Now you want the remainder, right? Divide through by (x+1)(x-3)=x²-2x-3.

Yes, but as eumyang already pointed out, you don't have to do this at all. In what way is this simpler than the original suggestion? It's certainly less general if you need to assume that f(x) is quadratic. Which you don't.

 

[Harrisonized]

Yes, but as eumyang already pointed out, you don't have to do this at all. In what way is this simpler than the original suggestion? It's certainly less general if you need to assume that f(x) is quadratic. Which you don't.

It's just another way to solve the same problem. I ended up writing the rest of it, because the OP posted this

thanks for help, but i cannot deduce a unique equation from this set of equations nor a set of rules

this question came from an exam paper so there should be some sort of reasonable answer, I'm not sure if my solution of 2x - 7 is correct, now I'm trying to plug in something to verify

and I wanted to show him that this method works as well.

 

[Dick]

It's just another way to solve the same problem.

Yes, it is and it works. Because you know you'll get the same answer no matter what form you assume for the two polynomials. It's just not simpler than the original approach and confused the OP. That's my only problem with it.