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Finding values for a and b for this polynomial

  • Thread starter chef99
  • Start date
75
4
the polynomial f(x) = ax4 - 3x3- 63x2+ 152x - b



has one of its zeros at x = 5 and passes through the point (-2, -560)



Question: Use this info to find the values of a and b

I am prepping for a test and this one question is really stumping me, I wondered if anyone would be able to help. For all of this type of question up till now where it asks to find a and b, I have been given a factor and the remainder. I know with these values I should set up two equations, one for a and one for b, then use them to solve each other. I know if a zero is x = 5 then a factor will be x = -5 but I have no idea what to do with the point given, is there something obvious I am missing? Any help on this would be greatly appreciated

[Moderator's note: Moved from a technical forum and thus no template.]
 
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10,812
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So the fact that it has a zero when x=5 means the point (5,0) is on the curve.

Plugging those in gives you equation in two unknowns a and b right?

What happens when you plug in the other point?

You get second equation in two unknowns a and b again.

Given that what is a and b?
 
32,626
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o the fact that it has zeros when x=5 means the point (0,5) is on the curve
No, it means that (5, 0) is on the curve.
The rest of the hint is fine.
 
10,812
4,349
You saw my post before I edited it. I had a dyslexic moment and then saw my mistake.
 
75
4
So the fact that it has a zero when x=5 means the point (5,0) is on the curve.

Plugging those in gives you equation in two unknowns a and b right?

What happens when you plug in the other point?

You get second equation in two unknowns a and b again.

Given that what is a and b?
Thank you for your help.
Ok so for the first equation (x-5) --> remainder is 0, so f(-5) = 0
and for the second equation (x+2) --> remainder of -560

Is that correct? I have tried following this and using the two equations to solve for the variables in each other but when I try to test it afterword, with my values of a and b f(-5) should = 0, but I am not getting zero as a value so either there is an error in my calculations or I am applying your advice incorrectly. I have included my work below.

(x-5) --> remainder is 0 so f(-5) = 0


f(x) = a(-5)4 - 3(-5)3- 63(-5)2+ 152(-5) - b

-625a -(-375)-1575-760 - b = 0

-625 - 2710 - b = 0

-625a - b = 2710


Now for the second equation;

(-2, -560) is a point so (x + 2) --> remainder of -560

a(2)4 - 3(2)3- 63(2)2+ 152(2) - b = -560

16a - 24 - 252 + 304 - b = -560

16a + 28 - b = -560

16a +588 = b


Now plugging 16a +588 = b into the first equation,

-625a - 16a +588 = 2710

a = 2710/-609


finally, solve for b,

16a +588 = b

16(2710/-609) +588 = b

b = 516.8


Checking for f(-5) = 0,

f(-5) = (2710/-609)(-5)4 - 3(-5)3- 63(-5)2+ 152(-5) - 516.8


= -5248

As you can see this is obviously not right as it should equal 0. Am I just making errors in my calculations or do I have the wrong idea entirely?
 
10,812
4,349
No that's not how to do it:

## f(5) = 0 ##
##0 = a(5)^4 - 3(5)^3- 63(5)^2+ 152(5) - b ##
 
33,236
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The other point has the same sign error.
It is (x,y)=(-2, -560), you have to plug in -2 for x.
 
75
4
No that's not how to do it:

## f(5) = 0 ##
##0 = a(5)^4 - 3(5)^3- 63(5)^2+ 152(5) - b ##
I have fixed this error, and the one pointed out by mfb, using f(5) remainder of 0 and f(-2) remainder of -560 but am still not getting f(5) = 0 when testing it afterward. I also keep getting fraction/decimal numbers, is that to be expected or should I be getting whole numbers? Perhaps I am screwing up the process later?
Here is what I got:

f(x) = a(5)4 - 3(5)3- 63(5)2+ 152(5) - b

625a -375-1575+760 - b = 0

625 - 1190 - b = 0

625a - b = 1190



and

a(-2)4 - 3(-2)3- 63(-2)2+ 152(-2) - b = -560

-16a + 24 - 252 - 304 - b = -560

-16a -532 - b = -560

b = -16a + 28

then using this value for b, -16a + 28, determine a:

625a - b = 1190

625a - (-16a + 28) = 1190

641a = 1162

a = 1162/641 <--- as you can see I'm getting a fraction, don't know if that's wrong


Using this to find b:

b = -16a + 28

b = -16(1162/641) + 28

b = -18592/641 + 28

b = 644/641


Then when I test these values for a and b to see if f(5) = 0,

f(5) = (1162/641)(5)4 - 3(5)3- 63(5)2+ 152(5) - (644/641)

f(5) = - 37184/641
f(5) = -58

This is obviously not right but I don't know where I'm messing up. Am I using the correct method to find a and b or is it arithmetic error I'm making?
 

SammyS

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This is obviously not right but I don't know where I'm messing up. Am I using the correct method to find a and b or is it arithmetic error I'm making?
You have some sign errors.

In particular, ##(-2)^4 \ne -16 ## .
 

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