# Finding values for a and b for this polynomial

#### chef99

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

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#### jedishrfu

Mentor
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?

• symbolipoint

#### Mark44

Mentor
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.

• jedishrfu

#### jedishrfu

Mentor
You saw my post before I edited it. I had a dyslexic moment and then saw my mistake.

• symbolipoint

#### chef99

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?

#### jedishrfu

Mentor
No that's not how to do it:

$f(5) = 0$
$0 = a(5)^4 - 3(5)^3- 63(5)^2+ 152(5) - b$

#### mfb

Mentor
The other point has the same sign error.
It is (x,y)=(-2, -560), you have to plug in -2 for x.

#### chef99

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

Staff Emeritus
Homework Helper
Gold Member
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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