# Polynomial factoring

1. Feb 18, 2017

### late347

1. The problem statement, all variables and given/known data
$\frac{3x^2+x+C}{x+5}$
find value of constant C such that the clause can be cancelled in some manner. What will be the cancelled form of the clause.
2. Relevant equations
-presumably C is a constant, and also an integer.
-polynomial factorization will be attempted
-cancelling of rational numbers keeps the value the same. In cancelling, you divide numerator and denominator by the same number (other number than 1 or 0).
3. The attempt at a solution

we know initially that we have to have something like
$\frac{(x+5)*(?? ~~??}{x+5}$
this is because there will be no further factorization for the divisor. For the clause to be cancellable, I think the (x+5) must be one of the factors in the upstairs of the division line.

and from an educated guess point of view, I think we have to get the 3x^2 term finished somehow. So, that 3x^2 is the product of something.
$\frac{(x+5)*(3x ~~??}{x+5}$

if the other integer is an unknown one. maybe it can be represented by letter r
(x+5)(3x+r) should equal what we originally had upstairs (3x^2 +x +C) and we know C will be an integer also
At this point I was honestly a little bit stumped and I had to review some older Khan academy problems about factoring. I'm still alittle bit unsure why the so-called analytical method of finding the factorization works...

I was fiddling around with the clause (x+5)*(3x+r)
and if you multiply it out, it comes out such as
3x^2 +xr+15x+5r

Well... looking at that clause in that form it looks like xr +15x=1x must be a true equation.
Because this would allow you to choose the r so that the factors are correct for the original clause's terms for x^1 terms

So, I suppose firstly r= 1-15
r= -14
C= 5*(-14) = -70

[(x+5)(3x-14)]/(x+5)
=3x-14

[(x+5)(3x-14)]= 3x^2 -14x +15x -70

2. Feb 18, 2017

### Staff: Mentor

Looks good, but what is your question?

3. Feb 18, 2017

### Ray Vickson

Your method works because when you expand $q(x) = (x+5)(3x+r)$ you get a second-degree polynomial in $x$ that must be the same as $p(x) = 3x^2+x+C$. That means that the coefficients in the two polynomials must be the same; that is, if $p(x) = q(x)$ for ALL $x$, then all the coefficients in $p(x)$ must be the same as all the coefficients in $q(x)$. That gives you two conditions for the two unknown constants $r$ and $C$.

BTW: a much faster and much easier way is to use the following general theorem: "The remainder when a polynomial $p(x)$ is divided by $(x-a)$ is equal to $p(a)$". So, in your case you have $a = -5$ and to have $p(x) = 2x^2 +x+C$ divisible by $(x+5)$ the remainder must equal zero. That means that you need $0 = p(-5) = 3(-5)^2 + (-5) + C = 70 - C,$ so $C = 70$.

4. Feb 18, 2017

### late347

thank you for the tip sir... sadly I never bothered to learn much about polynomial division. I was once taught about that in high school but I never learned how to do it properly. It's like a dark area for my math knowledge. I do know how to divide regular numbers in the so-called American long division algorithm, though.

5. Feb 18, 2017

### Staff: Mentor

https://en.wikipedia.org/wiki/Polynomial_long_division

6. Feb 18, 2017

### Staff: Mentor

7. Feb 18, 2017

### Ray Vickson

The result is incredibly easy to obtain. If we divide $p(x)$ by $(x-a)$ we have some quotient $q(x)$ and some remainder $r$ (a number). Basically, that means that
$p(x) = q(x) \, (x-a) + r.$
What do you get when you put $x = a?$

8. Feb 18, 2017

### late347

Well, you said originally that for the clause to cancellable... then the remainder will be zero when you do the cancellation thing.

when you have the thing set-up that way. it looks like it will become
p(a)=q(a) * 0 + r
<=> p(a)=r

so you were saying that essentially p(a) = 0, because r=0

Im going to sleep now because it is 01:47 so it's getting pretty late.