# Polynomial factoring

• late347

## Homework Statement

##\frac{3x^2+x+C}{x+5}##
find value of constant C such that the clause can be canceled in some manner. What will be the canceled form of the clause.

## Homework 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).

## 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

Looks good, but what is your question?

## Homework Statement

##\frac{3x^2+x+C}{x+5}##
find value of constant C such that the clause can be canceled in some manner. What will be the canceled form of the clause.

## Homework 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).

## The Attempt at a Solution

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...

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##.

late347
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##.

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.

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

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.

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?##

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?##
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.