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## 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