# Finding the Constant for Cancellation in Polynomial Factoring

• late347
In summary, to find the value of constant C such that the clause (3x^2+x+C)/(x+5) can be canceled in some manner, we can use the general theorem "The remainder when a polynomial is divided by (x-a) is equal to the polynomial evaluated at a." In this case, the remainder must be 0 in order for the clause to be cancelable. Therefore, setting x = -5 and equating the remainder to 0, we get C = 70. This means the canceled form of the clause is 3x-14. Another method to find the value of C is by setting up and solving a system of equations using the polynomial factorization method.
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?

late347 said:

## 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
Ray Vickson said:
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.

late347 said:
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

late347 said:
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?##

Ray Vickson said:
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.

## What is polynomial factoring?

Polynomial factoring is the process of breaking down a polynomial expression into its simplest form by finding its factors. This is useful for solving equations and understanding the behavior of the polynomial function.

## Why is polynomial factoring important?

Polynomial factoring is important because it allows us to solve equations, find the roots of a polynomial function, and understand the behavior of the function. It also helps to simplify complicated expressions and make them easier to work with.

## What are the steps involved in polynomial factoring?

The steps for polynomial factoring include: 1) Identifying the greatest common factor (GCF), 2) using the distributive property to factor out the GCF, 3) looking for special patterns such as the difference of squares or perfect square trinomials, and 4) using techniques such as grouping or trial and error to factor the remaining terms.

## What are some common mistakes to avoid when factoring polynomials?

Some common mistakes to avoid when factoring polynomials include: 1) Forgetting to check for a GCF, 2) not using the correct factoring technique for a given expression, 3) making errors when applying the distributive property, 4) not checking the final factored expression to ensure it is correct, and 5) not checking for extraneous solutions when solving equations.

## Can all polynomials be factored?

No, not all polynomials can be factored. Some polynomials, such as prime polynomials, have no factors other than 1 and itself. Additionally, some polynomials may not have rational factors, meaning they cannot be factored using whole numbers or fractions. In these cases, other techniques such as the quadratic formula may be used to solve equations or understand the behavior of the function.

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