Factoring expressions! I know the answers, I need help learning how to get answers.

I know the answers to these expressions, Its just that I don't know how to work them. I need help learing (Worried)

(fx)=3x/(7-x) Answer:7

(x^2-64)/(8-x) Answer: -x-8

(2x^2-5x-3)/(6x^3+3x^2+2x+1) Answer: (x-3)/(3x^2+1)

Please Help me! The reason I gave the answers is so that you guys know that you have work the problem correctly to the end. :)

Re: Factoring expressions! I know the answers, I need help learning how to get answer

I don't understand your notation for the first one. If , then if and only if . On the other hand, is not defined (division by zero). So, I don't understand what your answer: 7 means. Or, if f is a constant and the equals sign should be a plus sign (the equals sign and the plus sign share a key), then , then your answer should include an f.

For number 2, you should simplify the numerator first: . So, . For the denominator, you have . Then, you can cancel the from the numerator and denominator to get .

The third one:

We use the rational roots theorem to find factors of the numerator and denominator. For any polynomial, we only need to look at the coefficients of the highest power of x and of the lowest power of x to find rational roots. Possible roots of the numerator are . Possible roots of the denominator are . The ones that overlap are and . So, let's try all four and see which one works. Plug in values for . If is a factor of a polynomial, then plugging in will cause the polynomial to equal zero. Let's check:

, so is not a factor of the numerator.

, so is not a factor of the numerator.

, so is not a factor of the numerator.

, so is a factor of the numerator. Now let's check if it is a factor of the denominator.

, so is also a factor of the denominator.

The numerator becomes . This tells us that , so . So, the numerator is .

The denominator becomes . Equating coefficients, we have , , so , and , so . So, the denominator is . Cancelling, we get:

Now, you notice you can factor a 2 from both the numerator and denominator to get:

Re: Factoring expressions! I know the answers, I need help learning how to get answer

Quote:

Originally Posted by

**Zeyna18** I know the answers to these expressions, Its just that I don't know how to work them. I need help learing (Worried)

(fx)=3x/(7-x) Answer:7

It is not clear what you are asking. This does't appear to have anything to do with "factoring". If you just want to do then indicated division then

first, 3x/(7- x)= -(3x/(x- 7)). x divides into 3x three times with remainder 3x- 3(x- 7)= 21. .

I don't see what "7" could have to to with this.

Perhaps you are asking "for what value of x does this function NOT exist". In that case, since you cannot divide by 0, this fraction does not exist when 7- x= 0 which means x= 7.

Quote:

(x^2-64)/(8-x) Answer: -x-8

Here, **if** the problem is to do the indicated division, then we can use "factoring". Knowing that (x- a)(x+ a)= x^2- a^2 for any a, we can write x^2- 64= x^2- 8^2= (x- 8)(x+ 8). Dividing by x- 8 would just cancel the (x- 8) factor giving x+ 8. Dividing by 8- x= -(x- 8) gives -(x+ 8)= -x- 8.

Quote:

(2x^2-5x-3)/(6x^3+3x^2+2x+1) Answer: (x-3)/(3x^2+1)

Okay, you can factor here as well. The only way to factor "2x^2", with integer coefficients, is (2x)(x). The only ways to factor "-3" are either (3)(-1) or (-3)(1). So we know we must have one of (2x+3)(x- 1) or (2x- 3)(x+ 1) or (2x+ 1)(x- 3) or (2x-1)(x+ 3). Testing each of those:

(2x+ 3)(x- 1)= 2x^2+ 3x- 2x- 3= 2x^2+ x- 3.

(2x- 3)(x+ 1)= 2x^2- 3x+ 2x- 3= 2x^2- x- 3.

(2x+1)(x- 3)= 2x^2+ x- 3x- 3= 2x^2- 2x- 3.

(2x- 1)(x+ 3)= 2x^2- x+ 3x- 3= 2x^2+ 2x- 3.

NONE of those is equal to 2x^2- 5x- 3. It **cannot** factored with integer coefficients so it **cannot** reduce to (x- 3)/(3x^2+ 1)

Quote:

Please Help me! The reason I gave the answers is so that you guys know that you have work the problem correctly to the end. :)

I recommend you go back and recheck the questions. The questions and/or the answer you give make no sense.

Re: Factoring expressions! I know the answers, I need help learning how to get answer

Quote:

Originally Posted by

**HallsofIvy** Okay, you can factor here as well. The only way to factor "2x^2", with integer coefficients, is (2x)(x). The only ways to factor "-3" are either (3)(-1) or (-3)(1). So we know we must have one of (2x+3)(x- 1) or (2x- 3)(x+ 1) or (2x+ 1)(x- 3) or (2x-1)(x+ 3). Testing each of those:

(2x+ 3)(x- 1)= 2x^2+ 3x- 2x- 3= 2x^2+ x- 3.

(2x- 3)(x+ 1)= 2x^2- 3x+ 2x- 3= 2x^2- x- 3.

(2x+1)(x- 3)= 2x^2+ x- 3x- 3= 2x^2- 2x- 3.

(2x- 1)(x+ 3)= 2x^2- x+ 3x- 3= 2x^2+ 2x- 3.

NONE of those is equal to 2x^2- 5x- 3. It **cannot** factored with integer coefficients so it **cannot** reduce to (x- 3)/(3x^2+ 1)

So, it does reduce to as both the OP and I wrote.