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 $\displaystyle f(x) = \dfrac{3x}{7-x}$, then $\displaystyle f(x) = 7$ if and only if $\displaystyle x = 4.9$. On the other hand, $\displaystyle f(7) = \dfrac{21}{0}$ 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 $\displaystyle (fx) + \dfrac{3x}{7-x}$, then your answer should include an f.

For number 2, you should simplify the numerator first: $\displaystyle a^2-b^2 = (a+b)(a-b)$. So, $\displaystyle x^2-64 = x^2-8^2 = (x+8)(x-8)$. For the denominator, you have $\displaystyle 8-x = -1\cdot (x-8)$. Then, you can cancel the $\displaystyle (x-8)$ from the numerator and denominator to get $\displaystyle \dfrac{x+8}{-1} = -(x+8) = -x-8$.

The third one:

$\displaystyle \dfrac{2x^2-5x-3}{6x^3+3x^2+2x+1}$

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 $\displaystyle (x \pm 1),(x \pm 3),\left(x \pm \dfrac{1}{2}\right), \left(x \pm \dfrac{3}{2}\right)$. Possible roots of the denominator are $\displaystyle (x\pm 1), \left(x \pm \dfrac{1}{2}\right), \left(x \pm \dfrac{1}{3}\right), \left(x \pm \dfrac{1}{6}\right)$. The ones that overlap are $\displaystyle (x\pm 1)$ and $\displaystyle \left(x \pm \dfrac{1}{2}\right)$. So, let's try all four and see which one works. Plug in values for $\displaystyle x$. If $\displaystyle (x-k)$ is a factor of a polynomial, then plugging in $\displaystyle x=k$ will cause the polynomial to equal zero. Let's check:

$\displaystyle 2(1)^2-5(1)-3 = -6$, so $\displaystyle (x-1)$ is not a factor of the numerator.

$\displaystyle 2(-1)^2-5(-1)-3 = 5$, so $\displaystyle (x+1)$ is not a factor of the numerator.

$\displaystyle 2\left(\dfrac{1}{2}\right)^2-5\left(\dfrac{1}{2}\right)-3 = -5$, so $\displaystyle \left(x-\dfrac{1}{2}\right)$ is not a factor of the numerator.

$\displaystyle 2\left(-\dfrac{1}{2}\right)^2 - 5\left(-\dfrac{1}{2}\right)-3 = 0$, so $\displaystyle \left(x + \dfrac{1}{2}\right)$ is a factor of the numerator. Now let's check if it is a factor of the denominator.

$\displaystyle 6\left(-\dfrac{1}{2}\right)^3+3\left(-\dfrac{1}{2}\right)^2+2\left(-\dfrac{1}{2}\right)+1 = 0$, so $\displaystyle \left(x + \dfrac{1}{2}\right)$ is also a factor of the denominator.

The numerator becomes $\displaystyle \left(x+\dfrac{1}{2}\right)(ax+b) = ax^2+\left(b+\dfrac{a}{2}\right)x + \dfrac{b}{2} = 2x^2-5x-3$. This tells us that $\displaystyle a=2, \dfrac{b}{2}=-3$, so $\displaystyle b=-6$. So, the numerator is $\displaystyle \left(x+\dfrac{1}{2}\right)(2x-6)$.

The denominator becomes $\displaystyle \left(x+\dfrac{1}{2}\right)(cx^2+dx+e) = cx^3 + \left(\dfrac{c}{2} + d\right)x^2 + \left(\dfrac{d}{2}+e\right) + \dfrac{e}{2}$. Equating coefficients, we have $\displaystyle c = 6$, $\displaystyle \dfrac{e}{2} = 1$, so $\displaystyle e = 2$, and $\displaystyle \dfrac{c}{2}+d = 3+d = 3$, so $\displaystyle d=0$. So, the denominator is $\displaystyle \left(x+\dfrac{1}{2}\right)\left(6x^2+0x+2\right)$. Cancelling, we get:

$\displaystyle \dfrac{2x-6}{6x^2+2}$

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

$\displaystyle \dfrac{x-3}{3x^2+1}$

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. $\displaystyle \frac{3x}{7- x}= -3+ \frac{21}{7- x}$.

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.

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(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)

$\displaystyle (2x+1)(x-3) = 2x^2+x-6x-3 = 2x^3-5x-3$

So, it does reduce to $\displaystyle \dfrac{x-3}{3x^2+1}$ as both the OP and I wrote.