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

• Nov 10th 2013, 06:42 AM
Zeyna18
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)

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. :)
• Nov 10th 2013, 07:27 AM
SlipEternal
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 $f(x) = \dfrac{3x}{7-x}$, then $f(x) = 7$ if and only if $x = 4.9$. On the other hand, $f(7) = \dfrac{21}{0}$ 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 $(fx) + \dfrac{3x}{7-x}$, then your answer should include an f.

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

The third one:

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

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

$2(1)^2-5(1)-3 = -6$, so $(x-1)$ is not a factor of the numerator.
$2(-1)^2-5(-1)-3 = 5$, so $(x+1)$ is not a factor of the numerator.
$2\left(\dfrac{1}{2}\right)^2-5\left(\dfrac{1}{2}\right)-3 = -5$, so $\left(x-\dfrac{1}{2}\right)$ is not a factor of the numerator.
$2\left(-\dfrac{1}{2}\right)^2 - 5\left(-\dfrac{1}{2}\right)-3 = 0$, so $\left(x + \dfrac{1}{2}\right)$ is a factor of the numerator. Now let's check if it is a factor of the denominator.

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

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

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

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

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

$\dfrac{x-3}{3x^2+1}$
• Nov 10th 2013, 10:08 AM
HallsofIvy
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)

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

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:

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.
• Nov 10th 2013, 10:17 AM
SlipEternal
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)

$(2x+1)(x-3) = 2x^2+x-6x-3 = 2x^3-5x-3$
So, it does reduce to $\dfrac{x-3}{3x^2+1}$ as both the OP and I wrote.