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How to factor quadratic, cubic in this rational expression?
- Thread starter CB1101
- Start date
D
Deleted member 4993
Guest
Simplify:
x^2 - 2x - 8
x^3 + x^2 - 2x
I factored the numerator into (x+2)(x-4) by doing this:
x^2 - 2x - 8
= x^2 - 4x + 2x -8
= x(x-4)+2(x-4)
= (x+2)(x-4)
Now how would I go about factoring the denominator?
First thing you should do is to factor out x
x^3 + x^2 - 2 * x = x * (x^2 + x - 2)
Now you have a quadratic expression within (). Factorize further and continue...
POST EDITED
Perhaps, as someone else said, the bottom can have \(\displaystyle x\) factored out, and the top can be factored using the quadratic equation. (On the other hand, top (the quadratic) can also be factored in the traditional manner). Finally, you must cancel out any all common factors to get the final answer.
\(\displaystyle \dfrac{x^{2} - 2x - 8}{x^{3} + x^{2} - 2x}\)
\(\displaystyle \dfrac{(x - 4)(x + 2)}{x(x^{2} + x - 2)}\)
\(\displaystyle \dfrac{(x - 4)(x + 2)}{x(x^{2} + x - 2)}\) - :shock: Careful, another quadratic to factor, on top of the 1st one we did.
\(\displaystyle \dfrac{(x - 4)(x + 2)}{x (x + 2)(x - 1)}\) - Cancel out any common factors.
\(\displaystyle \dfrac{(x - 4)}{x(x - 1)}\) :shock: Assuming \(\displaystyle x\) does NOT equal \(\displaystyle 0\) or \(\displaystyle 1\) (Those numbers would cause \(\displaystyle 0\) in the denominator, and hence, cause the expression to be undefined.
Note: We figure out which numbers are excluded from the domain by setting \(\displaystyle x\) and \(\displaystyle (x -1)\) equal to \(\displaystyle 0\) (in seperate equations). Next, we solve for \(\displaystyle x\).
a.
\(\displaystyle x = 0\)
b.
\(\displaystyle (x - 1) = 0\)
\(\displaystyle x = 1\)
Finally - step by step on how I factored the quadratics:
a.
\(\displaystyle x^{2} - 2x - 8\) - What two numbers have a sum of \(\displaystyle -2\) and a product of \(\displaystyle -8\)?
\(\displaystyle x^{2} + 2x - 4x - 8\) - Put those two numbers in the middle (while connecting them to x at the same time). In your head, divide the equation into two halves.
\(\displaystyle x(x + 2) - 4(x + 2)\) - Pull out the GCF (greatest common factor) from each side.
\(\displaystyle (x - 4)(x + 2)\)
b.
\(\displaystyle x^{2} - x - 2\)
\(\displaystyle x^{2} + x - 2x - 2\)
\(\displaystyle x(x - 1) + 2(x - 1)\)
\(\displaystyle (x + 2)(x - 1)\)
Note the above equations can also be solved using the quadratic equation.
Perhaps, as someone else said, the bottom can have \(\displaystyle x\) factored out, and the top can be factored using the quadratic equation. (On the other hand, top (the quadratic) can also be factored in the traditional manner). Finally, you must cancel out any all common factors to get the final answer.
\(\displaystyle \dfrac{x^{2} - 2x - 8}{x^{3} + x^{2} - 2x}\)
\(\displaystyle \dfrac{(x - 4)(x + 2)}{x(x^{2} + x - 2)}\)
\(\displaystyle \dfrac{(x - 4)(x + 2)}{x(x^{2} + x - 2)}\) - :shock: Careful, another quadratic to factor, on top of the 1st one we did.
\(\displaystyle \dfrac{(x - 4)(x + 2)}{x (x + 2)(x - 1)}\) - Cancel out any common factors.
\(\displaystyle \dfrac{(x - 4)}{x(x - 1)}\) :shock: Assuming \(\displaystyle x\) does NOT equal \(\displaystyle 0\) or \(\displaystyle 1\) (Those numbers would cause \(\displaystyle 0\) in the denominator, and hence, cause the expression to be undefined.
Note: We figure out which numbers are excluded from the domain by setting \(\displaystyle x\) and \(\displaystyle (x -1)\) equal to \(\displaystyle 0\) (in seperate equations). Next, we solve for \(\displaystyle x\).
a.
\(\displaystyle x = 0\)
b.
\(\displaystyle (x - 1) = 0\)
\(\displaystyle x = 1\)
Finally - step by step on how I factored the quadratics:
a.
\(\displaystyle x^{2} - 2x - 8\) - What two numbers have a sum of \(\displaystyle -2\) and a product of \(\displaystyle -8\)?
\(\displaystyle x^{2} + 2x - 4x - 8\) - Put those two numbers in the middle (while connecting them to x at the same time). In your head, divide the equation into two halves.
\(\displaystyle x(x + 2) - 4(x + 2)\) - Pull out the GCF (greatest common factor) from each side.
\(\displaystyle (x - 4)(x + 2)\)
b.
\(\displaystyle x^{2} - x - 2\)
\(\displaystyle x^{2} + x - 2x - 2\)
\(\displaystyle x(x - 1) + 2(x - 1)\)
\(\displaystyle (x + 2)(x - 1)\)
Note the above equations can also be solved using the quadratic equation.
Last edited:
D
Deleted member 4993
Guest
Perhaps, as someone else said, the bottom can have \(\displaystyle x\) factored out, and the top can be factored using the quadratic equation. (On the other hand, top (the quadratic) can also be factored in the traditional manner). Finally, you must cancel out any all common factors to get the final answer.
\(\displaystyle \dfrac{x^{2} - 2x - 8}{x^{3} + x^{2} - 2x}\)
\(\displaystyle \dfrac{(x - 4)(x + 2)}{x(x^{2} + x - 2)}\)
\(\displaystyle \dfrac{(x - 4)(x + 2)}{x(x^{2} + x - 2)}\) - :shock: Careful, another quadratic to factor, on top of the 1st one we did.
\(\displaystyle \dfrac{(x - 4)(x + 2)}{(x + 2)(x - 1)}\) - Cancel out any common factors.... Incorrect - where did the factor 'x' go from the denominator. It should be:
\(\displaystyle \dfrac{(x - 4)(x + 2)}{x*(x + 2)(x - 1)}\)
\(\displaystyle \dfrac{(x - 4)}{x *(x -1)} \ \ [ for \ \ x \ne (-2)]\) - Answer
Finally - step by step on how I factored the quadratics:
a.
\(\displaystyle x^{2} - 2x - 8\) - What two numbers have a product of \(\displaystyle -8\) and a sum of \(\displaystyle 2\)?
\(\displaystyle x^{2} + 2x - 4x - 8\) - Put those two numbers in the middle (while connecting them to x at the same time). In your head, divide the equation into two halves.
\(\displaystyle x(x + 2) - 4(x + 2)\) - Pull out the GCF (greatest common factor) from each side.
\(\displaystyle (x - 4)(x - 2)\)
b.
\(\displaystyle x^{2} - x - 2\)
\(\displaystyle x^{2} - x - 2x - 2\)
\(\displaystyle x(x - 1) + 2(x - 1)\)
Note the above equations can also be solved using the quadratic equation.
.
Post has been edited to include x in the bottom, and factoring was redone. However \(\displaystyle x\) CANNOT equal \(\displaystyle 0\) or \(\displaystyle 1\).
Last edited:
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
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- 7,763
Most of the "rest" of what he said was just including the "x" in the denominator. But the crucial part was this: The original fraction was \(\displaystyle \frac{x^2- 2x- 8}{x^3+ x^2- 2x}= \frac{(x- 4)(x+ 2)}{x(x+2)(x- 1)}\) which is clearly NOT DEFINED for x= 0, x= 1 or x= 2. Of course, there is a factor of x+ 2 in both numerator and denominator so we can factor that getting \(\displaystyle \frac{x- 4}{x(x- 1}\). That last fraction is defined for all x except x= 0 and x= 1. Since the first fraction was not defined at x= - 2, but this one is, they are NOT equal there. What is true is that \(\displaystyle \frac{x^2- 2x- 8}{x^3+ x^2- 2x}= \frac{x- 4}{x(x- 1)}\) for all x except x= - 2. In general, \(\displaystyle \frac{a}{a}= 1\) except for a= 0.
Last edited by a moderator:
Most of the "rest" of what he said was just including the "x" in the denominator. But the crucial part was this: The original fraction was \(\displaystyle \frac{x^2- 2x- 8}{x^3+ x^2- 2x}= \frac{(x- 4)(x+ 2)}{x(x+2)(x- 1)}\) which is clearly NOT DEFINED for x= 0, x= 1 or x= 2. Of course, there is a factor of x+ 2 in both numerator and denominator so we can factor that getting \(\displaystyle \frac{x- 4}{x(x- 1}\). That last fraction is defined for all x except x= 0 and x= 1. Since the first fraction was not defined at x= 2, but this one is, they are NOT equal there. What is true is that \(\displaystyle \frac{x^2- 2x- 8}{x^3+ x^2- 2x}= \frac{x- 4}{x(x- 1)}\) for all x except x= 2. In general, \(\displaystyle \frac{a}{a}= 1\) except for a= 0.
The \(\displaystyle 0\) and \(\displaystyle 1\) are excluded from the domain, but I don't see the \(\displaystyle 2\) in all this. Khan even got \(\displaystyle -2\).
D
Deleted member 4993
Guest
Most of the "rest" of what he said was just including the "x" in the denominator. But the crucial part was this: The original fraction was \(\displaystyle \frac{x^2- 2x- 8}{x^3+ x^2- 2x}= \frac{(x- 4)(x+ 2)}{x(x+2)(x- 1)}\) which is clearly NOT DEFINED for x= 0, x= 1 or x= 2. Of course, there is a factor of x+ 2 in both numerator and denominator so we can factor that getting \(\displaystyle \frac{x- 4}{x(x- 1}\). That last fraction is defined for all x except x= 0 and x= 1. Since the first fraction was not defined at x=-2, but this one is, they are NOT equal there. What is true is that \(\displaystyle \frac{x^2- 2x- 8}{x^3+ x^2- 2x}= \frac{x- 4}{x(x- 1)}\) for all x except x= -2. In general, \(\displaystyle \frac{a}{a}= 1\) except for a= 0.
HoI had a small typo - I fixed it in this "quote"
But read his response carefully - try to understand what he explains.