It has been some time since I have studied math (and even then it was base-level math), and I know this will probably be easy for most, but I need help with Simplifying Algebra equations.

Simple equations like so I can do: 3x + 7 + 9x -4 would (I really hope) equal 12x + 3, right?

I now have an equation I wish to get my head around, and it is:

Basic Algebra - Simplify

Now I have done some searching for the guides on how to perform the above and have either ended up with base level answers which did not help the above, or more difficult examples I could not quite get my head around. I have attempted to get the answer then "work backwards" at it, but I cannot wrap my head around the following (Wolfram Alpha):

enter image description here

I would ask you, the wonderful users of this forum, to either help explain the above question, or provide me the ability to learn this independently.


After looking into the FOIL method, I referred to my genius co-worker for assistance- I think he is able to explain things to an idiot (me) and it works. Basically we went through the quadratic formula method (which I understand!) and my workout was as follows:

Quadratic Formula

This is how I have processed this equation (with great skill!):

Tier 1 Workout

I then apply the same method to the bottom and get the following: Tier 2 Working Out

With the top line, I understand we perform both 3+7 and 3-7 (which results in 5 and -2), however in the second option I assume (and this is merit to my lack of math knowledge), we performed the opposite function of minus due to the original sum being plus 5. Is this correct? Following on, performing both the addition and subtraction we get 6+4 which results in 10 and six minus 4 which results in two, which then when divided is 1. I am just attempting to understand why the 1 in this function is negative.

Can you please either highlight anything I am doing incorrectly, advise if there is a incorrect step or let me know if all is okay? Much appreciated!

  • $\begingroup$ The word that describes what WA is doing is "factoring", and the word that describes the type of problem you're solving is "rational expression". $\endgroup$ – Eric Stucky Dec 10 '14 at 1:09
  • $\begingroup$ @EricStucky I have updated my question with a procedure I believe I understand. Will you be able to advise me if all is correct? $\endgroup$ – DankyNanky Dec 10 '14 at 4:41
  • $\begingroup$ The -1 in the WA function should correspond to a +1 in your Quadratic Formula (QF) calculation. (Notice that the -2 in your QF calculation came from a +2 in the WA function) This is because the quadratic formula tells you for which $r$ we have $ar^2+br+c=0$, which means that $r$ is a "root" of the quadratic. The Remainder Theorem states that if $r$ is a root of a polynomial, then $x-r$ is a factor of the polynomial. Thus since you got $r=\frac{10}{2},\frac{2}{2}=5,1$, you get $(x-5)(x-1)$, as suggested by WA's calculation. $\endgroup$ – Eric Stucky Dec 10 '14 at 7:41
  • $\begingroup$ [Actually the Remainder Theorem is more general, but this is the form in which it is most commonly used.] $\endgroup$ – Eric Stucky Dec 10 '14 at 7:41

When expanding any factored expression of the form $(x+a)(x+b)$ it is useful to remember the FOIL rule, meaning, first-outer-inner-last, specifying all the terms that need to be multiplied to get the expansion. That is,

First: $x\cdot x = x^2$

Outer: $x\cdot b = bx$

Inner: $a\cdot x = ax$

Last: $a\cdot b = ab$

Now sum, to obtain $(x+a)(x+b) = x^2 + bx+ax+ab = x^2 + (a+b)x + ab$. This is the rationale behind factoring an expression of the form $x^2 + cx + d$, we want to find the $a$ and $b$ values in $(x+a)(x+b)$ such that $a+b=c$ and $ab = d$.

In your example, we are just doing that independently for the numerator and the denominator to get both of them in factored form. Then we can cancel the common terms and obtain the desired answer.

Response to update:

The quadratic formula is a fool-proof way of obtaining the roots of an expression of the form $ax^2+bx+c$ (roots meaning the values of $x$ where $ax^2+bx+c=0$). It's a useful equation that is worth committing to memory. With practice, you will be able to see what the roots of an equation are without applying the formula, but if the roots aren't nice numbers, the quadratic equation is a good way of finding them.

In finding the roots of the numerator, most of it is right. One small error is you wrote $3^2-4a10$, when in fact, it should be $3^2-(4)(-10) = 3^2+40 = 49$. You've written it down correctly two lines down though!

For the roots of the denominator, you are right, except for the last line, the roots are $x=(6+4)/2=5$ and $x=(6-4)/2=1$. Remember, roots are the points that make the equation equal to zero, so $(x-5)(x-1)=0$ only when $x=5$ or $x=1$.

With that you have arrived at your answer since the root $x=5$ is common between the numerator and denominator and thus the $(x-5)$ terms cancel.

Let me know if you want me to explain any steps further!

  • 1
    $\begingroup$ Thank you for that. I do understand the FOIL method (and am currently revisiting this to ensure it correct). Once I have looked into this I will either refine my question or follow up. Thanks! $\endgroup$ – DankyNanky Dec 10 '14 at 2:02
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    $\begingroup$ I have updated my "question" with the way I believe I am able to sort/understand this. As I suspected this has raised more questions for me to ask. I appreciate your answer and whilst I cannot say I fully understand the procedure I believe you correct. Can you please however review my "update" and tell me if all is well? $\endgroup$ – DankyNanky Dec 10 '14 at 4:40
  • $\begingroup$ @MichaelNancarrow I've updated my response, let me know if you have any questions regarding the steps. $\endgroup$ – Erik M Dec 10 '14 at 5:48
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    $\begingroup$ thank you so much for your assistance, you've been a great help! I do plan on practising this and will refine the equation to what I would call an acceptable standard. I may pester you at a later date to review these items but for the time being you've been more than helpful. Thanks for welcoming me to this forum! $\endgroup$ – DankyNanky Dec 10 '14 at 5:51
  • $\begingroup$ @MichaelNancarrow Not a problem, glad to help! $\endgroup$ – Erik M Dec 10 '14 at 5:53

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