Factoring quadratics - how to?

Question

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:

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):

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.

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Update

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:

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

I then apply the same method to the bottom and get the following:

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!

Answer 1

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.

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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!

Answer 2

I think factorization method would be easier in this case $$\frac{x^2-3x-10}{x^2-6x-5}$$ Then you simplify making it: $$\frac{x^2-5x+2x-10}{x^2-5x-1x+5}$$ After which you factorise it: $$\frac{x(x-5)+2(x-5)}{x(x-5)-1(x-5)}$$ Which equates to: $$\frac{(x-2) (x-5)}{(x-1)( x-5)} $$ Then $x-5$ cancels each other thereby making it: $$\frac{x+2}{x+1}$$