How to factor the trinomial: x^2 - x - 14 ?

Question

ericwise · Answer

I'm lost at the part where you need to find the corresponding product and sum values

ericwise · Answer

None of the factors seem to work out

jhannybean · Answer

Use the completing the square method,

nnesha · Answer

find two number if you multiply them you should get -14 and if you add or subtract them you should get -1
this is headphone method

ericwise · Answer

Hmm, I've forgotten what that one's all about @Jhannybean

ericwise · Answer

@Nnesha, yeah I considered that. But it seems that none of the factors 1, 2, 7, 14 have a sum or difference of -1

ericwise · Answer

well, except 1 and 2... doesnt work though for the product

ericwise · Answer

wow, all this attention... hi everyone! :D

anonymous · Answer

Did we teach you this the other day?

ericwise · Answer

nope, must've been another one @wio, I don't think I've addressed a completing the square factoring question before

jhannybean · Answer

That's one method to go about factoring, but i'm sure there are several others.

jhannybean · Answer

Do you understand what I did, @EricWise ?

ericwise · Answer

I do not seem to recognize what is happening in the second step

nnesha · Answer

i guess something wrong :(

anonymous · Answer

Where did $-1/2$ come from?

jhannybean · Answer

I think I've doubled it. Whoops.

jhannybean · Answer

Sorry, I misread the function.

jhannybean · Answer

Now in the completing the square method, you want to "complete" a quadratic function inside the parenthesis.

ericwise · Answer

We're assuming that the 16 is actually a 14 right? :)

jhannybean · Answer

Yes, lol

jhannybean · Answer

$$x^2-x-14$$

jhannybean · Answer

$$(x^2-x)-14$$

jhannybean · Answer

In order to do that, we find a new $c$ value, because if you compare the quadratic formula to what is inside the parenthesis, you will see that you have: $ax^2+bx$ whereas in order to complete a quadratic, you must fit the form :$ax^2+bx+c$

jhannybean · Answer

Therefore we're essentially completing a quadratic within the parenthesis.

jhannybean · Answer

$$c=\left(\frac{b}{2}\right)^2$$

jhannybean · Answer

Our $b$ is $-1$. $$c=\left(\frac{-1}{2}\right)^2 = \frac{1}{4}$$

jhannybean · Answer

With this $c$ value, we have completed our quadratic. rewrite the function. $$\left(x^2-x+\frac{1}{4}\right)-14-\frac{1}{4}$$

ericwise · Answer

oh ok... this might be ringing a bell

anonymous · Answer

$$
x^2 - x - 14 \
ax^2+bx+c
$$

jhannybean · Answer

now to keep our equation consistent with our original, we must also subtract a $-1/4$

jhannybean · Answer

Like what @wio posted above my post.

jhannybean · Answer

A little trick to remember is: $$c=\color{red}{\left(\frac{-1}{2}\right)}^2 = \frac{1}{4}$$ Our quadratic will always simplify to $x\pm \text{the red portion}$

ericwise · Answer

oh okay!

jhannybean · Answer

$$\left(x-\frac{1}{2}\right)^2 -\frac{57}{4}$$

jhannybean · Answer

that is your factored form.

jhannybean · Answer

What I did earlier, was I mistook your equation for $x^2-\dfrac{x}{2} -14$

ericwise · Answer

oh okay^^

jhannybean · Answer

Do you understand the completing the square method? :)

ericwise · Answer

So if this whole factored thing is all equal to zero, that means:|dw:1419568567188:dw|