Factoring Trinomials: 7x^2-15x+7x-105=0

[supermath]

Homework Statement

Solve the equation.

Relevant Equations

x(7x-8)=15

I do
7x^2-8x=15
7x^2-8x-15=0

I turn the problem into a Grouping(4 term) problem by multiplying 7*(-15) = 105
I find what numbers multiply to -105 and add to -8. -15 and 7 multiply to -105 and add to -8.
7x^2-15x+7x-105=0

I group them.
7x^2-15x | 7x-105=0

This is where i get lost, because i get
7x(x-1\2)*7(x-15)
The numbers in parenthesis are suppose to match.

 

[etotheipi]

Whilst I wouldn't advise your method (it's often faster to just do trial and error), if you want to do it this way you should have it written as $7x^2 + 7x - 15x -15 =0$. You are just decomposing the term pertaining to $x$ and shouldn't change the coefficients of $x^2$ or $x^0$.

 

[supermath]

Whilst I wouldn't advise your method (it's often faster to just do trial and error), if you want to do it this way you should have it written as $7x^2 + 7x - 15x -15 =0$. You are just decomposing the term pertaining to $x$ and shouldn't change the coefficients of $x^2$ or $x^0$.

Is the -15x from 7x and -8x? Shouldn't it be a -1x?

 

[etotheipi]

No, you start with $7x^2 - 8x - 15 = 0$. Then you say $-8x = 7x - 15x$. It follows that $7x^2 + 7x - 15x -15 = 0$. Now you can factorise the first and last pairs fairly easily.

 

[supermath]

No, you start with $7x^2 - 8x - 15 = 0$. Then you say $-8x = 7x - 15x$. It follows that $7x^2 + 7x - 15x -15 = 0$. Now you can factorise the first and last pairs fairly easily.

You've lost me. Is that Reverse FOIL? BTW, is the equation a Non-monic Trinomial? I'm looking through my notes to see which method to use, and Non-monic trinomials seem like the one to use.

 

[etotheipi]

You're getting lost in the names and the acronyms, and you're not thinking about the maths!

You start off with the equation $7x^2 - 8x - 15 = 0$. The easiest way to factorise this is by trial and error, but for sake of explanation we will use the method you are referring to.

Like you say, we look for two numbers whose product is $-105$ and whose sum is $-8$. These are, also as you say, $-15$ and $7$. Now that we have these two numbers ($-15$ and $7$), we can break the $-8x$ term in our original equation down, like so:

$7x^2 +7x - 15x - 15 = 0$

Now, what can you do? Perhaps $7x(x+1) - 15(x+1) = 0$?

 

[supermath]

You're getting lost in the names and the acronyms, and you're not thinking about the maths!

You start off with the equation $7x^2 - 8x - 15 = 0$. The easiest way to factorise this is by trial and error, but for sake of explanation we will use the method you are referring to.

Like you say, we look for two numbers whose product is $-105$ and whose sum is $-8$. These are, also as you say, $-15$ and $7$. Now that we have these two numbers ($-15$ and $7$), we can break the $-8x$ term in our original equation down, like so:

$7x^2 +7x - 15x - 15 = 0$

Now, what can you do? Perhaps $7x(x+1) - 15(x+1) = 0$?

Oh, gotcha. Isn't that what I did in my original post? I thought you are suppose to keep the -105 in the equation like this.
$7x^2 +7x | - 15x - 105 = 0$
And then group.

 

[etotheipi]

Oh, gotcha. Isn't that what I did in my original post? I thought you are suppose to keep the -105 in the equation like this.
$7x^2 +7x | - 15x - 105 = 0$
And then group.

No, because that would be a different equation.

 

[etotheipi]

It might help you to see why the method you state works in the first place. Any polynomial of degree 2 can always be written in the following form,

$P(x) = (ax + b)(cx + d)$

Now, we expand:

$P(x) = acx^2 + bcx + adx + bd = acx^2 + (bc + ad)x + bd$

You will notice then that we can always find only one unique set of two numbers, $p = bc$ and $q = ad$ such that their product $pq$ equals the product of the coefficients of $x^2$ and $x^0$, and their sum equals the coefficient of $x$. Importantly, these numbers also each share a common factor with the coefficients of $x^2$ and $x^0$. If we now split the $bx$ term into two terms $adx + bcx$, we can now factorise the first and last pairs, and then the whole thing.

 

[supermath]

Should I memorize the formula?

Also when I grouped using $7x^2 -15x + 7x - 15 = 0$ I get (x-15/7)(7x+7).
When I switched to the way you wrote it $7x^2 +7x - 15x - 15 = 0$, I get (x+1)(7x-15).
my homework helper software confirms the answer is (x+1)(7x-15).
My other question is, am I suppose to know which way to arrange the equation so I can factor it further? If I am suppose to know which way to arrange it, does this mean I don't actually understand the material?

 

[etotheipi]

You'd rearrange it depending on which pairs have a nice common factor that you can pull out. This should usually be fairly clear.

I would say learning a set of rules is a last resort. Best to try and understand the maths first!

Try and do lots of practice questions and you'll get the hang of it.

 

[supermath]

Thanks. I'm actually doing just that. Seems like I'm just not getting it.

 

[supermath]

You'd rearrange it depending on which pairs have a nice common factor that you can pull out. This should usually be fairly clear.

I would say learning a set of rules is a last resort. Best to try and understand the maths first!

Try and do lots of practice questions and you'll get the hang of it.

Do all the terms need to have a common factor to pull out.
In this problem $42x^2 +13x -42 = 0$.
1 is the only common factor.
So, I multiply 42*(-42)=-1764
36, -49 multiply to -1764 and add to -13.
I once again get stuck like in the original problem.
$7x^2 +36x - 49x - 42 = 0$ or $7x^2 -49x + 36x - 42 = 0$

 

[etotheipi]

They should add to $+13$. Your two numbers are $49$ and $-36$. You rewrite the equation as

$42x^2 + 49x - 36x - 42 = 0$
$7x(6x+7) -6(6x + 7) = 0$
$(7x - 6)(6x+7) = 0$

 

[supermath]

They should add to $+13$. Your two numbers are $49$ and $-36$. You rewrite the equation as

$42x^2 + 49x - 36x - 42 = 0$
$7x(6x+7) -6(6x + 7) = 0$
$(7x - 6)(6x+7) = 0$

$42x^2 + 49x - 36x - 42 = 0$
I keep trying to bring down the -36, and then dividing the -36 with the -42.
-36x-42=0
-36(x+42/36)
And then move on to the other group[##42x^2]

 

[etotheipi]

$42x^2 + 49x - 36x - 42 = 0$
I keep trying to bring down the -36, and then dividing the -36 with the -42.
-36x-42=0
-36(x+42/36)
And then move on to the other group[##42x^2]

The common factor that you take out here is necessarily an integer. So you can always use the highest common factor in each pair. For the $-36x - 42$, you can take out a factor of $6$.

 

[supermath]

Ok.
I made a mistake I wrote +13x. it's -13x.

In this previous problem 7x^2 + 7x - 15x -15 = 0
for the second group (-15x-15), did we factor out a -15 and arrive to -15(x+1)?

 

[etotheipi]

In this previous problem 7x^2 + 7x - 15x -15 = 0
for the second group (-15x-15), did we factor out a -15 and arrive to -15(x+1)?

That's right, yes!

You'll see that the process is really quite natural.

a) Find your two numbers $p$ and $q$
b) Rewrite the $bx$ term as two terms $px + qx$
c) Factorise each pair of terms, and finally factorise the whole thing

My post #14 shows all of these steps.

 

[supermath]

$42^2 -13x-42=0$
$42^2-49x | 36x-42=0$
$7x(6x-7) | +6(6x-7)$
$(6x-7)(7x+6)$
$x=7/6 x=-6/7$
$7/6, -6/7$ <--------- Answer

Thanks etotheipi

 

[etotheipi]

Okay, but just remember not to write the vertical "|" in the middle of your equation!

 

[supermath]

Yeah, that just helps me when grouping. Our homework and tests are online this semester. I have a notebook where I write the problems down, solve em, type the answers on the comp.

Side question, is there a link or thread on how to write codes or bb codes when posting.
For example, the double hashtag you use seems to make exponents more readable, and equations bold. I'm sure there is one for fractions, and other math symbols. Thanks.

 

[etotheipi]

Here you go : https://www.physicsforums.com/help/latexhelp/

 

[SammyS]

You've lost me. Is that Reverse FOIL? BTW, is the equation a Non-monic Trinomial? I'm looking through my notes to see which method to use, and Non-monic trinomials seem like the one to use.

Since you mentioned non-monic trinomials, may I then assume that for your notes have a much easier method for monic trinomials.
Say if you are to factor $u^2-8u-105$ , then you still get the same pair of numbers, $-15$ and $7$, such that $-15+7=-8$ and $(-15)\times(7)=-106$ . But with a monic you then immediately get $u^2-8u-105=(u+7)(u-15)$, without formally suing factoring by grouping.

If so, here's a trick so that you can treat a non-monic trinomial in a similar fashion.

Starting with the same non-monic you initially asked about:

Take $7x^2-8x-105$ and multiply through by $7$ as well as dividing the whole expression by $7$.

To aid in seeing how this works: When multiplying through by $7$, think of that as $\dfrac{(7x)^2-8(7x)-105}{7}$ rather than $\dfrac{49x^2-56x-105}{7}$ .

So for this, the $7x$ takes the place of $u$ above.

You then have: $\dfrac {(7x+7)(7x-15)} {7}$. Factor a $7$ out of $(7x+7)$ to get $\dfrac {7(x+1)(7x-15)} {7}$. Cancel the common factor, $7$, to get your result.

 

[supermath]

Since you mentioned non-monic trinomials, may I then assume that for your notes have a much easier method for monic trinomials.
Say if you are to factor $u^2-8u-105$ , then you still get the same pair of numbers, $-15$ and $7$, such that $-15+7=-8$ and $(-15)\times(7)=-106$ . But with a monic you then immediately get $u^2-8u-105=(u+7)(u-15)$, without formally suing factoring by grouping.

If so, here's a trick so that you can treat a non-monic trinomial in a similar fashion.

Starting with the same non-monic you initially asked about:

Take $7x^2-8x-105$ and multiply through by $7$ as well as dividing the whole expression by $7$.

To aid in seeing how this works: When multiplying through by $7$, think of that as $\dfrac{(7x)^2-8(7x)-105}{7}$ rather than $\dfrac{49x^2-56x-105}{7}$ .

So for this, the $7x$ takes the place of $u$ above.

You then have: $\dfrac {(7x+7)(7x-15)} {7}$. Factor a $7$ out of $(7x+7)$ to get $\dfrac {7(x+1)(7x-15)} {7}$. Cancel the common factor, $7$, to get your result.

So you still have to multiply the 7 * -15 to get -105?

 

[SammyS]

So you still have to multiply the 7 * -15 to get -105?

Yes, of course.

You still have the same "riddle" to solve. Find two numbers whose product is $-105$ and sum is $-8$. It's just that after that you can avoid the factoring by grouping.

It's simply an alternative method for factoring a trinomial. If it doesn't work for you, then don't use it.

 

[symbolipoint]

Why is this topic so long?
I see the original equation to solve: in pure text form, x(7x-8)=15

Distributive property is obvious to use, first.
7x^2-8x=15

7x^2-8x-15=0

Now if this is easily factorable, then try to set up these and test:

• (7x 3)(x 5)
• (7x 5)(x 3)
• (7x 1)(x 15)
• (7x 15)(x 1)

If any look good, check how the signs should work.

 

[Mark44]

Why is this topic so long?
I see the original equation to solve: in pure text form, x(7x-8)=15

Distributive property is obvious to use, first.
7x^2-8x=15

7x^2-8x-15=0

The OP is trying to use the technique for quadratic trinomials of the form $ax^2 + bx + c = 0$, in which you look for factors of ac that add up to b.

In this case, looking for factors of -105 that add up to -8. The pairs are (1, -105), (3, -35), (5, -21), and (7, -15). There are four more pairs, but all of them result in a positive sum.
In this case, the factors are 7 and -15, so you rewrite the equation as $7x^2 + 7x - 15x - 15 = 0$, and then group them in pairs as $7x(x + 1) - 15(x + 1) = 0$, resulting in $(7x - 15)(x + 1) = 0$.

 

[symbolipoint]

Thanks Mark44. Some of the discussion slipped by me because the topic is a little too long for what I saw in the original question posting.

 

[symbolipoint]

The OP is trying to use the technique for quadratic trinomials of the form $ax^2 + bx + c = 0$, in which you look for factors of ac that add up to b.

In this case, looking for factors of -105 that add up to -8. The pairs are (1, -105), (3, -35), (5, -21), and (7, -15). There are four more pairs, but all of them result in a positive sum.
In this case, the factors are 7 and -15, so you rewrite the equation as $7x^2 + 7x - 15x - 15 = 0$, and then group them in pairs as $7x(x + 1) - 15(x + 1) = 0$, resulting in $(7x - 15)(x + 1) = 0$.

The second example should seem clear now.
You would have began with 7x^2-8x-105=0;
and then want to factor the quadratic expression.

See some two-factor combinations for just the 105.
105=5*21=7*15 and there are others.
...
I started into but did not think all of this through, thoroughly yet. I'll work it a few minutes and see.
...
Now I checked. NO combination for just integers worked. I would check with general solution for quadratic equation.

For sure, not factorable by integers. Discriminant is 3004=4*751, and the 751 is a prime number.

 

[Mark44]

The second example should seem clear now.
You would have began with 7x^2-8x-105=0;

No, that equation isn't part of the problem. It arose from confusion on the part of the OP.
The equation in this problem is $7x^2 - 8x - 15 = 0$.

His misunderstanding is the part about multiplying 7 and -15 (to get -105), and thinking that this product needs to show up in the equation. Here I've quoted what he wrote quite a few posts back.

​

Oh, gotcha. Isn't that what I did in my original post? I thought you are suppose to keep the -105 in the equation like this. $7x^2 +7x | - 15x - 105 = 0$​

​

​

and then want to factor the quadratic expression.

See some two-factor combinations for just the 105.
105=5*21=7*15 and there are others.
...
I started into but did not think all of this through, thoroughly yet. I'll work it a few minutes and see.
...
Now I checked. NO combination for just integers worked. I would check with general solution for quadratic equation.

For sure, not factorable by integers. Discriminant is 3004=4*751, and the 751 is a prime number.