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Help with factoring trinomials

bkelly2442

New member
Jun 28, 2016
7
I really hope someone here can help me - I'm sure that's not going to be difficult for most. :)

I'm helping a friend's daughter factoring trinomials. I can get to the correct answer, but usually thru a system of trial and error and not a defined set of rules in every instance. Example

6x^2 + 39x - 21
3(2x^2 +13x-7)

Here's where I get stuck. I know that the factors of -7 are going to be 1, -7 or -1, 7. However, these don't add to 13. I suppose that has something to do with the factored out 3? Or the unkown? Also, when I do get the 2 sets of binomials, what is the rule to determine whether I use 1, -7 or -1, 7? What is the rule that determines if my sets are (2x-1)(x+7) or (x-1)(2x+7) or (2x+1)(x-7) or (x+1)(2x-7) or (2x-7)(x+1) and so on and so on and so on......
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,736
Recognizing that 3 is a factor of all coefficients is a good first step, as this will help keep things simple:

\(\displaystyle 6x^2+39x-21=3\left(2x^2+13x-7\right)\)

Now, for the quadratic factor, we see the leading coefficient (2) is prime, and so we know the factorization will take the following form:

\(\displaystyle 6x^2+39x-21=3\left(2x^2+13x-7\right)=3(2x\cdots)(x\cdots)\)

At this point we want to find two factors of the product of the first and last coefficients ($2(-7)=-14$) whose factors add to the middle coefficient ($13$), and we find:

\(\displaystyle (-1)(14)=-14\) and \(\displaystyle (-1)+(14)=13\)

So, we need $-1$ and $14$, and to get our 14, we will pair a 7 with the 2 and to get our -1, we will pair a -1 with our 1:

\(\displaystyle 6x^2+39x-21=3\left(2x^2+13x-7\right)=3(2x-1)(x+7)\)

When I say to pair the numbers, I am talking about the distributive property of binomials commonly called FOIL. So, the pairs are the two "outer" terms and the two "inner" terms.

Does that makes sense?
 

bkelly2442

New member
Jun 28, 2016
7
Mostly - here's where I don't have clarity

"Now, for the quadratic factor, we see the leading coefficient (2) is prime, and so we know the factorization will take the following form:

6x2+39x−21=3(2x2+13x−7)=3(2x⋯)(x⋯)

Because the coefficient (2) is prime it always holds the position in the first set of binomials (2x....)(x....)? It would never be (x.....)(2x....)? What if it was not a prime - (8x......)(x......) vs (x.....)(8x.....) I know there has to be a rule for this - just don't know what it is.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,736
Mostly - here's where I don't have clarity

"Now, for the quadratic factor, we see the leading coefficient (2) is prime, and so we know the factorization will take the following form:

6x2+39x−21=3(2x2+13x−7)=3(2x⋯)(x⋯)

Because the coefficient (2) is prime it always holds the position in the first set of binomials (2x....)(x....)? It would never be (x.....)(2x....)? What if it was not a prime - (8x......)(x......) vs (x.....)(8x.....) I know there has to be a rule for this - just don't know what it is.
The commutative law of multiplication says that $ab=ba$, so yes, it could be $(2x\cdots)(x\cdots)$ or $(x\cdots)(2x\cdots)$...I should have said one would have to to $(2x\cdots)$ and the other $(x\cdots)$.

Now, it the leading coefficient had been composite such as 8, then we would have to consider all factor pairs (1,8), (2,4). A prime number only has 1 factor pair, so that limits us to one choice.
 

bkelly2442

New member
Jun 28, 2016
7
Understood - but

"The commutative law of multiplication says that ab=ba, so yes, it could be (2x⋯)(x⋯) or (x⋯)(2x⋯)...I should have said one would have to to (2x⋯) and the other (x⋯)."

What is the rule that makes the placement of the (2x....)(x....) vs (x....)(2x...)? Or is it just that you have to try it one way and if that doesn't work, move them? And if the leading coefficient is not prime, and therefore has multiple factor pairs, do you have to try each pair each way until you get the right one?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,736
Understood - but

"The commutative law of multiplication says that ab=ba, so yes, it could be (2x⋯)(x⋯) or (x⋯)(2x⋯)...I should have said one would have to to (2x⋯) and the other (x⋯)."

What is the rule that makes the placement of the (2x....)(x....) vs (x....)(2x...)? Or is it just that you have to try it one way and if that doesn't work, move them? And if the leading coefficient is not prime, and therefore has multiple factor pairs, do you have to try each pair each way until you get the right one?
It will work either way...I simply choose to use that order.

If the leading coefficient is composite, then you may have to use some trial and error before you find the right combination. :)

Here is a link to a tutorial I wrote regarding factoring quadratics:

Factoring Quadratics
 

bkelly2442

New member
Jun 28, 2016
7
Wonderful - thanks so much Mark!
 

bkelly2442

New member
Jun 28, 2016
7
Ok, stuck again

24x^2 - 55x - 24

product of first and last coefficients (24(-24)= -576). Factors of -576 that ADD to middle coefficient (-55x) are -64, 9

However, if I check

(24x -64)(x+9)

24x^2 + 216x - 64x + 576

WHAT AM I DOING WRONG?????
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,736
Okay, you have done well in your identification of -576, -64 and 9.

So, what I would next look at are factors common to 24 and 9 which is 3...which means a form like:

\(\displaystyle (8x\cdots)(3x\cdots)\)

In order to get our -64 and 9, we need:

\(\displaystyle (8x+3)(3x-8)\)
 

bkelly2442

New member
Jun 28, 2016
7
WHAT????

Mike - you have been so helpful and patient, but I am completely lost now.

Let me see if I can explain it back to you in words I understand

After finding the 2 numbers (-64, 9) I THEN have to factor the existing numbers down to multiply to those 2 numbers???
 

bkelly2442

New member
Jun 28, 2016
7
What I need is a very simplistic step by step plan. I just can't understand going from one step to the next
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,736
WHAT????

Mike - you have been so helpful and patient, but I am completely lost now.

Let me see if I can explain it back to you in words I understand

After finding the 2 numbers (-64, 9) I THEN have to factor the existing numbers down to multiply to those 2 numbers???
My name is actually Mark, but this isn't the first time I have been called "Mike" instead. :p

Seriously though, after finding the two factors of -576 whose sum is 55 (which you did), you do in fact then need to look at the largest factors common to 24 and -64 (which is 8) OR 24 and 9 (which is 3). This will help you to determine which of the numerous factor pairs of 24 is the correct choice.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,736
What I need is a very simplistic step by step plan. I just can't understand going from one step to the next
Okay...we are given to factor:

\(\displaystyle 24x^2-55x-24\)

First, I observe that the greatest common divisor of 24 and 55 is 1, so there is no factor common to all 3 terms.

Next, I look at:

\(\displaystyle 24(-24)=-576\)

\(\displaystyle (-64)(9)=-576\) and \(\displaystyle (-64)+(9)=-55\)

I then see that 8 divides 64 and 3 divides 9 and $(3)(8)=24$

At this point, I know I have the form:

\(\displaystyle (8x\cdots)(3x\cdots)\)

Now, to get -64 I need to pair -8 with 8, like so:

\(\displaystyle (8x\cdots)(3x-8)\)

And to get 9 I need to pair 3 with 3 like so:

\(\displaystyle (8x+3)(3x-8)\)

And now we may conclude:

\(\displaystyle 24x^2-55x-24=(8x+3)(3x-8)\)
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Ok, stuck again

24x^2 - 55x - 24

product of first and last coefficients (24(-24)= -576). Factors of -576 that ADD to middle coefficient (-55x) are -64, 9
Good job! But your approach after that was incorrect. Those two numbers are what the middle term is broken up into, and then you factorise by grouping.

$\displaystyle \begin{align*} 24\,x^2 - 55\,x - 24 &= 24\,x^2 - 64\,x + 9\,x - 24 \textrm{ (do you see how } -64\,x + 9\,x = -55\,x\textrm{?)} \\ &= 8\,x\,\left( 3\,x - 8 \right) + 3\,\left( 3\,x - 8 \right) \textrm{ (here is where you find the highest common factor in each half)} \\ &= \left( 3\,x - 8 \right) \left( 8\,x + 3 \right) \textrm{ (and pull out the resulting factor)} \end{align*}$
 

soroban

Well-known member
Feb 2, 2012
409
Ok, stuck again

[tex]24x^2 - 55x - 24[/tex]

product of first and last coefficients (24(-24) = -576).

Factors of -576 that ADD to middle coefficient (-55x) are -64, 9

At this point, use those two numbers as coefficients of [tex]x[/tex]-terms.

[tex]\begin{array}{ccc}\text{So we have:} & 24x^2 - 64x + 9x -24\\
\text{Factor by grouping:} & 8x(3x-8) + 3(3x-8) \\
\text{Therefore:} & (3x-8)(8x+3) \end{array}[/tex]


Note: The order of those "middle terms" does not matter.

[tex]\begin{array}{ccc}\text{We have:} & 24x^2 + 9x - 64x - 24 \\
\text{Factor by grouping:} & 3x(8x+3) - 8(8x+3) \\
\text{Therefore:} & (8x+3)(3x-8) \end{array}[/tex]
 

BestMethod

New member
Dec 11, 2019
9
Let's factor y=(2x^2 +13x-7) this way;

If we multiply & divide the polynomial(let’s call it f) by a number,say m,f won’t be changed;

mf/m=(m/m)f=1*f=f

Now,one way here is to multiply & divide f by 2;

2(2x^2 +13x-7)/2 = (4x^2 +26x-14)/2 =

[(2x)^2+13(2x)-14]/2

Taking 2x=z;

y=(z^2+13z-14)/2

Here,let’s factor it using this famous identity,

(z+a)(z+b)=z^2+(a+b)z+ab

So,we should find a&b such that their sum be equal to 13,and their product to -14.

oh,they are a=-1 ,b=14.Then,

y=(z-1)(z+14)/2 = (2x-1)(2x+14)/2 =

(2x-1)(x+7)