Factoring Expressions: 28x^2 + 5xy-12y^2
- Thread starter Bebe
- Start date
Re: Factoring Expressions
Hello, Bebe!
There is a procedure for these problems . . .
Multiply the first coefficient by the last: \(\displaystyle \:28\,\times\,12\:=\:336\)
Because the last term is negative, we want a "difference".
. . [It is waa positive, we'd want a "sum".]
We want to factor the \(\displaystyle 336\) into two parts whose difference is \(\displaystyle 5\), the middle coefficient.
How do we find it?
Well, there's a primitive method that always works.
. . Divide \(\displaystyle 336\) by \(\displaystyle 1,\,2,\,3,\,4,\,\cdots\;\;\) (keep those that "come out even")
. . until we find the pair we want.
For example: \(\displaystyle \,336\,\div\,1 \:=\:336\;\;\Rightarrow\;\;336\:=\:1\cdot336\)
. . . . . . . . . . \(\displaystyle 336\,\div\,2 \:=\:168\;\;\Rightarrow\;\;336\:=\:2\cdot168\)
. . . . . . . . . . \(\displaystyle 336\,\div\,3\:=\:112\;\;\Rightarrow\;\;336\:=\:3\cdot112\)
And we will have a list of the complete factoring of \(\displaystyle 336:\)
. . \(\displaystyle \begin{array}{ccc}1&\cdot&336 \\ 2&\cdot&168\\ 3&\cdot&112\\ 4&\cdot&84 \\ 6&\cdot&56 \\8&\cdot&42\\12&\cdot&28\\14&\cdot&24\\16&\cdot&21\end{array}\)
Remember what pair we were looking for?
. . That's right . . . a difference of \(\displaystyle 5\)
It's the very last pair: \(\displaystyle \,16,\:21\)
Use these to "split" the middle term.
The middle term is \(\displaystyle +5xy\), so we will use \(\displaystyle +21xy\) and \(\displaystyle \,-16xy\)
The polynomial becomes: \(\displaystyle \:28x^2\,+\,21xy\,-\,16xy\,-\,12y^2\)
Factor "by grouping": \(\displaystyle \:7x(4x\,+\,3y)\,-\,4y(4x\,+\,3y)\)
Take out the common factor: \(\displaystyle \:\fbox{(4x\,+\,3y)(7x\,-\,4y)}\;\;\) . . . There!
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
To complete this lesson, suppose the last term is positive.
Example: \(\displaystyle \:6a^2\,-\,17ab\,+\,10b^2\)
"First times Last": \(\displaystyle \:6\cdot10\:=\:60\)
We want to factor \(\displaystyle 60\) into two parts whose sum is \(\displaystyle 17\).
So we factor \(\displaystyle 60\):
. . \(\displaystyle \begin{array}{ccc}1&\cdot&60\\2&\cdot&30\\3&\cdot&20\\4&\cdot&15\\5&\cdot&12\\6&\cdot&10\end{array}\)
The pair with a sum of 17 is: \(\displaystyle 5\) and \(\displaystyle 12\).
The middle term is \(\displaystyle \,-17ab\), so we will use \(\displaystyle \,-5ab\) and \(\displaystyle \,-12ab\).
The polynomial becomes: \(\displaystyle \:6a^2\,-\,5ab\,-\,12ab\,+\,10b^2\)
Factor by gourping: \(\displaystyle \:a(6a\,-\,5b)\,-\,2b(6a\,-\,5b)\)
Take out the common factor: \(\displaystyle \:\fbox{(6a\,-\,5a)(a\,-\,2b)}\)
You can, of course, check these answers yourself . . . right?
Hello, Bebe!
There is a procedure for these problems . . .
Multiply the first coefficient by the last: \(\displaystyle \:28\,\times\,12\:=\:336\)
Because the last term is negative, we want a "difference".
. . [It is waa positive, we'd want a "sum".]
We want to factor the \(\displaystyle 336\) into two parts whose difference is \(\displaystyle 5\), the middle coefficient.
How do we find it?
Well, there's a primitive method that always works.
. . Divide \(\displaystyle 336\) by \(\displaystyle 1,\,2,\,3,\,4,\,\cdots\;\;\) (keep those that "come out even")
. . until we find the pair we want.
For example: \(\displaystyle \,336\,\div\,1 \:=\:336\;\;\Rightarrow\;\;336\:=\:1\cdot336\)
. . . . . . . . . . \(\displaystyle 336\,\div\,2 \:=\:168\;\;\Rightarrow\;\;336\:=\:2\cdot168\)
. . . . . . . . . . \(\displaystyle 336\,\div\,3\:=\:112\;\;\Rightarrow\;\;336\:=\:3\cdot112\)
And we will have a list of the complete factoring of \(\displaystyle 336:\)
. . \(\displaystyle \begin{array}{ccc}1&\cdot&336 \\ 2&\cdot&168\\ 3&\cdot&112\\ 4&\cdot&84 \\ 6&\cdot&56 \\8&\cdot&42\\12&\cdot&28\\14&\cdot&24\\16&\cdot&21\end{array}\)
Remember what pair we were looking for?
. . That's right . . . a difference of \(\displaystyle 5\)
It's the very last pair: \(\displaystyle \,16,\:21\)
Use these to "split" the middle term.
The middle term is \(\displaystyle +5xy\), so we will use \(\displaystyle +21xy\) and \(\displaystyle \,-16xy\)
The polynomial becomes: \(\displaystyle \:28x^2\,+\,21xy\,-\,16xy\,-\,12y^2\)
Factor "by grouping": \(\displaystyle \:7x(4x\,+\,3y)\,-\,4y(4x\,+\,3y)\)
Take out the common factor: \(\displaystyle \:\fbox{(4x\,+\,3y)(7x\,-\,4y)}\;\;\) . . . There!
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
To complete this lesson, suppose the last term is positive.
Example: \(\displaystyle \:6a^2\,-\,17ab\,+\,10b^2\)
"First times Last": \(\displaystyle \:6\cdot10\:=\:60\)
We want to factor \(\displaystyle 60\) into two parts whose sum is \(\displaystyle 17\).
So we factor \(\displaystyle 60\):
. . \(\displaystyle \begin{array}{ccc}1&\cdot&60\\2&\cdot&30\\3&\cdot&20\\4&\cdot&15\\5&\cdot&12\\6&\cdot&10\end{array}\)
The pair with a sum of 17 is: \(\displaystyle 5\) and \(\displaystyle 12\).
The middle term is \(\displaystyle \,-17ab\), so we will use \(\displaystyle \,-5ab\) and \(\displaystyle \,-12ab\).
The polynomial becomes: \(\displaystyle \:6a^2\,-\,5ab\,-\,12ab\,+\,10b^2\)
Factor by gourping: \(\displaystyle \:a(6a\,-\,5b)\,-\,2b(6a\,-\,5b)\)
Take out the common factor: \(\displaystyle \:\fbox{(6a\,-\,5a)(a\,-\,2b)}\)
You can, of course, check these answers yourself . . . right?
\(\displaystyle \L\\28x^{2}+5xy-12y^{2}\)
One way(my favorite way) is to find two numbers when added equal the x term and when multiplied equal the y^2 term.
But, because of the 28 in the leading term you must use 28*(-12)=-336
So, what two numbers when added equal 5y and when multiplied equal -336y^2
You can solve the small system to find them:
\(\displaystyle \L\\ab=-336y^{2}\)...[1]
\(\displaystyle \L\\a+b=5y\)...[2]
Solve [2] for a:
\(\displaystyle \L\\a=5y-b\)
Sub into [1]:
\(\displaystyle \L\\(5y-b)b=-336y^{2}\)
Solve for b= 21y and -16y
Group:
\(\displaystyle \L\\(28x^{2}+21xy)-(16xy+12y^{2})\)
Factor out common terms:
\(\displaystyle \L\\7x(4x+3y)-4y(4x+3y)\)
See?. the terms in parentheses are the same. We're in business.
\(\displaystyle \L\\(7x-4y)(4x+3y)\)
Just as Skeeter said. EDIT: and Soroban
One way(my favorite way) is to find two numbers when added equal the x term and when multiplied equal the y^2 term.
But, because of the 28 in the leading term you must use 28*(-12)=-336
So, what two numbers when added equal 5y and when multiplied equal -336y^2
You can solve the small system to find them:
\(\displaystyle \L\\ab=-336y^{2}\)...[1]
\(\displaystyle \L\\a+b=5y\)...[2]
Solve [2] for a:
\(\displaystyle \L\\a=5y-b\)
Sub into [1]:
\(\displaystyle \L\\(5y-b)b=-336y^{2}\)
Solve for b= 21y and -16y
Group:
\(\displaystyle \L\\(28x^{2}+21xy)-(16xy+12y^{2})\)
Factor out common terms:
\(\displaystyle \L\\7x(4x+3y)-4y(4x+3y)\)
See?. the terms in parentheses are the same. We're in business.
\(\displaystyle \L\\(7x-4y)(4x+3y)\)
Just as Skeeter said. EDIT: and Soroban
The way these equations are "devised" is some mean teacher makes up
a multiplication; as example: (2x - y)(x + 3y).
This meanie then does the multiplication:
2x^2 + 6xy - xy - 3y^2
= 2x^2 + 5xy - 3y^2
He/she (probably she!) then asks you to factor 2x^2 + 5xy - 3y^2;
so your job is to dismantle that into the original (2x - y)(x + 3y).
A good way for you to practice these is make some up :wink:
a multiplication; as example: (2x - y)(x + 3y).
This meanie then does the multiplication:
2x^2 + 6xy - xy - 3y^2
= 2x^2 + 5xy - 3y^2
He/she (probably she!) then asks you to factor 2x^2 + 5xy - 3y^2;
so your job is to dismantle that into the original (2x - y)(x + 3y).
A good way for you to practice these is make some up :wink:
One way that may make this type of expresion easier is to get rid of the y, factor, then stick it back in.
Go ahead and factor: \(\displaystyle 28x^{2}+5x-12\)
You get \(\displaystyle (4x+3)(7x-4)\)
Now, just stick the y back in after the 3 and -4
\(\displaystyle (4x+3y)(7x-4y)\)
Go ahead and factor: \(\displaystyle 28x^{2}+5x-12\)
You get \(\displaystyle (4x+3)(7x-4)\)
Now, just stick the y back in after the 3 and -4
\(\displaystyle (4x+3y)(7x-4y)\)
