MOST FACTORED FORM
The most factored form is the most factored version of a rational expression.
Being able to find the most factored form is an essential skill when simplifying
the derivatives found using product rule or quotient rule.
The most factored form is usually accomplished by common factoring the
expression. But, any type of factoring may come into play.
How to Find the Most Factored Form
Set-Up: It’s easier to factor a rational expression if the coefficient is a fraction
out front and if all functions are moved to the middle. Also, change all roots
into rational exponent form. For example:
3 𝑥# + 1
5 𝑥−3 #
3 #
𝑥 +1
5
(/#
*#
𝑥−3
Functions in Common: As with common factoring, any functions that are in
common between terms should be taken out using the lowest exponent. For
example:
2𝑥 𝑥 + 3
è The GCF would be 𝑥 + 3
*#
−3 𝑥+3
*(
*#
Coefficients in Common: For numerators, take out the greatest common
factor between each numerator. For the denominators, take out the lowest
common denominator. For example:
4
𝑥 𝑥+3
3
*#
+
8
𝑥+3
9
*(
è The coefficient of the GCF would be: 4/9.
EXAMPLE: Simplify the following to its most factored form:
a)
4𝑥
3 𝑥+3
#
+
8
9(𝑥 + 3)
è First, set up the function:
=
4
𝑥 𝑥+3
3
*#
+
8
𝑥+3
9
*(
Then, common factor:
=
4
𝑥+3
9
*#
[3𝑥 + 2(𝑥 + 3)]
Factoring
Most Factored Form
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Then, simplify within the bracket:
=
4
𝑥+3
9
*#
(5𝑥 + 6)
Finally, express with positive exponents:
=
b)
𝑥+2
−
4(5𝑥 + 6)
9 𝑥+3 #
𝑥
𝑥
𝑥+2
è First, set up the function:
= 𝑥 *(/# 𝑥 + 2
(/#
− 𝑥 (/# 𝑥 + 2
*(/#
Then, common factor:
= 𝑥 *(/# 𝑥 + 2
*(/#
[ 𝑥 + 2 − 𝑥]
Then, simplify within the bracket:
= 2𝑥 *(/# 𝑥 + 2
*(/#
Finally, express with positive exponents:
=
2
𝑥(𝑥 + 2)
9
+ 𝑥# 𝑥 + 1
2
è First, set up the function:
c)
18𝑥 𝑥 + 1 2𝑥 − 3
= 18𝑥 𝑥 + 1
(/#
2𝑥 − 3
*(
*(
9
+ 𝑥# 𝑥 + 1
2
*(/#
*(/#
*(
2𝑥 − 3
*(
2𝑥 − 3
− 18𝑥 # 𝑥 + 1 2𝑥 − 3
− 18𝑥 # 𝑥 + 1
(/#
2𝑥 − 3
*#
*#
Then, common factor:
=
9
𝑥 𝑥+1
2
*(/#
2𝑥 − 3
*#
4(𝑥 + 1)(2𝑥 − 3) + 𝑥(2𝑥 − 3) − 4𝑥(𝑥 + 1) Then, simplify and factor within the bracket:
=
9
𝑥 𝑥+1
2
*
(
#
2𝑥 − 3
*#
6𝑥 # − 11𝑥 − 12 Finally, express with positive exponents:
=
9𝑥 6𝑥 # − 11𝑥 − 12
2 𝑥 + 1 2𝑥 − 3
#
Factoring
Most Factored Form
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PRODUCT RULE
Suppose that the function you must differentiate can be written as the
product of two functions, 𝑓(𝑥) and 𝑔(𝑥):
𝑦 =𝑓 𝑥 ∙𝑔 𝑥
Let 𝑓 ) 𝑥 and 𝑔′(𝑥) be the derivative of 𝑓(𝑥) and 𝑔(𝑥). The derivative of 𝑦 can
be found by taking:
“The derivative of the first function times the second PLUS the derivative
of the second function times the first.”
In another way:
𝑦 ) = 𝑓 ) 𝑥 ∙ 𝑔 𝑥 + 𝑔′(𝑥) ∙ 𝑓(𝑥)
𝑦 =𝑓 𝑥 ∙𝑔 𝑥
EXAMPLE: Find the derivative of the function:
𝑦 = 3𝑥 . + 2𝑥 − 1 𝑥 . + 5𝑥 + 4
SOLUTION: We can see that this can be expressed as the product of two
functions:
𝑓 𝑥 = 3𝑥 . + 2𝑥 − 1 and 𝑔 𝑥 = 𝑥 . + 5𝑥 + 4
The derivative of each function individually is:
𝑓 ) 𝑥 = 6𝑥 + 2 = 2(3𝑥 + 1) and 𝑔′ 𝑥 = 2𝑥 + 5
So the derivative of the original function 𝑦 is:
𝑦 ) = 𝑓 ) 𝑥 ∙ 𝑔 𝑥 + 𝑔) 𝑥 ∙ 𝑓 𝑥
𝑦′ = 2 3𝑥 + 1 𝑥 . + 5𝑥 + 4 + 2𝑥 + 5 3𝑥 . + 2𝑥 − 1
Now factor to achieve the most factored form:
𝑦 ) = 2 3𝑥 + 1 (𝑥 + 1)(𝑥 + 4) + 2𝑥 + 5 3𝑥 − 1 𝑥 + 1
𝑦 ) = 𝑥 + 1 2 3𝑥 + 1 𝑥 + 4 + 2𝑥 + 5 3𝑥 − 1
𝑦 ) = (𝑥 + 1)(6𝑥 . + 39𝑥 + 13)
Avoiding Product Rule
Product rule is a longer and more difficult process that can sometimes be
avoided in order to make less mistakes. If the original polynomial function is
easy to expand, do that first, then find the derivative. For example:
𝑓 𝑥 = 3𝑥 − 1 2𝑥 + 1 = 6𝑥 . + 𝑥 − 1
𝑓 ) 𝑥 = 12𝑥 + 1
Differentiation
Product Rule
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QUOTIENT RULE
Suppose that the function you must differentiate can be written as the
quotient of two functions, 𝑓(𝑥) and 𝑔(𝑥):
𝑦=
𝑓 𝑥
𝑔 𝑥
Let 𝑓 ) 𝑥 and 𝑔′(𝑥) be the derivative of 𝑓(𝑥) and 𝑔(𝑥). The derivative of 𝑦 can
be found by taking:
“The derivative of the top function times the bottom MINUS the
derivative of the bottom function times the top ALL OVER the bottom
squared.”
In another way:
𝑓 ) 𝑥 ∙ 𝑔 𝑥 − 𝑔′(𝑥) ∙ 𝑓(𝑥)
𝑦 =
𝑔 𝑥 .
𝑓 𝑥
𝑦=
𝑔 𝑥
)
EXAMPLE: Find the derivative of the function:
𝑦=
2𝑥 − 1
𝑥. + 1
SOLUTION: We can see that this can be expressed as the quotient of two
functions:
𝑓 𝑥 = 2𝑥 − 1 and 𝑔 𝑥 = 𝑥 . + 1
The derivative of each function individually is:
𝑓 ) 𝑥 = 2 and 𝑔′ 𝑥 = 2𝑥
So the derivative of the original function 𝑦 is:
𝑦) =
𝑓 ) 𝑥 ∙ 𝑔 𝑥 − 𝑔) 𝑥 ∙ 𝑓 𝑥
𝑔 𝑥 .
𝑦′ =
2 𝑥 . + 1 − 2𝑥(2𝑥 − 1)
𝑥. + 1 .
We can’t factor the numerator so we’ll expand and simplify:
2𝑥 . + 2 − 4𝑥 . + 2𝑥 −2𝑥 . + 2𝑥 + 2
𝑦 =
=
𝑥. + 1 .
𝑥. + 1 .
)
𝑦′ =
−2(𝑥 . − 𝑥 − 1)
𝑥. + 1 .
Differentiation
Quotient Rule
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Avoiding Quotient Rule
Quotient rule is a longer and more difficult process that can sometimes be
avoided in order to make less mistakes. If the denominator is a monomial,
split the denominator over each term in the numerator. For example:
𝑓 𝑥 =
𝑥 . − 2𝑥 + 𝑥 𝑥 . 2𝑥
𝑥
= . − . + . = 1 − 2𝑥 67 + 𝑥 68/. 𝑥.
𝑥
𝑥
𝑥
Now, we can use power rule:
3
𝑓 ) 𝑥 = 2𝑥 6. − 𝑥 6:/. 2
Simplify to its most factored form:
1 6:/.
𝑥
2𝑥 7/. − 3
2
2 𝑥−3
𝑓) 𝑥 =
2 𝑥:
𝑓) 𝑥 =
Differentiation
Product Rule
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