The first step is always to “pull out” as much as you can...
Multiply the following, using the distributive property:
3x(4x2+5x+2)=3x(4x2+5x+2)= size 12{3x \( 4x rSup { size 8{2} } +5x+2 \) ={}} {} ____________________
Now, you’re going to do the same thing backward.
- a. “Pull out” the common term of 4y4y size 12{4y} {} from the following expression.
-
16
y
3
+
4y
+
8y
=
4x
(
_________
)
16
y
3
+
4y
+
8y
=
4x
(
_________
)
size 12{"16"y rSup { size 8{3} } +4y+8y=4x \( "_________" \) } {}
- b. Check yourself, by multiplying
4y4y size 12{4y} {} by the term you put in parentheses.
- c. Did it work? _______________
For each of the following expressions, pull out the highest common factor you can find.
9xy+12x=9xy+12x= size 12{9 ital "xy"+"12"x={}} {}____________________
10x2+9y2=10x2+9y2= size 12{"10"x rSup { size 8{2} } +9y rSup { size 8{2} } ={}} {}____________________
100x3+25x2=100x3+25x2= size 12{"100"x rSup { size 8{3} } +"25"x rSup { size 8{2} } ={}} {}____________________
4x2y+3y2x=4x2y+3y2x= size 12{4x rSup { size 8{2} } y+3y rSup { size 8{2} } x={}} {}____________________
Next, look to apply our three formulae...
Factor the following by using our three formulae for
(x+y)2(x+y)2 size 12{ \( x+y \) rSup { size 8{2} } } {},
(x−y)2(x−y)2 size 12{ \( x - y \) rSup { size 8{2} } } {}, and
x2−y2x2−y2 size 12{x rSup { size 8{2} } - y rSup { size 8{2} } } {}.
x2−9=x2−9= size 12{x rSup { size 8{2} } - 9={}} {}______________
x2−10x+25=x2−10x+25= size 12{x rSup { size 8{2} } - "10"x+"25"={}} {}______________
x2+8x+16=x2+8x+16= size 12{x rSup { size 8{2} } +8x+"16"={}} {}______________
x2+9=x2+9= size 12{x rSup { size 8{2} } +9={}} {}______________
3x2−27=3x2−27= size 12{3x rSup { size 8{2} } - "27"={}} {}______________
Start by pulling out the common factor!
If all else fails, factor the "old-fashioned way"...
- a: x2+7x+10=x2+7x+10= size 12{x rSup { size 8{2} } +7x+"10"={}} {}______________
- b: Check your answer by multiplying back:
- a: x2−5x+6=x2−5x+6= size 12{x rSup { size 8{2} } - 5x+6={}} {}______________
- b: Check your answer by plugging a number into the original expression, and into your modified expression:
x2−6x+5=x2−6x+5= size 12{x rSup { size 8{2} } - 6x+5={}} {}______________
x2+8x+6=x2+8x+6= size 12{x rSup { size 8{2} } +8x+6={}} {}______________
x2−x−12=x2−x−12= size 12{x rSup { size 8{2} } - x - "12"={}} {}______________
x2+x−12=x2+x−12= size 12{x rSup { size 8{2} } +x - "12"={}} {}______________
x2+4x−12=x2+4x−12= size 12{x rSup { size 8{2} } +4x - "12"={}} {}______________
2x2+7x+12=2x2+7x+12= size 12{2x rSup { size 8{2} } +7x+"12"={}} {}______________
"This is the "main" book in Kenny Felder's "Advanced Algebra II" series. This text was created with a focus on 'doing' and 'understanding' algebra concepts rather than simply hearing about them in […]"