Question

Factor by grouping and solve:x3y2 - 8x3y + 16x3 + y5 - 8y4 + 16y3

Answer 1

@hartnn

Answer 2

@phi

Answer 3

try the obvious (as a first step)
it looks like you can factor x^3 out of the first 3 terms

Answer 4

so would it be x^3(y^2-8y+16+y^5-8y^4+16y^3)

Answer 5

close, you put the parens around the terms where you *took out* the x^3
you don't put the parens around everything.

Answer 6

try again. You almost have it. But you put the parens in the wrong places

Answer 7

(x^3)(y^2-8y+16+y^5-8y^4+16Y^3)

Answer 8

I guess you are missing an idea.

Here is another way to think about it.

if you have a( a +b) +c
and you distribute the "a" (multiply each term inside the parens by a)
you get
aa +ab + c

or
a^2 +ab + c
if you "undo that" (factor out the a)
you take an "a" out of a^2 and out of ab leaving a+b
we put the a out front: a(a+b) + c
notice we don't put the parens around the c
a(a+b+c) means aa +ab+ac and that is different from aa+ab+c

Answer 9

oh so x^3(y^3)-(8y)+(16)+y^5-8y^4+16y^3

Answer 10

there are 3 terms where you "took out" x^3
you put all 3 terms in one set of parens. Think like this: there is a package of terms where I took out the x^3. Put those terms in one "bag" and label the bag with x^3 on the outside to show we took out the x^3

Answer 11

in other words, you use one ( and one )

Answer 12

that you use to show the 3 terms where you took out the x^3

Answer 13

not clear ?

Answer 14

ok im so dumb x^3(y^2-8y+16)+y^5-8y^4+16y^3

Answer 15

exactly. As a way to check the answer, can you distribute the x^3 ?
what do you get ?

Answer 16

just like th original equation

Answer 17

Yes. The main idea is to see that

\[ x^3(y^2-8y+16) \]
and
\[ x^3y^2 - 8x^3y + 16x^3 \]
mean the something, and you can switch between them (factor or distribute)

that idea has to be clear in your mind.

Answer 18

*same thing (not something )

Answer 19

now back to the problem
\[ x^3(y^2-8y+16)+y^5-8y^4+16y^3 \]
do you see what you can factor out the last 3 terms ?

Answer 20

y^3

Answer 21

yes, and if you factor y^3 out of the last 3 terms, what does the whole expression become ?

Answer 22

x^3(y^2-8y+16) +y^3(y^2-8y+16)

Answer 23

looking good.
now, though (y^2-8y+16) looks complicated, it is one "thing"
and can be factored out of each term

Answer 24

2

Answer 25

4

Answer 26

we don't have to factor (y^2-8y+16) yet.

Instead, if you had this problem
\[ x^3 a + y^3 a \]
could you factor it ? what do you get ?

Answer 27

x^3 4 +y^3 4

Answer 28

I don't know how you are getting a 4.

We are working on your problem, but to make progress I want to do a simpler problem first
the simpler problem is
\[ x^3 a + y^3 a \]
what is in both terms that you can factor out ?

Answer 29

3

Answer 30

the only 3 we have in
\[ x^3 a + y^3 a \]
are the exponents (the little 3 in the upper right)
we don't factor out exponents... they are just a short-cut to write
\[ x\cdot x \cdot x \cdot a + y\cdot y \cdot y \cdot a \]

Answer 31

there is only one variable (letter) that you can factor out of each term in
\[ x\cdot x \cdot x \cdot a + y\cdot y \cdot y \cdot a \]

Answer 32

x and y

Answer 33

x is *only in the first term*
y is *only in the second term*
you need to find a variable that is in *both terms* that you can "take out"

Answer 34

a

Answer 35

yes, only "a" can be factored out. what do you get ?

Answer 36

x^3y^3(a)

Answer 37

almost.

you start with
\[ x^3 a + y^3 a \]
you keep the plus sign (you start with two terms and you still have two terms (in parens) when you are done.

Answer 38

x^3+y^3(a)

Answer 39

@phi

Answer 40

ok, but where does the ( ) go ? remember, the ( ) go around the terms where you took out the a. The ( ) show which terms used to have the a.

Answer 41

(x^3+Y^3)a

Answer 42

yes. Does that make sense ?

Answer 43

ys thansk

Answer 44

The reason we went over that is that this rule works even if "a" is complicated
for example, if a was (y^2-8y+16)
it works the same way.

can you factor
\[ x^3 (y^2-8y+16) + y^3 (y^2-8y+16) \]

Answer 45

do you see you can take out (y^2-8y+16) from both terms ?

Answer 46

yes @phi

Answer 47

what do you get ?

Answer 48

X^3+y^3

Answer 49

that is what you get after you take out (y^2-8y+16)
but you have to put that in parens.
and put (y^2-8y+16) on the outside of the parens
(otherwise you changed the problem)

Answer 50

in other words, your problem can be factored into
\[ ( x^3 + y^3) (y^2-8y+16) \]

Answer 51

y^2 -8y+16 factors into (y-4)^2 so the final answer is
\[ (x^3 + y^3) (y-4)^2 \]