Question

Can someone please walk me through this problem?
Factor by grouping, if possible , and check
10x^3 – 25x^2 +2x – 5

Answer 1

5x^2(2x-5)+(2x-5)
(2x-5)(5x^2+1)

Answer 2

How did you reach that?

Answer 3

By taking 5x^2 common from first two terms

Answer 4

Sorry for all the questions but I'm really having a hard time with this. How did you get that 5x^2 is the common factor

Answer 5

jst look into the equation for a moment you will get it by practising more and more

Answer 6

Would you add together the ^2 and the ^3 to get the 5 Sorry I'm confused. I've read the book like 3 times and looked at examples but it is just making the confusion worse

Answer 7

no you cant simply add the powers
Sorry man bt you should revise the basics again before solving these problems you have not got it truely

Answer 8

I can review them all day long and it will do me no good because I can't retain new information at all but I can't drop this class because I don't have the money to pay for it again

Answer 9

http://www.teacherschoice.com.au/sample_help_1_alg.htm

Answer 10

EvelynW: Do you know this rule?\[x^a*x^b=x^{a+b}\]

Answer 11

It looks familiar but not ringing any bells for me. I'm trying to figure it out without feeling like an even bigger idiot than I already do

Answer 12

don't worry - I'm here to help, so feel free to ask any questions where you are not sure of something

Answer 13

so just to refresh your memory:\[x^a=x*x*x...\qquad\text{('a' times)}\]and:\[x^b=x*x*x...\qquad\text{('b' times)}\]

Answer 14

therefore:
\(x^a*x^b=\) 'a' lots of x's times 'b' lots of x's = 'a+b' lots of x's
therefore:\[x^a*x^b=x^{a+b}\]does that make sense?

Answer 15

sure i guess

Answer 16

maybe an example will help:\[x^3=x*x*x\]
and:\[x^2=x*x\]therefore:\[(x^3)*(x^2)=(x*x*x)*(x*x)=x*x*x*x*x=x^5=x^{3+2}\]

Answer 17

ok

Answer 18

ok, so now lets look at your question...

Answer 19

you are given:\[10x^3 – 25x^2 +2x – 5\]now lets take one term at a time. the first term is:\[10x^3\]

Answer 20

now 10 can be written as 5*2, so we can rewrite the first term as:\[5(2x^3)\]

Answer 21

Ok, would I do that for each term or for 2 terms at a time

Answer 22

lets take it one step at a time...

Answer 23

next, \(x^3=x*x*x=(x*x)*x=x^2*x\), therefore we can rewrite this further as:\[10x^3=5(2x^3)=5(2x^2*x)=5x^2(2x)\]

Answer 24

following along so far?

Answer 25

Yeah, I think so. What happens to the exponent 3

Answer 26

remember I showed:\[x^3=x*x*x=(x*x)*x=x^2*x\]

Answer 27

so we split the \(x^3\) into a product of \(x^2\) and \(x\)

Answer 28

and then pulled the \(x^2\) out of the bracket

Answer 29

Ok, I see it now. Sorry I gotta have people point it out to me sometimes

Answer 30

no problems - we all have to start somewhere - as I said - don't worry - I've been there :)

Answer 31

ok, so to summarise the steps so far, we had your original equation as:\[10x^3 – 25x^2 +2x – 5\]and we re-wrote the first term as:\[10x^3=5x^2(2x)\]ready for the next step?

Answer 32

Yeah, I understand the first step :)

Answer 33

now lets take the second term as rewrite it using similar steps as follows:\[-25x^2=-(5*5)x^2=5(-5x^2)=5x^2(-5)\]

Answer 34

Ok, I get it

Answer 35

so now we use these re-written terms in the original equation to get:\[10x^3 – 25x^2 +2x – 5=5x^2(2x)+5x^2(-5)+2x-5\]

Answer 36

now can you see that the first two terms have a common factor of \(5x^2\)?

Answer 37

Yeah, I see that now

Answer 38

ok, so we take this common factor out to get:\[10x^3 – 25x^2 +2x – 5=5x^2(2x)+5x^2(-5)+2x-5\]\[\qquad=5x^2(2x-5)+2x-5\]now you should spot there is another common factor here: \(2x-5\)

Answer 39

Yeah, it's -5

Answer 40

no - not quite, let me re-write it slightly differently to help you...

Answer 41

Oh wait, it would be 2x right? The first thing I saw was -5 on both sides

Answer 42

\[10x^3 – 25x^2 +2x – 5=5x^2(2x)+5x^2(-5)+2x-5\]\[\qquad=5x^2(2x-5)+2x-5\]\[\qquad=5x^2(2x-5)+(2x-5)\]\[\qquad=(2x-5)*5x^2+(2x-5)*1\]\[\qquad=(2x-5)(5x^2-1)\]

Answer 43

sorry - I made a mistake in the last line

Answer 44

it should be:\[\qquad=(2x-5)(5x^2+1)\]

Answer 45

So I take the 5x^2 from the first part of it and put it in the location of the second 2x and put the 1 in place of the 5

Answer 46

think of it this way:\[a*b+a*c=a(b+c)\]in our case above \(a=(2x-5)\) and \(b=5x^2\) and \(c=1\)