Factor completely: x^5y^3-9x^4y^2+20x^3y

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flora33

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We are factoring using the GCF, and in the following problem, I would like to make sure I am on the right track:

x^5y^3-9x^4y^2+20x^3y
GCF: x^3y
x^3y (x^2y^2 - 9xy + 20)
Answer: (x^3y) (x^2y^2 - 9xy + 20)
 
flora33 said:
x^5y^3-9x^4y^2+20x^3y
x^3y (x^2y^2 - 9xy + 20)
Answer: (x^3y) (x^2y^2 - 9xy + 20)
Click to expand...
Now factor the polynomial inside the second parentheses, and you'll be done! :D

Eliz.
 
Re:

stapel said:
Now factor the polynomial inside the second parentheses, and you'll be done!
Click to expand...
(x^3y) (x^2y^2 - 9xy + 20)
(x^3y){xy(xy - 9 + 20)}
Answer: (x^3y)(xy)(xy + 11) Is this correct? I am not positive on this answer...
 
flora33 said:
(x^3y) (x^2y^2 - 9xy + 20)
(x^3y){xy(xy - 9 + 20)}
Answer: (x^3y)(xy)(xy + 11) Is this correct? I am not positive on this answer...
Click to expand...
Since "20" has no "xy" to "factor out", I'm not seeing how you got that x[sup:3cyxdo6y]2[/sup:3cyxdo6y]y[sup:3cyxdo6y]2[/sup:3cyxdo6y] - 9xy + 20 equalled xy(xy + 11)...? (Since -9 + 20 = 11, this must be your final simplification. But -9xy + 20 does not equal 11xy.)

I'm guessing you've never factored quadratics before...? (They're sometimes also called "trinomials".) To learn how to do this, try studying from some of the many great lessons available online:

. . . . .Google results for "factoring quadratics"

. . . . .Google results for "factoring trinomials"

Have fun! :D

Eliz.
 
That's a pretty strange factoring exercise for someone who hasn't had much experience.

In the last step, you might want to try this. Let a = xy.

Then, x[sup:2seeyis9]3[/sup:2seeyis9]y ( x[sup:2seeyis9]2[/sup:2seeyis9]y[sup:2seeyis9]2[/sup:2seeyis9] - 9xy + 20 becomes x[sup:2seeyis9]3[/sup:2seeyis9]y (a[sup:2seeyis9]2[/sup:2seeyis9] - 9a + 20)

That looks a bit easier to factor, doesn't it.

That turns into: x[sup:2seeyis9]3[/sup:2seeyis9]y ( a - 4)(a - 5)

Now substitute xy back in for a and you have your final answer: x[sup:2seeyis9]3[/sup:2seeyis9]y (xy - 4)(xy - 5)

Dale
 
masters said:
In the last step, you might want to try this. Let a = xy.

Then, x[sup:zk6cmfnm]3[/sup:zk6cmfnm]y ( x[sup:zk6cmfnm]2[/sup:zk6cmfnm]y[sup:zk6cmfnm]2[/sup:zk6cmfnm] - 9xy + 20 becomes x[sup:zk6cmfnm]3[/sup:zk6cmfnm]y (a[sup:zk6cmfnm]2[/sup:zk6cmfnm] - 9a + 20)

That looks a bit easier to factor, doesn't it.

That turns into: x[sup:zk6cmfnm]3[/sup:zk6cmfnm]y ( a - 4)(a - 5)

Now substitute xy back in for a and you have your final answer: x[sup:zk6cmfnm]3[/sup:zk6cmfnm]y (xy - 4)(xy - 5)
Click to expand...
Well, I worked with trinomials for one week prior to this, but they did not involve GCF. It was just polynomials and functions, and multiplying polynomials. This is also an online class, so I'm basically teaching myself because my teacher is mostly absentee... Anyway, thanks for the tip, that really does seem easier. I appreciate it!

While I'm at it, I have another problem:
8x^2 - 32x + 24
8(x^2 - 4x + 3)
(8)(x - 1)(x - 3) Am I doing this part right? I have troubles with factoring "completely". This is the part where you find the combination of numbers that are multiplied to equal the last constant and added to equal the second constant, right?


Flora
 
flora33 said:
This is also an online class, so I'm basically teaching myself because my teacher is mostly absentee... Anyway, thanks for the tip, that really does seem easier. I appreciate it!

While I'm at it, I have another problem:
8x^2 - 32x + 24
8(x^2 - 4x + 3)
(8)(x - 1)(x - 3) Am I doing this part right? I have troubles with factoring "completely". This is the part where you find the combination of numbers that are multiplied to equal the last constant and added to equal the second constant, right?
Click to expand...
You did this one just right, Flora....

AND...I would NEVER recommend an online class for anyone who doesn't feel very confident about dealing with the subject material in the course. You do NOT get individual help in an online course. I would strongly recommend that you find a course in a community college or technical school in your area where you can have the benefit of a REAL teacher in a REAL classroom, as well as the opportunity to work with other students in your class in study groups.

If you do not have the opportunity to enroll in a community college or technical school in your area, then I'd recommend that you PAY for face-to-face tutoring.
 
Mrspi said:
You did this one just right, Flora....

AND...I would NEVER recommend an online class for anyone who doesn't feel very confident about dealing with the subject material in the course. You do NOT get individual help in an online course. I would strongly recommend that you find a course in a community college or technical school in your area where you can have the benefit of a REAL teacher in a REAL classroom, as well as the opportunity to work with other students in your class in study groups.

If you do not have the opportunity to enroll in a community college or technical school in your area, then I'd recommend that you PAY for face-to-face tutoring.
Click to expand...
Thank you for confirming. Yes, I was sooo apprehensive about taking this algebra class online. I remember having such a struggle when I was a freshman in high school, which was 10 years ago so I knew this would be tough. At this point, I'm almost finished. I have 4 weeks of a 10 week course left. I have a baby who is 7 months old, so I can't take classes locally- plus my husband works for the government, so we are moving in a few weeks. I've had a tough time with this class, but overall, it hasn't been as bad as I'd imagined it would be. Lucikly, I have a lot of time during the day to study and find resources on line. We also have access to tutors with my college, but it is only minimal hours during the week.

I'm just hoping I can continue with my understanding of what we are doing each week and maintain my 90(ish) average for a grade!
When I move to Lexington and I take my next math courses, I will probably be in the market for a REAL tutor then! It definitely seems worth the money! This stuff has me pulling my hair out sometimes but it is so fulfilling when I figure something out because I basically did it on my own! Well, and of course with the help of all you wonderful math whizzes here! :wink:

Thanks again,
Flora
 
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