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#1 2024-05-20 06:46:22

mathxyz
Member
From: Brooklyn, NY
Registered: 2024-02-24
Posts: 1,053

Factor Completely

Factor Completely.


Problem 108; page 58.


9y^2 + 9y - 4


I think factor by grouping works here.


9(-4) = -36


12(-3) = -36 and 12 + (-3) = 9.


9y^2 + 12y - 3y - 4


9y^2 + 12y = group A


-3y - 4 = group B


Factor group A


GCF = 3y


3y(3y + 4)


Factor group B


GCF = -1


-1(3y + 4)


I now have 3y(3y + 4) - 1(3y + 4)


My answer: (3y - 1)(3y + 4).


You say?

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#2 2024-05-20 07:38:36

Bob
Administrator
Registered: 2010-06-20
Posts: 10,627

Re: Factor Completely

Correct.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#3 2024-05-20 09:47:28

mathxyz
Member
From: Brooklyn, NY
Registered: 2024-02-24
Posts: 1,053

Re: Factor Completely

Bob wrote:

Correct.

Bob

Very good. I have decided to include precalculus to my self-study journey. College Algebra and Precalculus have more or less the same topics except for trigonometry, polar equations, math induction and a few others. If I go through two textbooks ONE AT A TIME, I will never reach Calculus 1.

I can start Calculus with you tomorrow and do well but I also think that studying College Algebra and Precalculus SIMULTANEOUSLY will increase my math skills and provide the confidence I need to UNDERSTAND Calculus not just memorize a bunch of formulas.

You say?

P. S. I will give you the name of my precalculus textbook for you to freely download when I get home. By the way, I passed precalculus with an A minus at Lehman College in the Spring 1993 semester. So, precalculus is also a revisit to material learned long ago when I was young.

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