SOLUTION: Once again, I am not sure that I got the topic correct.
Here is the question: 9(2x+1)^3+6(2x+1)^2-15(2x+1)
the answer is: 12(2x+1)(3x+4)
I get the whole answer except the latte
Algebra.Com
Question 367621: Once again, I am not sure that I got the topic correct.
Here is the question: 9(2x+1)^3+6(2x+1)^2-15(2x+1)
the answer is: 12(2x+1)(3x+4)
I get the whole answer except the latter 4! I am just not sure!
Found 2 solutions by Theo, mananth:
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the problem is:
9 * (2x + 1)^3 + 6 * (2x+1)^2 - 15 * (2x+1)
(2x+1)^2 is equal to 4x^2 + 4x + 1
(2x+1)^3 is equal to 8x^3 + 12x^2 + 6x + 1
9 * (2x+1)^3 is equal to 72x^3 + 108x^2 + 54x + 9
6 * (2x+1)^2 is equal to 24x^2 + 24x + 6
15 * (2x + 1) is equal to 30x + 15
your expression becomes:
72x^3 + 108x^2 + 54x + 9 + 24x^2 + 24x + 6 - (30x + 15)
remove parentheses to get:
72x^3 + 108x^2 + 54x + 9 + 24x^2 + 24x + 6 - 30x - 15
bring your like terms together to get:
72x^3 + 108x^2 + 24x^2 + 54x + 24x - 30x + 9 + 6 - 15
combine like terms to get:
72x^3 + 132x^2 + 48x
this is your original answer after you have multiplied all factors together.
you can factor out an x to get:
x * (72x^2 + 132x + 48)
you can factor out a 12 to get:
12 * x * (6x^2 + 11x + 4)
6x^2 + 11x + 4 factors out to be (2x+1) * (3x + 4) because when you multiply those factors together, you get 6x^2 + 11x + 4.
your solution is therefore equal to:
12 * x * (2x + 1) * (3x + 4)
if you multiply all of those together, you will get back to your original equation of 72x^3 + 132x^2 + 48x which was derived by multiplying all the original factors together.
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
9(2x+1)^3+6(2x+1)^2-15(2x+1)
...
(2x+1)(9(2x+1)^2+6(2x+1)-15)
..
(2x+1)(9(2x+1)^2+9(2x+1)-3(2x+1)-15))
let (2x+1)=a
a(9a^2+6a-15)
3a(3a^2+2a-5)
3a(3a^2+5a-3a-5)
3a((a(3a+5)-1(3a+5))
3a(3a+5)(a-1)
resubstitute the value of a
3a(3(2x+1)+5)(2x+1-1)
3(2x+1)(6x+3+5)(2x)
6x(2x+1)(6x+8)
12x(2x+1)(3x+4)
...
m.ananth@hotmail.ca
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