I NEED HELP
2007-10-17 09:07:13 · 8 answers · asked by longboarder870 2 in Science & Mathematics ➔ Mathematics
Using the distributive property, you would break it down as such (also called the FOIL method - First Outer Inner Last - where you mulitply the first term in each binomial, then the outer terms (first term in one, last term in other), then the inner terms (last term in one, first in other) and finally the last term in each binomial:
(2x+1)(3x+4) = (6x2+8x+3x+4) = 6x2+11x+4 | where 6x2 is 6x squared
having simplified the first two binomial factors into a trinomial, now you distribute the last binomial
(x+3)(6x2+11x+4) = (6x3+11x2+4x+18x2+33x+12) | where 6x3 is 6x cubed
combine like terms
(6x3+29x2+37x+12) is the answer
2007-10-17 09:22:10 · answer #1 · answered by ExpertGamer 3 · 0⤊ 1⤋
Always just take them separately, and start with the low coefficients: (3x+4)(x+3) = 3x^2+9x+4x+12 = 3x^2+13x+12.
(3x^2+13x+12)(2x+1)=6x^3+26x^2+24x
3x^2+13x+12
=6x^3+29x^2+37x+12
The little trick of putting one set of multiplication through and then putting the next below it to align like exponents is helpful.
2007-10-17 16:15:11 · answer #2 · answered by JP 3 · 0⤊ 1⤋
(2x+1)(3x+4)(x+3)
2x² +x +6x +3 * (3x+4) <-- expand the 1st and 3rd brackets
2x² +7x +3 * (3x+4) <-- add like terms
6x³ + 21x² +9x + 8x² +28x +12 <-- expand the "2nd" bracket
6x³ + 29x² +37x +12 <-- add like terms
ANS: 6x³ + 29x² +37x +12
2007-10-17 16:19:41 · answer #3 · answered by David F 5 · 1⤊ 0⤋
(2x + 1)(3x² + 13x + 12)
6x³ + 26x² + 24x + 3x² + 13x + 12
6x³ + 29x² + 37x + 12
2007-10-18 05:08:21 · answer #4 · answered by Como 7 · 0⤊ 0⤋
(2x+1)[(3x+4)(x+3)]
(2x+1)[3x^2+9x+4x+12]
(2x+1)[3x^2+13x+12]
=>6x^3+26x^2+24x+3x^2+13x+12
==>6x^3+29x^2+37x+12
I am sure that mine is rite... others have made small mistakes in adding the terms.. plz chek
2007-10-17 16:15:57 · answer #5 · answered by freeky 2 · 0⤊ 1⤋
What's the problem? Do you need to rewrite this in expanded form?
6x³ + 29x² + 45x + 12
2007-10-17 16:14:39 · answer #6 · answered by Dave 6 · 1⤊ 0⤋
well u have to make it this
2x+1(times)3x+4(times)x+3
then
2x(times)3x(times)x+1+4+3
5x+8 is your final answer
2007-10-17 16:16:11 · answer #7 · answered by ♥Marisa♥ 2 · 0⤊ 1⤋
(2x+1)(3x+4)(x+3)
(6x^2+8x+3x+4)(x+3)
6x^3+8x^2+3x^2+4x+18x^2+24x+9x+12
6x^3+29x^2+36x+12
2007-10-17 16:16:27 · answer #8 · answered by Runa 7 · 0⤊ 0⤋