2007-09-18 10:39:29 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
expand than simplify tht. i think its a trinomial
2007-09-18 10:45:11 · update #1
12 x ³ - 4x ² - 16 x - 21x ² + 7 x + 28
12 x ³ - 25 x ² - 9 x + 28
OR
12 x ³ - 4 x ² - 16 x
""""""- 21 x ² """7 x"""28
12 x ³ - 25x ² - 9 x + 28
2007-09-18 21:35:05 · answer #1 · answered by Como 7 · 1⤊ 0⤋
Let's see, if I remember 7th grade correctly this should be the same as:
(12x^2-4x-16x)-(21x^2+7x+28)
which comes to:
-9x^2-27x-28
First you multiply the first number, 4x by all the numbers in the next bracket and then you do the same for second number, 7. And condense the formula from there.
I wish I could give you a better explanation, but I'm not to good with explaining math. Your probably just best asking a teacher.
I'm probably wrong. Go with the calc lady, that one brought some skool back to memory. I think I forgot about the squared numbers become cubed in certain instances.
2007-09-18 17:52:55 · answer #2 · answered by Anonymous · 0⤊ 1⤋
In this problem you are multiplying the binomial(the first parentheses) by the trinomial (the second parentheses)
1. distribute the 4x to all 3 terms in the trinomial
2.and the -7 to all three terms also
3. simplify
1) 4x(3x^2-x-4) =12x^3-4x^2-16x
2) -7(3x^2-x-4) =-21x^2+7x+28
3) (12x^3-4x^2-16x)+(-21x^2+7x+28)
**add like terms**
12x^3-25x^2-9x+28
2007-09-18 17:49:26 · answer #3 · answered by CaroR 5 · 0⤊ 0⤋
Use the distributive property, as in 2(3+6) = 2*3 + 2*6. For multiples it would be (x+y)(4+5)= x(4+5)+y(4+5) =x*4+x*5+y*4+y*5 = 9x + 9y.
So for your problem it would be 4x times all the terms in the second set of parenthesis, and -7 times all the terms in the second set of parenthesis, and add the like terms (x^3, x^2, x, constants).
2007-09-18 17:48:21 · answer #4 · answered by rannug311 2 · 0⤊ 1⤋
Simple! first u do the (4x-7) whatever x is u do 4 times x then u minus that from 7 now it's time for the (3x^2-x-4) first u do 3 times whatever x is that ^ sign i don't kinw what the **** that is so................. ask someone else!! so i guess it's not simple!
2007-09-18 17:59:11 · answer #5 · answered by da_man_iz_bac 2 · 0⤊ 1⤋
(4x-7)(3x^2-x-4)
=4x(3x^2 - x - 4) - 7(3x^2 - x -4)
=12x^3 - 4x^2 - 16x - 21x^2 + 7x +28
=12x^3 - 25x^2 - 9x +28.
2007-09-18 18:11:10 · answer #6 · answered by Twiggy 7 · 0⤊ 0⤋
you can only mutiply like terms,
(4x-7)(3x^2-x-4) mutiply the 4x and -x then the -7 and -4
3x^2-4x+28
2007-09-18 17:47:51 · answer #7 · answered by Anonymous · 0⤊ 1⤋
12x cubed -25 x squared -9x +28
2007-09-18 17:45:24 · answer #8 · answered by Azalian 5 · 0⤊ 2⤋
Well, what are you asking? You can do a lot with that equation...
2007-09-18 17:44:32 · answer #9 · answered by Anonymous · 0⤊ 2⤋