20x^2 + 31x + 12
and
-45x^3 + 42x^2 - 9x
please help!
2007-11-28 08:52:38 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
The first one is (5x + 4)(4x + 3).
For the second one, notice that each term includes an x, so you can factor that out straight away. In fact, you may as well factor out the minus sign too, to get:
-x(45x^2 - 42x +9)
Now factorise the bracket to get:
-x(15x - 9)(3x -1).
Notice that the middle factor has a common factor of 3. Take this out to get:
-3x(5x - 3)(3x-1)
2007-11-28 08:56:22 · answer #1 · answered by _asv_ 3 · 0⤊ 0⤋
20x² + 31x + 12 = 0
The middle term is + 31x
Find the sum of the middle term
Multiply the first term 20 times the last term 12 equals 240 and factor
factors of 240
1 x 240
2 x 120
3 x 80
4 x 60
5 x 48
6 x 40
8 x 30
10 x 24
12 x 20
15 x 16. . .←. .Use these factors
+ 16 and + 15 satisfy the sum of the middle term
insert + 16x and + 15x into the equation
20x² + 31x + 12 = 0
20x² + 16x + 15x + 12 = 0
Group factor
(20x² + 16x) + (15x + 12) = 0
4x(5x + 4) + 3(5x + 4) = 0
(4x + 3)(5x + 4) = 0
- - - - - - - - s-
2007-11-28 09:27:26 · answer #2 · answered by SAMUEL D 7 · 0⤊ 0⤋
Hi,
20x^2 + 31x + 12
(4x+3)(5x+4)
This is not an equation, so I assume you only wanted the factors. When you do one like this, you might try factoring the smallest number first.
and
-45x^3 + 42x^2 - 9x
-3x(15x²-14x+3)
-3x(5x-3)(3x-1)
FE
2007-11-28 09:04:58 · answer #3 · answered by formeng 6 · 0⤊ 0⤋
Look for factors of 20(12) that add to 31: 15 & 16, and group:
20x^2 + 15x + 16x + 12
5x(4x + 3) + 4(4x + 3)
(4x + 3)(5x + 4)
Yes!
-45x^3 + 42x^2 -9x
= -3x(15x^2 - 14x + 3)
= -3x(5x - 3)(3x - 1)
that's it!
2007-11-28 09:00:16 · answer #4 · answered by Marley K 7 · 0⤊ 0⤋
Find 2 numbers that added give you 31 & multiplied give you 240 (20x12).
You do this by finding out the prime factors of 240.
240 = 2x2x2x2x3x5 = 16 & 15
You perform the following: (31x = 16x + 15x)
20x^2 + 16x + 15x + 12
4x (5x + 4) + 3 (5x + 4)...regrouping
(5x + 4) (4x + 3)...& you're done
Try the second one...
2007-11-28 09:04:03 · answer #5 · answered by achain 5 · 0⤊ 0⤋
20x^2 + 31x + 12
20x^2 + 15x + 16x + 12
5x(4x + 3) + 4(4x + 3)
(5x + 4)(4x + 3)
-45x^3 + 42x^2 - 9x
-3x(15x^2 - 14x + 3)
-3x(15x^2 - 9x - 5x + 3)
-3x[3x(5x - 3) - 1(5x - 3)
-3x(3x - 1)(5x - 3)
2007-11-28 09:10:28 · answer #6 · answered by Anonymous · 0⤊ 0⤋
(5x+4)(4x+3)
-3x(5x+1)(3x-3)
2007-11-28 09:26:45 · answer #7 · answered by itsok2bphat 2 · 0⤊ 0⤋
(5x+4)(4x+3)
-3x(5x+1)(3x-3)
2007-11-28 08:57:44 · answer #8 · answered by Ms. Exxclusive 5 · 0⤊ 0⤋