Factor out the trinomial and factor out all common nomomial first 21x^3 - 26x^2 + 8x, I think the answer is -x(7x + 4)(3x -4) please help
2007-05-26 04:36:07 · 7 answers · asked by delphiniums 1 in Science & Mathematics ➔ Mathematics
the answer is
x*(3x-2)(7x-4)
2007-05-26 04:42:14 · answer #1 · answered by Mamad 3 · 0⤊ 1⤋
Common factor first
21x³ - 26x² + 8x
x(21x² - 26x + 8)
The middle term inside the parenthesis is - 26x
Find the sum of the middle term
multiply the first tern 21 times the last term 8 equals 168 and factor.
factors of 168 are
1 x 168
2 x 84
3 x 56
4 x 46
6 x 28
7 x 24
8 x 21
12 x 14. . . <=. . use these factors
- 14 and - 12 satisfy the sum of the middle term
inserty - 14x and - 12x into the equation
x(21x² - 26x + 8)
Group factor
x(21x² - 14x - 12x + 8)
x[7x(3x - 2) - 4(3x - 2)]
x(7x - 4)(3x - 2)
- - - - - - - - - - -s-
2007-05-26 12:10:26 · answer #2 · answered by SAMUEL D 7 · 0⤊ 0⤋
21x^3 - 26x^2 + 8x can be factored as:
x (21x^2 - 26x + 8)
21 can be factored as 7 and 3. 26 can be obtained as 14 - 12. 8 can be factored as 4 and 2.
If we take 7x - 4 and 3x - 2 = 21x^2 -14x -12x + 8 which is 21x^2 - 26x + 8 and is OK.
So, we write the factors as:
x (7x - 4) (3x - 2)
The factors chosen by you are not correct for the following reason:
-x(7x + 4)(3x -4) = -x ( 21x^2 - 28x +12x -16) = -x (21x^2 - 16x -16) which works out to - 21x^3 +16x^2 + 16x and that is not the expression originally given.
2007-05-26 11:51:48 · answer #3 · answered by Swamy 7 · 0⤊ 0⤋
21x^3 - 26x^2 + 8x =
x(21x^2 - 26x + 8) =
x(21x^2 - 12x - 14x + 8) =
x[ (21x^2 - 12x) + (- 14x + 8) ] =
x[ 3x(7x - 4) - 2(7x - 4) ]
x(3x -2)(7x - 4)
2007-05-26 11:56:55 · answer #4 · answered by Ana 4 · 0⤊ 0⤋
21x^3 - 26x^2 + 8x=x(21x^2-26x+8)
=x(21x^2-14x-12x+8)
x(7x3x-2)-4(3x-2)
=x(3x-2)(7x-4).answer
2007-05-26 11:51:28 · answer #5 · answered by Anonymous · 0⤊ 0⤋
= x.(21x² - 26x + 8)
= x.(7x - 4).(3x - 2)
2007-05-27 02:44:37 · answer #6 · answered by Como 7 · 0⤊ 0⤋
x(3x-2)(7x-4)
2007-05-26 11:45:41 · answer #7 · answered by Anonymous · 0⤊ 0⤋