Someone remind me...
take the 12 to the other side
- |x-5| ≤ -3
divide by -1, and when you do that you have to invert the ≤
3 ≤ |x - 5|
now find the critical values
3 = x - 5
and
3 = -(x-5)
solve equations to find critical values, then set up your number line and solve
- |x-5| ≤ -3
divide by -1, and when you do that you have to invert the ≤
3 ≤ |x - 5|
now find the critical values
3 = x - 5
and
3 = -(x-5)
solve equations to find critical values, then set up your number line and solve
im trying to help with homework, yet no idea anymore
factor
12x ^4 – 17x ^3 + 6x ^2
i remember having to multiply the outer ones together, and taking the common multiples or something..
factor
12x ^4 – 17x ^3 + 6x ^2
i remember having to multiply the outer ones together, and taking the common multiples or something..
x^2 (12x^2 - 17x - 6)
ok, it factors to
(ax + b)(cx - d)
because the first - in 12x^2 - 17x - 6 means ax*d + b*cx must be negative, so obviously there must be one -... but they cant both be - because b * d must be - from the - 6... so one has to be positive and one must be negative
and reading that over you probably arent following at all
but im going with it anyways because im on a roll
so then its gotta be either
(4x + 2)(3x - 3) or
(4x + 3)(3x - 2) or
(4x + 1)(3x - 3) or
(4x + 6)(3x - 2) or
(3x + 2)(4x - 3) or
(3x + 3)(4x - 2) or
(3x + 1)(4x - 6) or
(3x + 6)(4x - 1) or
(12x + 2)(x - 3) or
(12x + 3)(x - 2) or
(12x + 1)(x - 6) or
(12x + 6)(x - 1) or
(x + 2)(12x - 3) or
(x + 3)(12x - 2) or
(x + 1)(12x - 6) or
(x + 6)(12x - 1) or
(6x + 2)(2x - 3) or
(6x + 3)(2x - 2) or
(6x + 1)(2x - 6) or
(6x + 6)(2x - 1) or
(2x + 2)(6x - 3) or
(2x + 3)(6x - 2) or
(2x + 1)(6x - 6) or
(2x + 6)(6x - 1)
because of the 12, and the x coefficients are multiples of 12
and because of the 6, and the a and d numbers are multiples of 6
wait, none of the work... fawk!
ok, it factors to
(ax + b)(cx - d)
because the first - in 12x^2 - 17x - 6 means ax*d + b*cx must be negative, so obviously there must be one -... but they cant both be - because b * d must be - from the - 6... so one has to be positive and one must be negative
and reading that over you probably arent following at all
but im going with it anyways because im on a roll
so then its gotta be either
(4x + 2)(3x - 3) or
(4x + 3)(3x - 2) or
(4x + 1)(3x - 3) or
(4x + 6)(3x - 2) or
(3x + 2)(4x - 3) or
(3x + 3)(4x - 2) or
(3x + 1)(4x - 6) or
(3x + 6)(4x - 1) or
(12x + 2)(x - 3) or
(12x + 3)(x - 2) or
(12x + 1)(x - 6) or
(12x + 6)(x - 1) or
(x + 2)(12x - 3) or
(x + 3)(12x - 2) or
(x + 1)(12x - 6) or
(x + 6)(12x - 1) or
(6x + 2)(2x - 3) or
(6x + 3)(2x - 2) or
(6x + 1)(2x - 6) or
(6x + 6)(2x - 1) or
(2x + 2)(6x - 3) or
(2x + 3)(6x - 2) or
(2x + 1)(6x - 6) or
(2x + 6)(6x - 1)
because of the 12, and the x coefficients are multiples of 12
and because of the 6, and the a and d numbers are multiples of 6
wait, none of the work... fawk!



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