How do you reduce this
2007-11-04 07:39:33 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
do the multiplications first
16x4=64
12x2=24
8x3=24
do the add and subtract next
64-49=15
24-24=0
reduce the remaining parenthesis by any common multiples (14x- 21)= 7(2x-3)
the remaining product is15/7(2x-3)
2007-11-04 07:48:27 · answer #1 · answered by rack922 2 · 0⤊ 0⤋
answer 12x^2 + 14x + 12 = 18 12x^2 + 14x - 6 = 0 Dividing the equation by 2, we've 6x^2 + 7x - 3 = 0 Factoring (3x - a million)(2x + 3) = 0 fixing for the x (quadratic equations have 2 x's) 3x - a million = 0 x = a million / 3 We have been given the 1st x! Now for the 2nd 2x + 3 = 0 x = -3 / 2 So, x = a million / 3 , -3 / 2 There you go!
2016-12-15 16:25:58 · answer #2 · answered by ? 4 · 0⤊ 0⤋
Long division
8x3 - 12x2 + 14x - 21 into 16x4 + 0x3 + 0x2 + 0x - 49
2x+3 remainder 64x2 - 83x +14
2007-11-04 07:46:11 · answer #3 · answered by Anonymous · 0⤊ 0⤋
16x^4 - 49 = (4x^2-7)(4x^2+7)
8x^3 - 12x^2 + 14x - 21 = 4x^2(2x - 3) + 7(2x - 3)
= (4x^2 + 7)(2x - 3)
so (16x^4 - 49) / (8x^3 - 12x^2 + 14x - 21) =
(4x^2-7)(4x^2+7) / (4x^2 + 7)(2x - 3) =
(4x^2 - 7)/(2x - 3)
2007-11-04 07:52:45 · answer #4 · answered by PeterT 5 · 0⤊ 0⤋
(64-49) / (24-24+14x -21)
(15)/(14x-21)
2007-11-04 07:46:43 · answer #5 · answered by mikenap92 1 · 0⤊ 0⤋
(4x^2-7)(4x^2+7)
---------------------------
(2x-3)(4x^2+7)
the 4x^2+7's cancel out
4x^2-7
----------
2x-3
2007-11-04 07:52:18 · answer #6 · answered by Dave aka Spider Monkey 7 · 0⤊ 0⤋