I need to see how these are done I am not understanding the factoring process for solving equations so please show me step by step so I can apply this to other problems in the future.
2007-06-12 00:59:34 · 9 answers · asked by Mothra 1 in Science & Mathematics ➔ Mathematics
When x=0 the equation is true therefore 0 is a solution.
Then suppose there are other solutions that is not = 0, since you can divide by a nonzero go ahead and divide
(5x^3 - 320 x )/x= 0/x
5x^2 - 320 = 0 //0 divided by anything = 0
5x^2 = 320 //add 320 to each side
x^2 = 64 //divide each side by 5
x = + - 8 //take square root of each side and note that root 64 = +8 & -8
therefore your solutions are 0, 8, -8
Key point always try to divide out an x but ony if it is non zero so you will have to treat the case of x=0 seperatly as above.
HTH PeterVincent
2007-06-12 01:11:18 · answer #1 · answered by PeterVincent 2 · 0⤊ 0⤋
The equation: 5x^3 - 320x = 0
Dividing by x throughout,
x [ 5x^2 - 320 ] = 0
So x = 0, or 5x^2 - 320 = 0
For 5x^2 - 320 = 0, rearranging gives
x^2 = 320/5 = 64
x = sqrt[64] = 8 or -8
Therefore x = 0, x = 8, or x = -8
P.S. It's a cubic equation, so it has three solutions.
2007-06-12 01:05:54 · answer #2 · answered by darrenfoong1 2 · 0⤊ 0⤋
take 5*x common from both 5*x^3 and -320*x
=> 5*x(x^2 - 64) = 0
the product of two things is zero hence
either 5*x = 0 = > x = 0
or
x^2 - 64 = 0
=>
(x+8)(x-8) = 0 [ from the formula (a^2 - b^2 = (a + b)(a - b)) ]
here 8^2 = 64
again the multiplication of both things
=> x + 8 =0 => x = -8
or x - 8 = 0 => x = 8
therefore the possible solutions for x are
x = -8,0,8
remember that the number of solutins for any equation with all powers with whole numbers is the highest power of the variable
here the variable is x
2007-06-12 01:08:59 · answer #3 · answered by vicky 2 · 0⤊ 0⤋
start up by factoring out an x x(x^2+5x-6)=0 then element (x^2+5x-6) to (x+6)(x-a million) so x(x+6)(x-a million)=0 set each and every term =0 x=0 x+6=0 x-a million=0 then remedy for x x=0 x=-6 x=a million and those final 3 numbers are your answer!
2016-10-17 00:20:00 · answer #4 · answered by ja 4 · 0⤊ 0⤋
5x^3 - 320 x = 0 ?
5x^3= 320
^3 = 320\5 = 64
^3 = 64
2007-06-12 01:32:19 · answer #5 · answered by Anonymous · 0⤊ 0⤋
first take out x
x(5x^2-320)=0
The note that 320/5=8^2
so
5x(x^2-64)=5x(x+8)(x-8)=0
There is no sure fire technique and most equations can't be factored. You just try some guesses.
2007-06-12 01:12:27 · answer #6 · answered by meg 7 · 0⤊ 0⤋
5x.(x² - 64) = 0
5x = 0 , x² = 64
x = 0, x = ± 8
2007-06-12 04:23:42 · answer #7 · answered by Como 7 · 0⤊ 0⤋
we have:5x^3 - 320x = 0
so
5x ( x^2 - 64) = 0
so x can have the values of :
x=0
x=8
x=-8
2007-06-12 01:05:42 · answer #8 · answered by s_jmp 2 · 0⤊ 0⤋
5x(x^2-64)=0
5x(x-8)(x+8)=0
x=0;8;-8
2007-06-12 01:36:11 · answer #9 · answered by Dave aka Spider Monkey 7 · 0⤊ 0⤋