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4X3+28X2-240X=0 BY completing the square

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6 Answers

Hi Roseland;
 
4X3+28X2-240X=0
Let's factor-out the 4...
4(x3+7x2-60x)=0
We can eliminate the 4 as it is the parenthetical equation which must equal zero.
x3+7x2-60x=0
Let's factor-out the x...
x(x2+7x-60)=0
x=0 is one solution.
x2+7x-60=0
is another solution which will generate two results.
For the FOIL...
FIRST must be (x)(x)=x2
OUTER and INNER must add-up to 7x.
LAST must be (60)(1) or (1)(60) or (2)(30) or (30)(2) or (4)(15) or (15)(4) or (5)(12) or (12)(5) or (6)(10) or (10)(6) and only one number must be negative to render the sum of +7 and product of -60...
(x+12)(x-5)=0
Let's FOIL...
FIRST...(x)(x)=x2
OUTER...(-5)(x)=-5x
INNER...(12)(x)=12x
LAST...(12)(-5)=-60
x2-5x+12x-60=0
x2+7x-60=0
(x+12)(x-5)=0
Either or both parenthetical equation(s) must equal zero...
x+12-0, x=-12
x-5=0, x=5
THESE ARE YOUR RESULTS.
First you must look for a greatest common factor. In this case, each term has 4x so we begin by factoring that out to get:
 
4x(x+ 7x - 60)=0
 
Since the product is equal to 0, either 4x = 0 or x+ 7x - 60= 0. 
 
If 4x = 0, then we divide by 4 on both sides and obtain x = 0. 
 
If x+ 7x -60=0 then we can factor this to obtain (x + 12)(x-5) = 0. 
 
Then we know that either x = 0 (as described above), or x + 12 = 0, or x - 5=0. So we ultimately obtain the answer: x=0, or x=-12, or x=5. 

4x^3+28x^2-240x=0 x(4x^2+28x-240)=0 4x(x^2+7x-60)=0 use the formula below: a=1 b=7 c=-60 x=(-b±v(b^2-4ac))/2a so we put the numbers in formula: -7±v(49-4x1x(-60)) -7±v(49+4x1x60) then you sovle the eqution and the answers are x=5 x=-12 x=0

 4 X^3 + 28 X^2 - 240 X = 0
 
    4X ( X^2 + 7X - 60 ) = 0                               This quadratic is factorable , ( X+ a ) ( X + b )
                                                                            Where ab = -60   a + b = 7
     Break 7x = 12X- 5X                                           a = 12  b = -5  is the answer.
 
    4x ( X^2 + -5X + 12X - 60 ) =0
   
      4X[( X ( X - 5 ) + 12 ( X - 5 ) ] = 0
  
       Factoring by grouping
      
         4X ( X -5) ( 12 + X ) =0
 
         X = 0    (X -5)= 0   X = 5      (X  +12) =0    X = - 12
 
         Completing Square:
   
            4x ( X^2 + 7X - 60)
           
            4X ( X^2 + 2 . 7X/2  + 49/ 4 - 49/4 - 60 ) = 0
 
             4X ( (  X + 7/2 ) ^2 - [(2940- 60)/4 ] = 0
 
             X = 0
 
                     ( X + 7/2 )  ^2 - 289/ 4 =0
 
 
                     X + 7/ 2 = ± √(289/4) = ±17 /2
 
                        X = -7/ 2 + 17/ 2   X = 10/2 = 5         X = -7/2 - 17/2 = -24/ 2 = -12
 
                   Quadratic formula derives from completing square of aX^2 + bX + c.    
 
 

0 = 4x^3 + 28x^2 - 240x

0 = x^3 + 7x^2 - 60x

0 = x(x^2 + 7x - 60)

(x + k)^2 = x^2 + 2kx + k^2
So we could replace the pattern on the right with the one on the left.

0 = x(x^2 + 2(7/2)x + (7/2)^2 - (7/2)^2 - 60)

0 = x( (x + 7/2)^2 - (7/2)^2 - 60)

0 = x( (x + 7/2)^2 - 49/4 - 240/4)

0 = x( (x + 7/2)^2 - 289/4)

Using Zero Product Property:

x = 0 or (x + 7/2)^2 - 289/4 = 0

(x + 7/2)^2 = 289/4

x + 7/2 = ±17/2

x = -7/2 ± 17/2 = —12, 5

So solutions are x = -12, 0, 5

check:

4(-12)^3 + 28(-12)^2 - 240(-12) =? 0
= -6912 + 4032 + 2880 = 0 check!

4(0)^3 + 28(0)^2 - 240(0) = 0 check!

4(5)^3 + 28(5)^2 - 240(5) =? 0
= 500 + 700 - 1200 = 0 check!

Woodbridge completing the square tutors