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

I know the answer is x=5 or x=-12. but could you show me how to work it out. than you. solve this equation by completing the square.

### 6 Answers by Expert Tutors

Vivian L. | Microsoft Word/Excel/Outlook, essay composition, math; I LOVE TO TEACHMicrosoft Word/Excel/Outlook, essay comp...
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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.
Rachel W. | Hispanic Studies Major, Mathematics MinorHispanic Studies Major, Mathematics Mino...
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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.
Xin L. | Chinese-Breakthrough with XinChinese-Breakthrough with Xin
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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

Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...
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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.

Shelly J. | Excellent Maths Tutoring for academic successExcellent Maths Tutoring for academic su...
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Hi Roseland,

4x³+28x²-240x=0

4x(x²+7x-60)=0

4x(x²+12x-5x-60)=0

4x(x(x+12)-5(x+12))=0

4x(x-5)(x+12)=0

x=0  0r x=5 or x=-12
Steve S. | Tutoring in Precalculus, Trig, and Differential CalculusTutoring in Precalculus, Trig, and Diffe...
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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!