^{3}+28X

^{2}-240X=0

^{3}+7x

^{2}-60x)=0

^{3}+7x

^{2}-60x=0

^{2}+7x-60)=0

^{2}+7x-60=0

^{2}

^{2}

^{2}-5x+12x-60=0

^{2}+7x-60=0

**x=-12**

**x=5**

**THESE ARE YOUR RESULTS.**

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.

Tutors, sign in to answer this question.

Hi Roseland;

4X^{3}+28X^{2}-240X=0

Let's factor-out the 4...

4(x^{3}+7x^{2}-60x)=0

We can eliminate the 4 as it is the parenthetical equation which must equal zero.

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

Let's factor-out the x...

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

x=0 is one solution.

x^{2}+7x-60=0

is another solution which will generate two results.

For the FOIL...

FIRST must be (x)(x)=x^{2}

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)=x^{2}

OUTER...(-5)(x)=-5x

INNER...(12)(x)=12x

LAST...(12)(-5)=-60

x^{2}-5x+12x-60=0

x^{2}+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**

Rachel W. | Hispanic Studies Major, Mathematics MinorHispanic Studies Major, Mathematics Mino...

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^{2 }+ 7x - 60)=0

Since the product is equal to 0, either 4x = 0 or x^{2 }+ 7x - 60= 0.

If 4x = 0, then we divide by 4 on both sides and obtain x = 0.

If x^{2 }+ 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

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

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...

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

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!

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