Solving an equation...?

Question

How do you solve something like..

(x)(x + 5)(x - 2) = 30

Answer 1

Factor by grouping

x(x^2-2x+5x-10)-30=0

x^3 - 2x^2 + 5x^2 - 10x - 30 = 0

x^3 + 3x^2 - 10x - 30 = 0

x^2(x + 3) + 10(x + 3) = 0

(x^2 + 10)(x + 3) = 0

Does that help? Can you figure it out from here?

Answer 2

First, you must expand the equation (complete the multiplication).

x^3 + 3x^2 - 10x = 30

Then, set the polynomial equal to zero

x^3 + 3x^2 -10x - 30 = 0

Next, using either long division of polynomials or synthetic division, find a binomial factor of your equation.  This is a strategy you should have learned.

The resulting factors will be

(x + 3)(x^2 - 10) = 0

(You can check this by multiplying.)

Then, using the zero-product principle,

x = -3 and x = +/-sqrt 10

Check these answers in the original equation.

Answer 3

okay, here we go, solving a cubic equation:

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1.possible way of solving it:

x³ + 3x² -10x - 30 = 0

(x + 3)(x² - 10) = 0

=> x1 = -3, x2,3 = +,- 3,16227766

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2.way of solving it, is the more formal one, I was taught back at the uni ;-) But I would not teach it to my students now, for this would confuse them too much, I guess. But anyway, here we go:

(x)(x + 5)(x - 2) = 30

(x² + 5x) (x - 2) = 30

x³ - 2x² + 5x² - 10x = 30

x³ + 3x² - 10x - 30 = 0

(=> ax³ + bx² + cx + d = 0)

now we calculate p and q:

p = (3ac - b²)/(3a²)

q = (2b³)/(27a³) - (bc)/(3a²) + (d)/(a)

hence using y³ + py + q = 0

with p = -13 and q = -38

=> y³ -13y - 38 = 0

next step: calculating the discriminant:

D = (q/2)² + (p/3)³

D = 279 17/27 => D > 0

u = [(3rd root of) -q/2 + {(square root of) D}]

v = [(3rd root of) -q/2 - {(square root of) D}]

u = 3,293409864

v = 1,315758898

now let's get to the three solutions:

y1 = u + v

-> y1= 4,609168762

y2,3 = - (u+v)/2 +,- (u-v)/(square root of)3i

-> y2 = -0,591888404

-> y3 = -4,0172803858

(*phew* typing this is really taking ages...)

And not FINALLY, let's substitute is back to find x1,2,3:

x1,2,3 = y1,2,3 - (b/3a)

...don't want to sound arrogant, but I'll leave it up to you, to do the last step ;-) Should take you about 20 seconds to get all three solutions!

(Btw - sorry, if I mistyped something, I did not proof-read it now...)

Hope that was helping.

Cheers,

Aleks

  

Answer 4

(x)(x + 5)(x - 2) = 30

(x^2 + 5x)(x - 2) = 30

x^3 - 2x^2 + 5x^2 - 10x = 30

x^3 + 3x^2 -10x -30 = 0

x^2(x + 3) - 10(x + 3) = 0

(x^2 -10)(x +3) = 0

x + 3 =0 or x^2 - 10 = 0

so x= -3 or x= +sqrt(10) or x= -sqrt(10)