# Solve for x in each of the following, x E C. (x+1)(x+5)(x+3) = -3the answer is x = -2, -7+- sqare root of 13 divided by 2

*print*Print*list*Cite

### 3 Answers

(x+1)(x+5)(x+3)= -3

First let us open brackets:

==> (x+1)*(x^2 + 8x + 15) = -3

==> (x^3 + 8x^2 + 15x + x^2 + 8x + 15 = -3

==> x^3 + 9x^2 + 23x + 18 = 0

Now let us factor"

==> (x+2)(x^2 + 7x + 9) = 0

==> **x1= -2**

**==> x2= [-7+sqrt(49-36)]/2 = (-7+ sqrt13)/2**

**==> x3= (-7-sqrt13)/2**

There is another method of solving the problem.

We notice that the expression is a product of 3 factors:

(x+1)(x+5)(x+3)

Let's consider -3 as a result of a product of 3 factors, too.

-3 = -1*1*3

Now, let's re-write the expression:

(x+1)(x+5)(x+3) = 1*(-1)*3

We'll put the factor (x+1) = 1

x+1 = 1

We'll subtract 1 both sides:

x = 0

x+5 = -1

We'll subtract 5 both sides, in order to isolate x:

x = -6

x+3 = 3

x = 0

If, we'll substitute the values into the expression (x+1)(x+5)(x+3) = -3, they verify it.

Another way of solving, would be to remove the brackets:

We'll calculate the product of the first and the second factor:

(x+1)(x+5) = x^2 + 5x + x + 5 = x^2 + 6x + 5

Now, we'll multiply the result by (x+3):

(x^2 + 6x + 5)(x+3) = x^3 + 3x^2 + 6x^2 + 18x + 5x + 15

We'll combine like terms:

x^3 + 9x^2 + 23x + 15 + 3 = 0

x^3 + 9x^2 + 23x + 18 = 0

Since -2 is a solution, we'll divide x^3 + 9x^2 + 23x + 18 by (x+2) and the reminedr will be 0.

x^3 + 9x^2 + 23x + 18 = (x+2)(ax^2 + bx + c)

We'll remove the brackets from the right side:

x^3 + 9x^2 + 23x + 18 = ax^3 + bx^2 + cx + 2ax^2 + 2bx + 2c

We'll combine like terms from the right side and we'll factorize:

x^3 + 9x^2 + 23x + 18 = ax^3 + x^2(b+2a) + x(c+2b) + 2c

Now, the polynomial from the left side is identically with the polynomial from the right side, if and only if the coefficients are equal:

a = 1

b+2a = 9 => b + 2 = 9 => b = 7

c + 2b = 23 => c + 14 = 23 => c = 9

The polynomial ax^2 + bx + c = x^2 + 7x + 9

Now, we'll determine the roots of the quadratic:

x^2 + 7x + 9 = 0

x1 = [-7+sqrt(49-36)]/2

**x1 = (-7+sqrt13)/2**

**x2 = (-7-sqrt13)/2**

**So, the roots of (x+1)(x+5)(x+3) = -3 are {-2 ; (-7+sqrt13)/2 ; (-7-sqrt13)/2}.**

(x+1)(x+5)(x+3) = -3

Put x = -2 , then (-2+1)(-2+5)(-2+3) = (-1)(3)(1) = RHS = -3.

Expanding the left and rewriting, we get:

x^3+(1+5+3)x +(1*5+5*3+3*1)x^2+1*5*3 = -3

x^3+9x+ 23x^2+15+3 = 0.........(1)

x^3+9x^2+24x+18 = 0.

As we know x=-2 should be satisfying the equation at (1):

So (-2)^3+9(-2)^2+23(-2)+18 = -8+36-46+18 = 28-28 = 0.

Therfore (x+2) is a factor of x^2+9x^2+23x+18 .

Therefore x^3 +9x^2+23x^2+18 = (x+2){x^2+kx+(18/2)}

x^3+9x^2+23x+18 = (x+2)(x^2+kx+9). Should be an identity.

So the each coefficients of x^3 , x^2,x a should be same on both sides.

x^3 and constant terms are already agreeing.

So coefficient os x's are equal on both sides. So,

23 = 2k+9.

Thefore k = (23-9)/2 = 14/2 = 7.

Therefore the other factor of x^3+9x^2+23x+18 is x^2+7x+9 = 0

So (x^3+23x^2+23x+18)= 0 implies

(x+2)(x^2+7x+9) = 0.

So x+2 = 0 gives x = -2 as already known.

x^2+7x+9= 0 is a quadatic equation like ax^2+bx+c = 0 whose roots are x1 = {-b+sqrt(b^2-4ac)}/2 Or x2 = {-b-sqrt(b^2-4ac)}/2

Here in x^2 +7x+9 =0, a= 1, b=7 and c = 9.

So the roots are :

x1 = {-7+sqrt(7^2-4*1*9)}/2 = {-7+sqrt(14-36)

x1 = {-7+sqrt(49-36)}/2 = {3.5 + 0.5sqrt(13) }

x2 = {-3.5 -0.5sqrt(13)}