Question

`(x+1) (x+3) (x+6) (x+4) = 72..` Find x?

Answer 1

`x=0`

Explanation:

The given problem
`(x+1)(x+3)(x+6)(x+4)=72`

you can use FOIL to expand the problem into the multiplication of two polynomials

`<=>`

`(x^2+4x+3)(x^2+10x+24)=72`

`<=>`Further simplification

`x^4+10x^3+24x^2+4x^3+10x^2+96x+3x^2+30x+72=72`

There are a lot of terms here, and one would be tempted to combine like terms to simplify further... but there is only one term which does not include `x` and that term is `72`

`therefore x=0`

Answer 2

`:. x=0, x=-7, x=(-7+-isqrt23)/2.`

Explanation:

`(x+1)(x+3)(x+6)(x+4)=72.`

`:. {(x+1)(x+6)}{(x+3)(x+4)}=72.`

`:. (x^2+7x+6)(x^2+7x+12)=72.`

`:. (y+6)(y+12)=72,.........[y=x^2+7x].`

`:. y^2+18y+72-72=0, i.e., y^2+18y=0.`

`:. y(y+18)=0.`

`:. y=0, or, y+18=0.`

`:. x^2+7x=0, or, x^2+7x+18=0.`

`:. x=0, or, x=-7, or, x=[-7+-sqrt{7^2-4(1)(18)}]/(2*1),`

`:. x=0, x=-7, x=(-7+-isqrt23)/2.`

Answer 3

`x_1=-7` and `x_2=0`. From first one, them are `x_3=(7+sqrt(23)*i)/2` and `x_4=(7-sqrt(23)*i)/2` .

Explanation:

I used difference of squares identity.

`(x+1)* (x+6)* (x+3)* (x+4)=72`

`(x^2+7x+6)* (x^2+7x+12)=72`

`(x^2+7x+9)^2-3^2=72`

`(x^2+7x+9)^2=81`

`(x^2+7x+9)^2-9^2=0`

`(x^2+7x+9+9)* (x^2+7x+9-9)=0`

`(x^2+7x+18)* (x^2+7x)=0`

`(x^2+7x+18)* x*(x+7)=0`

From second and third multiplier, roots of equations are `x_1=-7` and `x_2=0`. From first one, them are `x_3=(7+sqrt(23)*i)/2` and `x_4=(7-sqrt(23)*i)/2` .