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

Jul 31, 2017

$x = 0$

#### Explanation:

The given problem
$\left(x + 1\right) \left(x + 3\right) \left(x + 6\right) \left(x + 4\right) = 72$

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

$\iff$

$\left({x}^{2} + 4 x + 3\right) \left({x}^{2} + 10 x + 24\right) = 72$

$\iff$Further simplification

${x}^{4} + 10 {x}^{3} + 24 {x}^{2} + 4 {x}^{3} + 10 {x}^{2} + 96 x + 3 {x}^{2} + 30 x + 72 = 72$

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 $72$

$\therefore x = 0$

Jul 31, 2017

$\therefore x = 0 , x = - 7 , x = \frac{- 7 \pm i \sqrt{23}}{2.}$

#### Explanation:

$\left(x + 1\right) \left(x + 3\right) \left(x + 6\right) \left(x + 4\right) = 72.$

$\therefore \left\{\left(x + 1\right) \left(x + 6\right)\right\} \left\{\left(x + 3\right) \left(x + 4\right)\right\} = 72.$

$\therefore \left({x}^{2} + 7 x + 6\right) \left({x}^{2} + 7 x + 12\right) = 72.$

$\therefore \left(y + 6\right) \left(y + 12\right) = 72 , \ldots \ldots \ldots \left[y = {x}^{2} + 7 x\right] .$

$\therefore {y}^{2} + 18 y + 72 - 72 = 0 , i . e . , {y}^{2} + 18 y = 0.$

$\therefore y \left(y + 18\right) = 0.$

$\therefore y = 0 , \mathmr{and} , y + 18 = 0.$

$\therefore {x}^{2} + 7 x = 0 , \mathmr{and} , {x}^{2} + 7 x + 18 = 0.$

$\therefore x = 0 , \mathmr{and} , x = - 7 , \mathmr{and} , x = \frac{- 7 \pm \sqrt{{7}^{2} - 4 \left(1\right) \left(18\right)}}{2 \cdot 1} ,$

$\therefore x = 0 , x = - 7 , x = \frac{- 7 \pm i \sqrt{23}}{2.}$

Jul 31, 2017

${x}_{1} = - 7$ and ${x}_{2} = 0$. From first one, them are ${x}_{3} = \frac{7 + \sqrt{23} \cdot i}{2}$ and ${x}_{4} = \frac{7 - \sqrt{23} \cdot i}{2}$ .

#### Explanation:

I used difference of squares identity.

$\left(x + 1\right) \cdot \left(x + 6\right) \cdot \left(x + 3\right) \cdot \left(x + 4\right) = 72$

$\left({x}^{2} + 7 x + 6\right) \cdot \left({x}^{2} + 7 x + 12\right) = 72$

${\left({x}^{2} + 7 x + 9\right)}^{2} - {3}^{2} = 72$

${\left({x}^{2} + 7 x + 9\right)}^{2} = 81$

${\left({x}^{2} + 7 x + 9\right)}^{2} - {9}^{2} = 0$

$\left({x}^{2} + 7 x + 9 + 9\right) \cdot \left({x}^{2} + 7 x + 9 - 9\right) = 0$

$\left({x}^{2} + 7 x + 18\right) \cdot \left({x}^{2} + 7 x\right) = 0$

$\left({x}^{2} + 7 x + 18\right) \cdot x \cdot \left(x + 7\right) = 0$

From second and third multiplier, roots of equations are ${x}_{1} = - 7$ and ${x}_{2} = 0$. From first one, them are ${x}_{3} = \frac{7 + \sqrt{23} \cdot i}{2}$ and ${x}_{4} = \frac{7 - \sqrt{23} \cdot i}{2}$ .