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

3 Answers
256
Jul 31, 2017

#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#

Ratnaker Mehta
Jul 31, 2017

#:. 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.#

Cem Sentin · MathFact-orials.blogspot.com
Jul 31, 2017

#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# .

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