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Algebra Level 3

The numbers $a,b,c$ and $d$ are the solutions for the equation
$(x-5)(x-7)(x+6)(x+4)=504$ find the value of $abcd$

1 solution

Angel Daniel Lopez Flores
Jan 5, 2015

$(x-5)(x-7)(x+6)(x+4)=504$

$(x^2-x-20)(x^2-x-42)=504$ let $x^2-x=u$ so $u^2-62u+336=0$ factorising $(u-6)(u-56)=0$ then $(x^2-x-6)(x^2-x-56)=0$ $(x^2-x-6)=0=(x-3)(x+2)$

$(x^2-x-56)=0=(x-8)(x+7)$

The values for $x$ are $3,-2,8,-7$ And the value of $abcd$ = $\boxed{336}$

Nice solution. Another method would be to use Vieta's theorem, where the constant in the expansion would be $abcd = (-5)(-7)(6)(4) - 504 = 336$ .

Brian Charlesworth - 6 years, 5 months ago

I like the kitten

Jon Haussmann - 1 year, 7 months ago