Solve the following equation in the real number system.   x^3+7x^2+14x+8=0

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beckden | High School Teacher | (Level 1) Educator

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By the rational roots theorem the roots are +-p/q where p is a factor of 8 and q is a factor of one.  This gives us `+-1` , `+-2` , `+-4` , or `+-8` .  If we graph it, we can see that possible factors are -1, -2 and -4.   So we can try one of these (-1 is the easiest).

`f(-1)=(-1)^3+7(-1)^2+14(-1)+8=-1+7-14+8=0` (we could have used synthetic substitution but it does not show well in html.

This means `f(-1)=0` so `x=-1` or `x+1=0` is a factor.  Now we are left with the quadradic `x^2+6x+8` which we can easily factor into `(x+4)(x+2)` .  So our factored equation is `(x+1)(x+2)(x+4)=0` and by the zero product property either `x+1=0` , `x+2=0` or `x+4=0` .  This gives us a solution of `x=-1` , `x=-2` , or `x=-4` as our answer.

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