How do you find the zeros of $x^3 - 7x + 6 = 0$ ?

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Answer 1 · 2016-06-14T06:51:28

$x^3-7x+6=(x-1)(x-2)(x+3)$

We use the property that if $alpha$ is zero of $f(x)$ i.e. $f(alpha)=0$ , then $(x-alpha)$ is a factor of $f(x)$ .

As here $f(x)=x^3-7x+6$ and $f(1)=1^3-7*1+6=1-7+6=0$

$(x-1)$ is a factor of $x^3-7x+6$ .

Now dividing $x^3-7x+6$ by $(x-1)$ , we get

$x^2+x-6$ , whose discriminant is $1^2-4*1*(-6)=25=5^2$ , hence factors can be found by splitting middle term. Hence,

$x^2+x-6=x^2+3x-2x-6=x(x+3)-2(x+3)=(x-2)(x+3)$

Hence, $x^3-7x+6=(x-1)(x-2)(x+3)$

How do you find the zeros of x^3 - 7x + 6 = 0 x^3 - 7x + 6 = 0?