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

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

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Answer 1 · 2016-06-14T06:51:28

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

Explanation:

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

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

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

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

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

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

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

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

1 Answer

Explanation:

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Science

Math

Humanities

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How do you find the zeros of x3−7x+6=0 x^3 - 7x + 6 = 0?

1 Answer

Explanation:

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