How do I find the zeros of the equation given that one of the zeros is x = 2 ?
x^3 - 7x + 6 = 0

Question

How do I find the zeros of the equation given that one of the zeros is x = 2 ?
x^3 - 7x + 6 = 0

anonymous · Answer

Since one of the zeroes is 2, you know that you can factor the given function so that it looks like this:
$$\text{If }p(x)=x^3-7x+6, \text{ then }p(x)=(x-2)q(x).$$

anonymous · Answer

To find $q(x)$, try dividing $p(x)$ by $x-2$, either by long division or synthetic division.

anonymous · Answer

q(x) is quotient?
synthetic division gives me
$$x^3 + 2x^2 -3x$$

anonymous · Answer

that means my equation is
$$p(x) = (x-2)(x^3+2x^2-3x)$$
or am I completely off?

anonymous · Answer

Your result from synthetic division is a bit off, yeah.

The way you have $p(x)$, if you were to distribute the $x-2$, you'd get another polynomial of degree higher than the given polynomial, which can't be true.

anonymous · Answer

Your only mistake there was that you should have gotten $q(x)=x^2+2x-3$.

So, you have $x^3-7x+6=(x=2)(x^2+2x-3)$.

anonymous · Answer

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where's my mistake? (I don't want to make it again)