Find zeros of f(x) = (x^3) - 7x + 6?

Answer 1

beacuse this is a cubic equation this has got 3 zeros.
it has got atleast one real root. may be 3
if root is rational it is factor of 6, +/-1 +/-2, +/-3

f(1) = 1-7+6=0
so f(x-1) is a root

x^3-7x +6 = BY SYNTHETIC DIVISION
x^2(x-1) + x^2 - 7x +6
= x^(x-1) + (x-1)(x-6)
= (x-1)(x^2+x -6)
the quadratic can be factored sum is +1 produbct -6 so 2 -3

= (x-1)(x-2)(x+3)

zeros are 1 2 -3

Answer 2

I think in this question you are supposed to spot the pattern (1, 0, -7, 6) of the numbers - called the "coefficients" of the terms - and realise that they could add to zero. Specifically, when you set x = 1, f(x) evaluates to zero, therefore (x - 1) is an exact factor. If the pattern of numbers was wrong for this, you would need to try other values of x.

Synthetic division of x^3 - 7x + 6 by (x - 1) gives (x^2 + x - 6) whose zeroes are easier to find.

Answer 3

There's probably a better way to do this, but I did it by guess-and-check and division.

By guess-and-check, I found that f(2)=0. This means that (x-2) is a factor, so we can divide by that. The result of division is (x²+2x-3). This can be factored into (x-1)(x+3), with the final factored equation being (x-2)(x-1)(x+3)=0. Therefore, the roots (or zeroes) of f(x) are 2, 1, and -3.

Answer 4

You may be familiar with the quadratic equation?
Well there is an equation for a cubic equation. i dont know it because i'm about to learn it. Are you taking pre-calc or something? The only way you can find zeroes is by solving for x, but i don't know to do that. I'm sorry.