# How do you find the number of possible positive real zeros and negative zeros then determine the rational zeros given f(x)=x^3-7x-6?

Oct 23, 2017

Rational zeros of $f \left(x\right)$ are $x = - 1 , x = - 2 \mathmr{and} x = 3$

#### Explanation:

$f \left(x\right) = {x}^{3} - 7 x - 6$ . Here constant number is $- 6$ and leading

Coefficient is $1$ . Factors of $6$ are $1 , 2 , 3 , 6 ,$ and factors of $1$

Are $1 \therefore$ Possible rational zeros are $\pm \frac{1 , 2 , 3 , 6}{1} \therefore$

Possible rational zeros are $\pm \left(1 , 2 , 3 , 6\right)$ On checking,

#f (1)= -12 , f(-1) =0 , f(2) = -12, f(-2)=0 , f(3)=0 :.

Zeros are $x = - 1 , x = - 2 \mathmr{and} x = 3$ hence

$\left(x + 1\right) , \left(x + 2\right) \mathmr{and} \left(x - 3\right)$ are factors.

$\therefore f \left(x\right) = {x}^{3} - 7 x - 6 = \left(x + 1\right) \left(x + 2\right) \left(x - 3\right)$

Rational zeros of $f \left(x\right)$ are $x = - 1 , x = - 2 \mathmr{and} x = 3$ [Ans]