f(x) = 2x^3 – x^2 – 7x + 6 How many real zeros are there?
Try to find a zero by finding out where the function changes sign.
x | 0 | 1 |
f(x)| 6 | 0 |
From above we see that there is a zero at x = 1.
This means that (x – 1) is a factor of the cubic expression 2x^3 – x^2 – 7x + 6.
Taking out the factor gives us,
(x – 1)(2x^2 + x – 6) = 0
The quadratic expression now factorises.
(x – 1)(2x – 3)(x + 2) = 0
The solutions then are,
Three real zeros: x = 1, x = 3/2, x = -2