Question

Degree and real roots of (x-3)(x+1)(x-1)=0

Answer 1

Explanation:

at first multiply the factors

`({x}^(2 ) + x \: - 3x \: - 3)(x - 1) = 0`

`{x}^(3) - {x}^(2) - 3 {x}^(2) - \: 3x - {x}^(2) - x \: + 3x + 3 = 0`

`{x}^(3) - 5 {x}^(2) - x + 3 = 0`

Here we see the highest power is 3 so the degree of the equation is 3

As for the real roots, the factors each as in (x-1), (x-3) & (x+1) can be considered zero each to find the real roots because for the value of x to be zero any of the multiplied factors must be zero.

either, x-1=0

=> x = 1

or, x+1=0

=> x = -1

or, x-3=0

=> x = 3

real roots of x are: 1, -1 and 3