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

Step-by-step explanation:at first multiply the factors[tex] ({x}^{2 } + x \: - 3x \: - 3)(x - 1) = 0[/tex][tex] {x}^{3} - {x}^{2} - 3 {x}^{2} - \: 3x - {x}^{2} - x \: + 3x + 3 = 0[/tex][tex] {x}^{3} - 5 {x}^{2} - x + 3 = 0[/tex]Here we see the highest power is 3 so the degree of the equation is 3As 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 = 1or, x+1=0=> x = -1or, x-3=0=> x = 3real roots of x are: 1, -1 and 3