The kind of roots does the equation f(x) is option (C) 3 real roots is the correct answer.
A polynomial equation is a sum of constants and variables. A polynomial is an expression consisting of indeterminate's (also called variables) and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables.
For the given situation,
The equation [tex]f(x) = x^3 - x^2 - x + 1[/tex]
The roots of the polynomial equation can be found as follows,
Step 1:
Substitute the value of x in which the equation f(x) equals zero.
⇒ Put [tex]x = 1[/tex], the equation becomes
⇒ [tex]f(1) = 1^3 - 1^2 - 1 + 1[/tex]
⇒ [tex]f(1) = 0[/tex]
Thus [tex](x - 1)[/tex] is the one root of the equation.
Step 2:
The other roots can be found by framing the quadratic equation on dividing the equation [tex]f(x) = x^3 - x^2 - x + 1[/tex] by [tex](x - 1)[/tex]
On dividing [tex]f(x) = x^3 - x^2 - x + 1[/tex] by [tex](x - 1)[/tex] we get the quotient,
⇒ [tex](x^{2} -1)[/tex]
Step 3:
Now factorize the quadratic equation, [tex](x^{2} -1)[/tex]
It can be expand using the identity,
⇒ [tex](x^{2} -1^{2} ) = (x+1)(x-1)[/tex] [∵ (a^2 - b^2) = (a+b)(a-b) ]
Thus the factors of the equation are [tex](x-1)(x+1)(x-1)[/tex]. So the roots are [tex]1,-1,1[/tex].
Hence we can conclude that the kind of roots does the equation f(x) is option (C) 3 real roots is the correct answer.
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