Find the value of p for which the polynomial 3x^3 -x^2 + px +1 is exactly divisible by x-1, hence factorise the polynomial

ANSWER[tex]p = - 3[/tex]The completely factored form is[tex](x + 1)(x - 1)( 3x - 1)[/tex]EXPLANATIONThe given polynomial expression is [tex]3 {x}^{3} - {x}^{2} + px + 1[/tex]Let [tex]f(x) = 3 {x}^{3} - {x}^{2} + px + 1[/tex]According to the Remainder Theorem, if f(x) is exactly divisible by x-1, then the remainder is zero.This implies that:[tex]f(1) = 0[/tex][tex]3 {(1)}^{3} - {(1)}^{2} + p(1)+ 1 = 0[/tex][tex]3 - 1 + p + 1 = 0[/tex][tex]3 + p = 0[/tex][tex]p = - 3[/tex]When we substitute the value of p back into the function, f(x) we get:[tex]f(x) = 3 {x}^{3} - {x}^{2} - 3x + 1[/tex]We now perform long division as shown in the attachment.We can factor the function to get:[tex]f(x) =(x - 1)( 3 {x}^{2} + 2x - 1)[/tex]We now split the middle term of the quadratic term and factor it completely to obtain:[tex]f(x) =(x - 1)( 3 {x}^{2} + 3x - x - 1)[/tex][tex]f(x) =(x - 1)( 3x(x + 1) - 1(x + 1))[/tex][tex]f(x) =(x - 1)( 3x - 1)(x + 1) [/tex][tex]f(x) =(x + 1)(x - 1)( 3x - 1)[/tex]