Question

Show that (x-1) is a factor of p(x) = 3x^3 - x^2 - 3x - 4 and factorise p(X)

Answer 1

$p(x) = 3 {x}^{3} - {x}^{2} - 3x - 4 \\ g(x) = (x - 1) \\ let \: g(x) = 0 \\ (x - 1) = 0 \\ x = 1 \\ p(1) = 3 {(1)}^{3} - {(1)}^{2} - 3(1) - 4 \\ \: \: \: \: \: \: \: \: \: = 3 - 1 - 3 - 4 \\ \: \: \: \: \: \: \: \: \: = - 1 - 4 \\ \: \: \: \: \: \: \: \: \: = - 5 \\ so, \: (x - 1) \: is \: not \: a \: factor \: of \: p(x)$
hope it helps.........
Plz Mark As Braililest.........

Answer 2

ANSWER✔

$\large\underline\bold{GIVEN,}$

$\sf\dashrightarrow p(x) = 3x^3 - x^2 - 3x - 4$

$\sf\dashrightarrow g(x)=x-1$

$\large\underline\bold{TO\:FIND,}$

$\sf\dashrightarrow factorize\:p(x)\:and\:show\:that\:g(x)\:is\:a\:factor\:of\:p(x).$

$\large\underline\bold{SOLUTION,}$

TAKING, g(x)=0

$\sf\implies x-1=0$

$\sf\implies x=1$

putting the value of x in p(x), WE GET,

$\sf\therefore p(x)=0$

$\sf\implies p(1) = 3(1)^3 - (1)^2 - 3(1) - 4$

$\sf\implies 3-1-3-4=0$

$\sf\implies 2-3-4=0$

$\sf\implies -1-4=0$

$\sf\implies -5=0$

$\sf\therefore L.H.S \neq R.H.S$

$\large\underline\bold{\therefore \: g(x) \:is\:not\:a\:factor\:of\:p(x).}$

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