Question

p(x)=x^3-4x^2+x+6,g(x)=x-3​

Answer 1

Answer: Given

p(x) = x^3-4x^2+x+6

g(x) = x-3

Now

let g(x) be 0

Then, g(x) = x-3=0

= x= 3

Now putting value of x

(3)^3-4*3^2+3+6

27-36+9

36-36

0

hence, x-3 is factor of x^3-4x^2-x+6

Answer 2

$\huge{\underline{\tt{\black{P(x)={x}^{3}-{4x}^2+x+6}}}}$

$\huge{\underline{\tt{\blue{Solution-}}}}$

$\longrightarrow \sf{g(x) = x -3}$

$\longrightarrow \sf{g(x) = 0}$

$\longrightarrow \sf{x -3 = 0}$

$\longrightarrow \sf{x = 3}$

$\longrightarrow \sf{P(+3) = {x}^{3}-{4x}^{2}+x+6}$

$\longrightarrow \sf{P(3) = {3}^{3}-4{(3)}^{2}+3+6}$

$\longrightarrow \sf{P(3) = 27-36+9}$

$\longrightarrow \sf{P(3) = 36 -36}$

$\longrightarrow \sf{P(3) = 0}$

$\longrightarrow \sf{ x -3 = is\:a\:factor\:of\{x}^{3}-{4x}^{2}+x+6}$