Question

f(x)=x[tex]x^{3 } -x^{2} +1 g(x)=(x-1)

Answer 1

For finding the solution of this polynomial ; we need to firstly find the value of g(x) and then the value we get for x needs to be substituted in the polynomial f(x) and then simplify the polynomial with necessary operations.

Step 1 : Finding the value of g(x)

Firstly we need to make the equation :

$\longrightarrow{\sf{g(x) = x - 1 = 0}}$

Now we have to transpose -1 to another side and thus it becomes +1 :

$\longrightarrow{\sf{g(x) = x = 0 + 1}}$

$\longrightarrow{\sf{g(x) = x = 1}}$

So , the value of x is 1 .

Step 2 : Substituting the value of x in polynomial f(x) makes it :

$\longrightarrow{\sf{f(x) = {(x)}^{3} - {(x)}^{2} + 1}}$

Substituting the value of x :

$\longrightarrow{\sf{f(1) = {(1)}^{3} - {(1)}^{2} + 1}}$

Step 3 : Simplification :

$\longrightarrow{\sf{f(1) = {(1)}^{3} - {(1)}^{2} + 1}}$

$\longrightarrow{\sf{f(1) = 1 - 1 + 1}}$

$\longrightarrow{\sf{f(1) = 0 + 1}}$

$\large{\boxed{\implies{\sf{f(1) = 1}}}}$

So the value of the given polynomial is 1.

Answer 2

$\sf \red{g(x) = x - 1}$

$\implies \sf{x - 1 = 0}$

$\implies \sf{x = 0 + 1}$

$\implies \sf{x = 1}$

_____________.....

$\sf \red{f(x) = {x}^{3} - {x}^{2} + 1 }$

$\implies \sf{ {1}^{3} - {1}^{2} + 1}$

$\implies \sf{ \cancel1 - \cancel1 + 1}$

$\implies\sf {1}$

The answer is 1....