Using remainder value theorem, the value of f(3) is 3
Solution:
Given, function is [tex]f(x)=-x^{3}+4 x^{2}-2 x[/tex]
We have to find the value of f(3) by using remainder theorem.
Now, we have to divide f(x) with x – 3, because we want value of f(3)
If we want f(a), then we have to divide with x – a.
Now, first let us factorize the f(x)
[tex]\begin{array}{l}{\text { Then, } f(x)=-x^{3}+4 x^{2}-2 x} \\\\ {=-x\left(x^{2}-4 x+2\right)} \\\\ {=-x\left(x^{2}-4 x+3-1\right)} \\\\ {=-x\left(x^{2}-3 x-x+3-1\right)} \\\\ {=-x(x(x-3)-1(x-3)-1)} \\\\ {=-x((x-3)(x-1)-1)} \\\\ {=-x(x-3)(x-1)+x}\end{array}[/tex]
Now, let us divide the f(x) with x – 3
[tex]\begin{array}{l}{\rightarrow \frac{-\mathrm{x}(\mathrm{x}-3)(\mathrm{x}-1)+\mathrm{x}}{x-3}=\frac{-x(x-3)(x-1)}{x-3}+\frac{x}{x-3}} \\\\ {=-\mathrm{x}(\mathrm{x}-1)+\frac{x-3+3}{x-3}} \\\\ {=-\mathrm{x}(\mathrm{x}-1)+1+\frac{3}{x-3}}\end{array}[/tex]
Now, multiply f(x) with (x -3) ⇒ (-x(x -1) + 1)(x -3) + 3
This is in the form of:
[tex]\text {quotient } \times \text { divisor }+\text { remainder.}[/tex]
Hence, the remainder is 3.
Therefore the value of f(3) is 3
