Question

Find $p\left(3\right)\text{and}p\left(-1\right)$ for the function $p\left(x\right)=x^4-3x^3+2x^2-5x+1$.

Short answer

The value of $p\left(3\right)=4\text{and}p\left(-1\right)=12$.

Step-by-step solution

Step 1. Write down the given information.

The given function is $p\left(x\right)=x^4-3x^3+2x^2-5x+1$.

Step 2. Calculation.

Evaluating $p\left(3\right)\text{and}p\left(-1\right)$by plugging $x=3\text{and}x=-1$in given function $p\left(x\right)=x^4-3x^3+2x^2-5x+1$.

$\begin{array}{c}p\left(x\right)=x^4-3x^3+2x^2-5x+\mathrm{1....}\left(\text{Given}\right)\\ p\left(3\right)={\left(3\right)}^4-3{\left(3\right)}^3+2{\left(3\right)}^2-5\left(3\right)+1\\ =4\\ \text{Now,}p\left(-1\right)={\left(-1\right)}^4-3{\left(-1\right)}^3+2{\left(-1\right)}^2-5\left(-1\right)+1\\ =12\end{array}$

Step 3. Conclusion.

The value of $p\left(3\right)=4\text{and}p\left(-1\right)=12$.