Question

If $p\left(x\right)=3x^2-2x+5\text{and}r\left(x\right)=x^3+x+1$ then find $r\left(x+1\right)-r\left(x^2\right)$.

Short answer

The value of $r\left(x+1\right)-r\left(x^2\right)=-x^6+x^3+2x^2+4x+2$.

Step-by-step solution

Step 1. Write down the given information.

The given functions are $p\left(x\right)=3x^2-2x+5\text{and}r\left(x\right)=x^3+x+1$.

Step 2. Calculation.

Evaluating $r\left(x+1\right)-r\left(x^2\right)$by plugging $x=x+1\text{and}x=x^2$in given function $r\left(x\right)=x^3+x+1$and then subtracting $r\left(x^2\right)$from $r\left(x+1\right)$.

Therefore,

$\begin{array}{c}r\left(x\right)=x^3+x+1....\left(\text{Given}\right)\\ r\left(x+1\right)={\left(x+1\right)}^3+\left(x+1\right)+1\\ =x^3+1+3x^2+3x+x+1+1\\ =x^3+3x^2+4x+\mathrm{3....}\left(1\right)\\ \text{Now,}r\left(x^2\right)={\left(x^2\right)}^3+\left(x^2\right)+1\\ =x^6+x^2+\mathrm{1....}\left(2\right)\end{array}$

Subtracting (2) from (1) gives,

$\begin{array}{c}r\left(x+1\right)-r\left(x^2\right)=\left(x^3+3x^2+4x+3\right)-\left(x^6+x^2+1\right)\\ =-x^6+x^3+2x^2+4x+2\end{array}$

Step 3. Conclusion.

The value of $r\left(x+1\right)-r\left(x^2\right)=-x^6+x^3+2x^2+4x+2$.