Question

If $p\left(x\right)=2x^3+6x-12\text{and}q\left(x\right)=5x^2+4$ then find $3p\left(a\right)-q\left(a+1\right)$.

Short answer

The value of $3p\left(a\right)-q\left(a+1\right)=6a^3-5a^2+8a-45$.

Step-by-step solution

Step 1. Write down the given information.

The given functions are $p\left(x\right)=2x^3+6x-12\text{and}q\left(x\right)=5x^2+4$.

Step 2. Calculation.

Evaluating $3p\left(a\right)-q\left(a+1\right)$ by plugging $x=a$ in $p\left(x\right)=2x^3+6x-12$ and multiply by 3 and then subtract $q\left(a+1\right)$ by plugging $x=a+1$ in $q\left(x\right)=5x^2+4$. Therefore,

role="math" localid="1648111658019" $\begin{array}{c}p\left(x\right)=2x^3+6x-12....\left(\text{Given}\right)\\ 3p\left(x\right)=3\left(2x^3+6x-12\right)\\ 3p\left(a\right)=3\left(2{\left(a\right)}^3+6\left(a\right)-12\right)\\ =6a^3+18a-36....\left(1\right)\\ \text{Now,}q\left(x\right)=5x^2+4\\ q\left(a+1\right)=5{\left(a+1\right)}^2+4\\ =5a^2+10a+9....\left(2\right)\end{array}$

Subtract (2) from (1) gives,

$\begin{array}{c}3p\left(a\right)-q\left(a+1\right)=\left(6a^3+18a-36\right)-\left(5a^2+10a+9\right)\\ =6a^3-5a^2+8a-45\end{array}$

Step 3. Conclusion.

The value of $3p\left(a\right)-q\left(a+1\right)=6a^3-5a^2+8a-45$.