Question

Let p(x)=3x – 2 and q(x) = 2x + 5. Finda. p(x) – Q(x) =b. p(x) + q(x) =c. p(x)q(x) =

Answer 1

We are given the following functions;

$\begin{gathered} p(x)=3x-2 \\ q(x)=2x+5 \end{gathered}$

We shall find the values of the following;

$p(x)-q(x)$

We substitute for the values of each function and we'll have;

$\begin{gathered} p(x)-q(x)=(3x-2)-(2x+5) \\ p(x)-q(x)=3x-2-2x-5_{} \end{gathered}$

Take note that the negative sign in front of the parenthesis affects both values. Hence negative +5 becomes -5. We now have;

$\begin{gathered} p(x)-q(x)=3x-2-2x-5 \\ p(x)-q(x)=3x-2x-5-2 \\ p(x)-q(x)=x-7 \end{gathered}$

(B):

$p(x)+q(x)$

Just like in part (a), we shall substitute for the value of each of the functions, as follows;

$\begin{gathered} p(x)+q(x)=(3x-2)+(2x+5) \\ p(x)+q(x)=3x-2+2x+5 \\ p(x)+q(x)=3x+2x+5-2 \\ p(x)+q(x)=5x+3 \end{gathered}$

(C):

$p(x)q(x)$

In this case, we calculate the product of both functions, and we start by substituting for the value of each;

$\begin{gathered} p(x)q(x)=(3x-2)(2x+5) \\ p(x)q(x)=3x\cdot2x+3x\cdot5-2\cdot(+2x)-2\cdot(+5) \end{gathered}$

We can now simplify the above and we'll have;

$\begin{gathered} p(x)q(x)=6x^2+15x-4x-10 \\ p(x)q(x)=6x^2+11x-10 \end{gathered}$

ANSWER:

$\begin{gathered} (a)=x-7 \\ (b)=5x+3 \\ (c)=6x^2+11x-10 \end{gathered}$