We are given the following functions;
[tex]\begin{gathered} p(x)=3x-2 \\ q(x)=2x+5 \end{gathered}[/tex]
We shall find the values of the following;
[tex]p(x)-q(x)[/tex]
We substitute for the values of each function and we'll have;
[tex]\begin{gathered} p(x)-q(x)=(3x-2)-(2x+5) \\ p(x)-q(x)=3x-2-2x-5_{} \end{gathered}[/tex]
Take note that the negative sign in front of the parenthesis affects both values. Hence negative +5 becomes -5. We now have;
[tex]\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}[/tex]
(B):
[tex]p(x)+q(x)[/tex]
Just like in part (a), we shall substitute for the value of each of the functions, as follows;
[tex]\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}[/tex]
(C):
[tex]p(x)q(x)[/tex]
In this case, we calculate the product of both functions, and we start by substituting for the value of each;
[tex]\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}[/tex]
We can now simplify the above and we'll have;
[tex]\begin{gathered} p(x)q(x)=6x^2+15x-4x-10 \\ p(x)q(x)=6x^2+11x-10 \end{gathered}[/tex]
ANSWER:
[tex]\begin{gathered} (a)=x-7 \\ (b)=5x+3 \\ (c)=6x^2+11x-10 \end{gathered}[/tex]
