Answer
$a\quad 3x^{2}+2x-8$
$b.\quad 4n^{2}-31n+42$
$c.\quad 4p^{3}-10p+6p-15$
Work Step by Step
a.
a. $\begin{array}{llllll}
& & \text{First} & \text{Outer} & \text{Inner} & \text{Last}\\
(3x-4)(x+2) & = & (3x)(x) & +(3x)(2) & (-4)(x) & (-4)(2)\\
& = & 3x^{2} & +6x & -4x & -8\\
& = & 3x^{2} & +2x & & -8
\end{array}$
The simpler product is $3x^{2}+2x-8$.
b. $\begin{array}{llllll}
& & \text{First} & \text{Outer} & \text{Inner} & \text{Last}\\
(n-6)(4n-7) & = & (n)(4n) & +(n)(-7) & +(-6)(4n) & +(-6)(-7)\\
& = & 4n^{2} & -7n & -24n & +42\\
& = & 4n^{2} & -31n & & +42
\end{array}$
The product is $4n^{2}-31n+42$.
c. $\begin{array}{llllll}
& & \text{First} & \text{Outer} & \text{Inner} & \text{Last}\\
(2p^{2}+3)(2p-5) & = & (2p^{2})(2p) & +(2p^{2})(-5) & +(3)(2p) & +(3)(-5)\\
& = & 4p^{3} & -10p^{2} & +6p & -15
\end{array}$
The product is $4p^{3}-10p+6p-15$.
