## Algebra 1: Common Core (15th Edition)

$a\quad 3x^{2}+2x-8$ $b.\quad 4n^{2}-31n+42$ $c.\quad 4p^{3}-10p+6p-15$
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 $4p^{3}-10p+6p-15$.