# How do you multiply (5x^3-x^2+x-4) (x+1)?

Jul 4, 2015

Use distributivity to find:

$\left(5 {x}^{3} - {x}^{2} + x - 4\right) \left(x + 1\right) = 5 {x}^{4} + 4 {x}^{3} - 3 x - 4$

#### Explanation:

$\left(5 {x}^{3} - {x}^{2} + x - 4\right) \left(x + 1\right)$

$= \left(5 {x}^{3} - {x}^{2} + x - 4\right) \cdot x + \left(5 {x}^{3} - {x}^{2} + x - 4\right) \cdot 1$

$= 5 {x}^{4} - {x}^{3} + {x}^{2} - 4 x + 5 {x}^{3} - {x}^{2} + x - 4$

$= 5 {x}^{4} - {x}^{3} + 5 {x}^{3} + {x}^{2} - {x}^{2} - 4 x + x - 4$

$= 5 {x}^{4} + \left(- 1 + 5\right) {x}^{3} + \left(1 - 1\right) {x}^{2} + \left(1 - 4\right) x - 4$

$= 5 {x}^{4} + 4 {x}^{3} - 3 x - 4$

Alternatively, look at each power of $x$ in descending order and total up the relevant products of the coefficients:

${x}^{4}$ : $5 \cdot 1 = 5$

${x}^{3}$ : $\left(5 \cdot 1\right) + \left(- 1 \cdot 1\right) = 4$

${x}^{2}$ : $\left(- 1 \cdot 1\right) + \left(1 \cdot 1\right) = 0$

$x$ : $\left(1 \cdot 1\right) + \left(- 4 \cdot 1\right) = - 3$

$1$: $- 4 \cdot 1 = - 4$

Hence: $5 {x}^{4} + 4 {x}^{3} - 3 x - 4$