# How do you multiply (4x-5)(2x^5 + x3 - 1)?

Jul 26, 2015

For each power of $x$ from ${x}^{6}$ down to ${x}^{0}$, pick out the pairs that multiply to give that power and add them to find:

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

$= 8 {x}^{6} - 10 {x}^{5} + 4 {x}^{4} - 5 {x}^{3} - 4 x + 5$

#### Explanation:

Looking at each power of $x$ in turn from ${x}^{6}$ down to ${x}^{0}$, pick out the pairs that multiply to give a term with that power of $x$:

${x}^{6}$ : $4 x \cdot 2 {x}^{5} = 8 {x}^{6}$

${x}^{5}$ : $- 5 \cdot 2 {x}^{5} = - 10 {x}^{5}$

${x}^{4}$ : $4 x \cdot {x}^{3} = 4 {x}^{4}$

${x}^{3}$ : $- 5 \cdot {x}^{3} = - 5 {x}^{3}$

${x}^{2}$ : none

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

$1$ : $- 5 \cdot - 1 = 5$

$8 {x}^{6} - 10 {x}^{5} + 4 {x}^{4} - 5 {x}^{3} - 4 x + 5$
Normally when multiplying two polynomials in this way you would have two pairs to multiply and add for most of the powers of $x$, but a similar approach works.