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

Jun 27, 2015

Consider each power of $x$ in descending order and total up the possible combinations of coefficients to find:

$\left(x - 1\right) \left(x + 1\right) \left(x - 3\right) \left(x - 5\right) = {x}^{4} - 8 {x}^{3} + 14 {x}^{2} + 8 x - 15$

#### Explanation:

Given $f \left(x\right) = \left(x - 1\right) \left(x + 1\right) \left(x - 3\right) \left(x - 5\right)$, consider each power of $x$ in descending order and total up the combinations of coefficients:

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

${x}^{3}$ : $\left(- 1 \cdot 1 \cdot 1 \cdot 1\right) + \left(1 \cdot 1 \cdot 1 \cdot 1\right) + \left(1 \cdot 1 \cdot - 3 \cdot 1\right) + \left(1 \cdot 1 \cdot 1 \cdot - 5\right) = - 1 + 1 - 3 - 5 = - 8$

${x}^{2}$ : $\left(- 1 \cdot 1 \cdot 1 \cdot 1\right) + \left(- 1 \cdot 1 \cdot - 3 \cdot 1\right) + \left(- 1 \cdot 1 \cdot 1 \cdot - 5\right) + \left(1 \cdot 1 \cdot - 3 \cdot 1\right) + \left(1 \cdot 1 \cdot 1 \cdot - 5\right) + \left(1 \cdot 1 \cdot - 3 \cdot - 5\right) = - 1 + 3 + 5 - 3 - 5 + 15 = 14$

$x$ : $\left(- 1 \cdot 1 \cdot - 3 \cdot 1\right) + \left(- 1 \cdot 1 \cdot 1 \cdot - 5\right) + \left(- 1 \cdot 1 \cdot - 3 \cdot - 5\right) + \left(1 \cdot 1 \cdot - 3 \cdot - 5\right) = 3 + 5 - 15 + 15 = 8$

$1$ : $- 1 \cdot 1 \cdot - 3 \cdot - 5 = - 15$

So $f \left(x\right) = {x}^{4} - 8 {x}^{3} + 14 {x}^{2} + 8 x - 15$

Check:

$f \left(2\right) = {2}^{4} - 8 \cdot {2}^{3} + 14 \cdot {2}^{2} + 8 \cdot 2 - 15$

$= 16 - 64 + 56 + 16 - 15$

$= 9$

$\left(2 - 1\right) \left(2 + 1\right) \left(2 - 3\right) \left(2 - 5\right) = 1 \cdot 3 \cdot - 1 \cdot - 3 = 9$