# How do you solve (x-8)(x-1)=0?

May 24, 2015

One way of seeing this is using some logic and math.

If you have a product that equals zero, as in this case, you must agree that at least one of the factors is zero.

Thus, either $x - 8$ and/or $x - 1$ equals zero.

Thus, ${x}_{1} = 8$ and ${x}_{2} = 1$.

Another way is distributing these factors. To distribute, you need to multiply each term of the first by all the terms in the second, as follows:

$\left(x - 8\right) \left(x - 1\right)$

$\left(x \cdot x\right) + \left(x \cdot \left(- 1\right)\right) + \left(\left(- 8\right) \cdot x\right) + \left(\left(- 8\right) \left(- 1\right)\right)$
${x}^{2} - x - 8 x + 8$
${x}^{2} - 9 x + 8$

And now let's find its roots. I'll use Bhaskara here.

$\frac{9 \pm \sqrt{81 - 4 \left(1\right) \left(8\right)}}{2}$
$\frac{9 \pm 7}{2}$
${x}_{1} = \frac{9 + 7}{2} = 8$
${x}_{2} = \frac{9 - 7}{2} = 1$

:)