Question

use factoring to solve this equation
x^2-x=6

Answer 1

quadratic equation

Answer 2

Before you can factor, you need to subtract 6 from both sides.
\[x^{2}-x-6=0\]

Answer 3

Now you want two numbers that multiply to 6 but are different by 1. Can you think of two numbers that work for this?

Answer 4

2 and 3

Answer 5

x=3 is a root.
(x^2-x-6)/(x-3)=x+2
Thus:
x^2-x-6 = (x-3)(x+2) = 0
x1 = 3 and x2 = -2

Answer 6

Very good. So let's rewrite the middle term\[x^{2}-3x+2x-6=0\]and then regroup\[(x^{2}-3x) +(2x-6)\]

Answer 7

We can factor an x out of the first group and a 2 out of the second and then we'll see something nice

Answer 8

\[x(x-3)+2(x-3)\]Now, factor out the x-3 from both sets\[(x-3)(x+2)\]and we're done!