$x^2-x-6=x^2+2x-3x-6=x(x+2)-3(x+2)\\\\=\boxed{(x+2)(x-3)}$

The factors are: (x+2) and (x-3).

$x^2-x-6=x^2+\underbrace{2x-3x}_{=-x}-6\\\\=\underbrace{x\cdot\boxed{x}}_{=x^2}+\underbrace{2\cdot\boxed{x}}_{=2x}+\underbrace{\fbox{-3}\cdot x}_{=-3x}+\underbrace{\fbox{-3}\cdot2}_{=-6}=\boxed{x}(x+2)+\fbox{-3}(x+2)\\\\=(x+2)(x+(-3))=\boxed{(x+2)(x-3)}$

8 0

3 years ago

Can anyone work through this. I have no idea how to do it. Please help

Tpy6a [65]

That seem really hard try math papa

5 0

3 years ago

Solve. h - 18 1/2 = -7 3/4

barxatty [35]

Make the denominators the same:

h - 18 $( \frac{2}{2}) ( \frac{1}{2}) = \frac{2}{4}$ = -7 $\frac{3}{4}$

bring the 18 1/2 to the other side:

h=11 1/4

7 0

3 years ago

8 (n + 2) = __n + __

8 (n + 2) = n +