Answer:
[tex]2x\left(x-2\right)\left(x+3\right)\left(x+2\right)[/tex]
Step-by-step explanation:
Lets go ahead and take a step by step approach to solving your problem.
First we must start by factoring [tex]x^2+x-6[/tex].
To do this, we must break the expression into groups! Although I will also provide a definition to help you understand! For [tex]ax^2+bx+c[/tex] we need to find u, v > [tex]u\cdot \:v=a\cdot \:c[/tex] and [tex]u+v=b[/tex] which we will group into [tex]\left(ax^2+ux\right)+\left(vx+c\right)[/tex].
The values we have are...[tex]a=1,\:b=1,\:c=-6[/tex] and [tex]u\cdot v=-6,\:u+v=1[/tex].
Now we need the factors of 6 which are 1, 2, 3 and 6. We also need the negative factors which you get simply by multiplying the positives by -1 or just reversing them.
Now we need to check for every two factors if u + v = 1.
[tex]\mathrm{Check}\:u=1,\:v=-6: \:u\cdot v=-6,\:u+v=-5 \Rightarrow \mathrm{False}[/tex]
[tex]\mathrm{Check}\:u=2,\:v=-3: \:u\cdot v=-6,\:u+v=-1 \Rightarrow \mathrm{False}[/tex]
[tex]\mathrm{Check}\:u=3,\:v=-2: \:u\cdot v=-6,\:u+v=1 \Rightarrow \mathrm{True}[/tex]
[tex]\mathrm{Check}\:u=6,\:v=-1: \:u\cdot v=-6,\:u+v=5 \Rightarrow \mathrm{False}[/tex]
Therefore [tex]u=3,\:v=-2[/tex].
Now we want to [tex]\mathrm{Group\:into\:}\left(ax^2+ux\right)+\left(vx+c\right)[/tex]. Which is [tex]\left(x^2-2x\right)+\left(3x-6\right)[/tex].
Now we must factor x from [tex]x^2-2x[/tex].
Lets apply the exponent rule of [tex]a^{b+c}=a^ba^c[/tex] which means [tex]x^2=xx[/tex].
Therefore we now have xx - 2x. Lets factor out the common term of x to get x(x - 2).
Now factor 3 out of [tex]3x-6[/tex].
Rewrite 6 as 3 * 2. [tex]3x-3\cdot \:2[/tex].
Now factor out the common term of 3 > [tex]3\left(x-2\right)[/tex].
[tex]x\left(x-2\right)+3\left(x-2\right)[/tex] > Factor out the common term of x - 2 > [tex]\left(x-2\right)\left(x+3\right)[/tex].
Now lets factor [tex]2x^2+4x[/tex].
Apply the previous exponent rule of [tex]a^{b+c}=a^ba^c[/tex] > [tex]x^2=xx[/tex] > [tex]2xx+4x[/tex].
Now rewrite 4 as 2 * 2 > [tex]2xx+2\cdot \:2x[/tex].
Now factor out the common term of 2x to get 2x(x + 2).
Combine it all and we get [tex]2x\left(x-2\right)\left(x+3\right)\left(x+2\right)[/tex].
Hope this helps!
