Answer:
The solutions are:
[tex]x=-2,\:x=-3,\:x=-1,\:x=4[/tex]
Step-by-step explanation:
To find the solutions to the equation [tex]\left(x^2+5x+6\right)\left(x^2-3x-4\right)=0[/tex] you need to:
Break the expression into groups
[tex]x^2+5x+6=\left(x^2+2x\right)+\left(3x+6\right)[/tex]
Factor out [tex]x[/tex] from [tex](x^2+2x)[/tex]
[tex]x^2+2x=x\left(x+2\right)[/tex]
Factor out 3 from [tex]3x+6[/tex]
[tex]3x+6=3\left(x+2\right)[/tex]
[tex]x^2+5x+6=x\left(x+2\right)+3\left(x+2\right)\\\\\mathrm{Factor\:out\:common\:term\:}x+2\\\\x^2+5x+6=\left(x+2\right)\left(x+3\right)[/tex]
[tex]x^2-3x-4=\left(x^2+x\right)+\left(-4x-4\right)\\\\x^2-3x-4=x\left(x+1\right)-4\left(x+1\right)\\\\x^2-3x-4=\left(x+1\right)\left(x-4\right)[/tex]
Therefore
[tex]\left(x^2+5x+6\right)\left(x^2-3x-4\right)=\left(x+2\right)\left(x+3\right)\left(x+1\right)\left(x-4\right)=0[/tex]
Using the Zero Factor Theorem:
