Answer: (-3, 0) and (-1, 0)
Step-by-step explanation:
[tex]-4x^2-16x-12[/tex]
[tex]\mathrm{Rewrite\:}-16\mathrm{\:as\:}-4\cdot \:4[/tex]
[tex]\mathrm{Rewrite\:}-12\mathrm{\:as\:}-4\cdot \:3[/tex]
[tex]=-4x^2-4\cdot \:4x-4\cdot \:3[/tex]
[tex]\text{Factor out common term -4}[/tex]
[tex]=-4\left(x^2+4x+3\right)[/tex]
[tex]x^2+4x+3[/tex]
[tex]=\left(x^2+x\right)+\left(3x+3\right)[/tex]
[tex]=x\left(x+1\right)+3\left(x+1\right)[/tex]
[tex]\mathrm{Factor \ out \ common \ term \ x+1}[/tex]
[tex]=\left(x+1\right)\left(x+3\right)[/tex]
[tex]=-4\left(x+1\right)\left(x+3\right)[/tex]
Roots:
x + 3 = 0 and x + 1 = 0
x = 0 - 3 x = 0 - 1
x = -3 x = -1
Therefore, the zeroes of the function are (-3, 0) and (-1, 0)
