Answer:
[tex]2(3x-2)(x+2)=6x^2-8x-8[/tex]
The factored form of [tex]6x^2-8x-8=0[/tex] is [tex]2(3x-2)(x+2)=0[/tex]
Step-by-step explanation:
[tex]6x^2-8x-8=0[/tex]
The way the quadratic equation was given, we can't have a factored form in the format: [tex](ax-b)(cx+d)[/tex]
First, divide both sides by 2
[tex]3x^2-4x-4=0[/tex]
Now, it is about thinking. From the equation, we will get something in the format: [tex](ax-b)(cx+d)[/tex]
Let's expand this: [tex](ax-b)(cx+d) = acx^2+adx-bcx-bd[/tex]
From here, we can give some values for those variables, based on the quadratic equation [tex]3x^2-4x-4=0[/tex]:
[tex](3x-b)(x+d) = 3\cdot 1\cdot x^2+3dx-b\cdot 1 \cdot x-bd= \boxed{3x^2+3dx-bx-bd}[/tex]
Once we want the middle term to be -4 and bd to be 4, we can easily evaluate the other variables.
[tex](3x-2)(x+2) = 3\cdot 1\cdot x^2+3(2)x-(2)\cdot 1 \cdot x-(2)(2)= \boxed{3x^2+6x-4x-4}[/tex]
Therefore,
[tex](3x-2)(x+2)=3x^2-4x-4[/tex]
But we are not ready yet!
This is the factored form of [tex]3x^2-4x-4=0[/tex], to get the factored form of the problem equation, just multiply the factored form we got by 2.
[tex]2(3x-2)(x+2)=6x^2-8x-8[/tex]
