following the quadratic formula got me confused. Can 3x^2-3x-6x-6x=0 work? If so, i can factor
3x(x-3) but what about (-6x-6x) Can you show me my mistake please?
3x(x-3) but what about (-6x-6x) Can you show me my mistake please?
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3x^2 - 9 = 6x
Bring 6x to the left
3x^2 - 6x - 9
factor 3 out
3(x^2 - 2x - 3)
3(x-3)(x+1)
That is your factored form.
I think the mistake you made is you put -6x two times.
Hope that helps! Good luck!
Bring 6x to the left
3x^2 - 6x - 9
factor 3 out
3(x^2 - 2x - 3)
3(x-3)(x+1)
That is your factored form.
I think the mistake you made is you put -6x two times.
Hope that helps! Good luck!
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3x^2-9=6x
i.e. 3x^2-6x-9=0
i.e. 3(x^2-2x-3)=0
i.e. x^2-2x-3 =0 [dividing both sides by 3]
i.e. x^2-3x+x-3=0
i.e. x(x-3)+1(x-3)=0
i.e. (x+1)(x-3)=0
i.e. 3x^2-6x-9=0
i.e. 3(x^2-2x-3)=0
i.e. x^2-2x-3 =0 [dividing both sides by 3]
i.e. x^2-3x+x-3=0
i.e. x(x-3)+1(x-3)=0
i.e. (x+1)(x-3)=0
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3x^2 - 9 = 6x
3x^2 - 6x - 9 = 0
(3x - 9)(x + 1) = 0
3x - 9 = 0
3x = 9
x = 3
x + 1 = 0
x = -1
3x^2 - 6x - 9 = 0
(3x - 9)(x + 1) = 0
3x - 9 = 0
3x = 9
x = 3
x + 1 = 0
x = -1