How would you find the x-intercepts of y=3x^2-6X by factoring?
Thanks

y=3x^2-6X

So, to find X-intercepts, y must equal 0

So,

y=3x^2-6X
0=3x^2-6X
0=3x(x-2)

3x=0
x = 0

AND

x-2=0
x=2

Therefore, x-intercepts are 0 and 2.

Now to check. Sub the values of x into equation, y must equal 0

y=3x^2-6X
y=3(0)^2-6(0)
y = 0-0
y =0

AND

y=3x^2-6X
y=3(2)^2-6(2)
y = 12 - 12
y = 0

Therefore, the x-intercepts are 0 and 2

Thank you. So out of curiosity would 3x^2-4=0 find the same intercepts? Like would you factor it out to 3x(x-2) or 3x(x+2)?

Quote:

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Originally Posted by phoenirius

Thank you. So out of curiosity would 3x^2-4=0 find the same intercepts? Like would you factor it out to 3x(x-2) or 3x(x+2)?

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Your welcome.

3x^2-4=0 would NOT give you the same X-intercepts.

3x(x-2) or 3x(x+2) is NOT the factor of 3x^2-4=0

Rather,
3x(x-2) = 3x^2 – 6x
3x(x+2) = 3x^2 + 6x

To solve 3x^2-4=0 you need to use the quadratic equation.