Answer:
Need help i have 5min left proving angles
1 answer:
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The answer is C.
To solve these types of equations you first need to make sure that you equation is equal to 0. In this case all of the numbers are already on the same side so no work needs to be done there. Then we can get a, b and c values for the quadratic equation by looking at the coefficients.
a = 4 (number attached to x^2)
b = -6 (number attached to x)
c = 1 (number with no variable attached)
Now we can put this into the quadratic equation.
To solve these types of equations you first need to make sure that you equation is equal to 0. In this case all of the numbers are already on the same side so no work needs to be done there. Then we can get a, b and c values for the quadratic equation by looking at the coefficients.
a = 4 (number attached to x^2)
b = -6 (number attached to x)
c = 1 (number with no variable attached)
Now we can put this into the quadratic equation.
Answer:
{-2/3, 2/3, 3}
Step-by-step explanation:
The equation can be factored by grouping.
f(x) = (9x^3 -27x^2) +(-4x +12)
f(x) = 9x^2(x -3) -4(x -3)
f(x) = (9x^2 -4)(x -3)
We recognize the first factor as the difference of squares, so know it can be further factored to give ...
f(x) = (3x -2)(3x +2)(x -3)
The zeros are the values of x that make these factors be zero:
3x -2 = 0 ⇒ x = 2/3
3x +2 = 0 ⇒ x = -2/3
x -3 = 0 ⇒ x = 3
The zeros of the polynomial are x = -2/3, 2/3, 3.
