To algebraically find the zeroes of this function, we must factor by grouping
y = ax^2 -bx +c
y = 6x^2 - 11x + 3
Multiply a and c.
6 x 3 = 18
Identify b, b= -11
We need to think of 2 numbers that when added, will give us the b value -(11) and when multiplied, will give us (ac) 18.
-9 + -2 = -11
-9 • -2 = 18
Now, we rewrite the equation using these 2 numbers.
y = 6x^2 - 9x - 2x + 3
Now, we put brackets around the first 2 and last 2 terms.
y = (6x^2 - 9x) (-2x + 3)
Now, simplify the two sets of brackets individually. You will find that the things left inside the brackets will be the exact same.
y = -3x(2x + 3) -1(2x + 3)
Attach the 2 coefficients together and rewrite.
y = (-3x - 1)(2x + 3)
Since we are solving for zeroes, make y = 0
0 = (-3x - 1) (2x + 3)
Both of these binomials equal 0, we must evaluate them separately.
0 = -3x - 1
3x = -1
x = -1/3
Second one:
0 = 2x + 3
-3 = 2x
x = -3/2
zeroes: (-1/3) , (-3/2)