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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)







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