hi. i have to factor this expression.
3y^2-16y-12
i got (y-4)(y+1)
i know it not right but i don't know what i did wrong.
http://www.jiskha.com/display.cgi?id=1427052839
thank you
To factor the expression 3y^2-16y-12, you need to find two binomial factors that, when multiplied together, result in the original expression.
Here's how you can factor the expression correctly:
Step 1: Multiply the coefficient of the quadratic term (3) by the constant term (-12).
3 * (-12) = -36
Step 2: Find two numbers whose product is -36 and whose sum is equal to the coefficient of the linear term (-16). In this case, those numbers are -18 and +2.
-18 * 2 = -36
-18 + 2 = -16
Step 3: Rewrite the original expression, replacing the linear term (-16y) with the sum of the two numbers found in step 2.
3y^2 - 18y + 2y - 12
Step 4: Group the terms in pairs:
(3y^2 - 18y) + (2y - 12)
Step 5: Factor out the greatest common factor from each group:
3y(y - 6) + 2(y - 6)
Step 6: Notice that (y - 6) is common to both groups. Factor it out:
(y - 6)(3y + 2)
So, after correctly factoring the expression 3y^2-16y-12, you should end up with (y - 6)(3y + 2) rather than (y - 4)(y + 1).