hello.i have to factor this expression.
3y^2-16y-12
i did my work and i got (y-4)(y+1)
is it right?
(3 y + 2)(y - 6)
thank you!
You are welcome.
can you show how u got that answer please???
Sure, fooled around with factors of 12
3 and 4 did not work, 1 and 12 did not work, so tried 6 and 2
I wrote
(3 y - ) (y + )
and tried factors
soon saw needed
(3 y + ) (y - )
then fooled with 3 and 4 and then 6 and 2 until it worked
i messed up on my math . i used -12 and 3. now i know why i messed up. thank you again.:)
No problem :)
To factor the expression 3y^2 - 16y - 12, we can use the method of trial and error or the quadratic formula. Let's go through the steps to factor it using the trial and error method:
First, multiply the coefficient of the quadratic term (3) by the constant term (-12). The result is -36.
Next, we need to find two numbers whose product is -36 and whose sum is equal to the coefficient of the linear term (-16).
After some trial and error, we find that -18 and 2 fit the criteria. (-18 * 2 = -36, -18 + 2 = -16)
Now, we rewrite the middle term (-16y) using these numbers: -18y + 2y.
3y^2 - 18y + 2y - 12
Next, we group the terms and factor them separately:
(3y^2 - 18y) + (2y - 12)
Factor out the greatest common factor from each group:
3y(y - 6) + 2(y - 6)
Notice that both groupings now have a common factor, (y - 6).
(y - 6)(3y + 2)
So, the factored form of the expression 3y^2 - 16y - 12 is (y - 6)(3y + 2).
It looks like you made a mistake in your factored expression. The correct factorization is (y - 6)(3y + 2), not (y - 4)(y + 1).