How do I factor
-3p^2+66p+9
see my other post. On this one, divide out the factor 3 first.
I took out the three to get 3(-p^2+31) but my answer is not making any sense with the question I am being asked because of my p value must be between 18 and 28. HELP
Is this a different problem? You should be factoring -3p^2+84p-288 if it's the same one as before.
Start by taking out a common -3.
I didn't think you were going to answer the other question.
Now I have -3(p^2-28p+96)
I am now stuck I eventually need this to be set to equal zero
This is for -3p^2+84p-288, but I am also stuck on the other problem
To factor the quadratic expression -3p^2 + 66p + 9, you are looking for two binomial factors that can be multiplied together to give this expression.
Step 1: Check if there is a common factor you can factor out from all three terms. In this case, there is a common factor of 3.
-3p^2 + 66p + 9 can be written as 3(-p^2 + 22p + 3).
Step 2: Look for two binomial factors that can be multiplied together to give -p^2 + 22p + 3. To do this, find two numbers that multiply to give the constant term (3) and add up to give the coefficient of the middle term (22). In this case, those numbers are 1 and 3.
-p^2 + 22p + 3 can be rewritten as -p^2 + 3p + 19p + 3.
Step 3: Group the terms in pairs and factor out the common factors from each pair:
-p^2 + 3p + 19p + 3 can becomes -p(p - 3) + 19(p - 3).
Step 4: Notice that both groups now have a common factor of (p - 3), so factor it out:
-p(p - 3) + 19(p - 3) can be rewritten as (p - 3)(-p + 19).
So, the factored form of -3p^2 + 66p + 9 is (p - 3)(-p + 19).