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s^2(t – u) – 9t^2(t – u)

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2 Answers

s^2(t – u) – 9t^2(t – u) =

Take out Greatest Common Factor of (t - u):

(t – u)(s^2 - (3t)^2) =

Use Difference of Squares next:

(t – u)(s + 3t)(s - 3t)

check:

(t – u)(s + 3t)(s - 3t) =

(t – u)(s(s - 3t) + 3t(s - 3t)) =

(t – u)(s^2 - 3st + 3st - (3t)^2)) =

(t – u)(s^2 - (3t)^2)) =

(t – u)s^2 - (t – u)(3t)^2) =

s^2(t – u) - 9t^2(t – u)   √
This is a polynomial with 2 terms. The minus sign in front of the 9 indicates subtraction. Everything before that minus sign is the first term and everything after that minus sign is the second term.
 
Each term consists of a series of things being multiplied together. When no other operation is clearly given, assume that multiplication is the operation. Let us consider the first term.
 
s^2(t - u) means s to the second power times some quantity in parentheses. In this case, the thing in parentheses is t - u.
 
Let us consider the second term.
 
9t^2(t - u) means that we multiply 9 times t to the second power times some quantity in parentheses. Again, the thing in parentheses is t - u
 
Both terms involve multiplying 2 or 3 different things together. Which of those things shows up in both terms?
 
s^2 only shows up in the first term so that won't be a common factor between both terms.
Likewise, 9 and t^2 only show up in the second term. However, (t-u) shows up in both terms and can be factored out.
 
(t-u)(s^2 - 9t^2) is the factored form.
 
Another way of approaching this problem is to make a new rule and do substitution. For example, I am going to decide that I don't want to write (t-u) so much, so I set up this rule: A = (t-u). Now, with this rule I can replace (t-u) with A every time it appears in the problem:
 
s^2A - 9 t^2 A
 
From this we can see that A appears in both terms and can be factored out:
 
A(s^2 - 9 t^2)
 
Finally, we go back to our rule and replace the A with what it really represents:
 
(t - u) (s^2 - 9 t^2) which is the same answer.
 
To check your work, use the distributive property to multiply your answer out an see if you get the original problem.
 
I hope that helps.
 
Good luck,
Robert