Fill the blank with a constant, so that the resulting expression can be factored as the product of two linear expressions:
$\[3ab - 12a + 7b + \underline{~~~~}\]$
+738 Hi guest!
\( 3ab - 12a + 7b + \underline{~~~~}\)
So let's start off by noticing the signs. Since all the signs are positive except for -12a, that means that \(a\) has to be multiplied with a negative number.
This is what we have so far: \((\text{_}a+\text{_})(\text{_}b-\text{_})\) (I'm using the blank spaces to represent numbers we don't know yet)
Now, we see that there is a \(7b\) term. What do we have to multiply \(b\) by to get \(7b\)? It's \(7\). So we can put \(7\) in the blank on the right of \(a\).
Now we have: \((\text{_}a+7)(\text{_}b-\text{_})\)
We also know that there is a \(3ab\) term. Since there is \(-12a\) term, we have to multiply \(a\) by 3 since -12a is divisible by 3.
Now we have: \((3a+7)(\text{_}b-\text{_})\)
The only term we have to deal with is the \(-12a\) term. What do we have to multiply \(3a\) by to get \(-12a\)? It's \(4\)! So we can put the 4 on the blank.
Now we have: \((3a+7)(b-4)\)
All we have to do now is multiply it all out!
3a*b=3ab
3a*4=-12a
7*b=7b
7*-4=-28
Adding all the terms together, we see that the quadratic is: \(3ab-12a+7b-28\)
So the blank is \(\boxed{-28}\)
I hope this helped you, guest!
If you have any questions, don't hesitate to ask!
:)
