Factoring... again... (1 of 11)
\[12x^2-5x-3\] a=12 b=-5 c=-3 ac=-36 (1)(36) (2)(18) (3)(12) (4)(9) (6)(6)
Looks good so far :)
\[-9+4=-5/] \[-9*4=-36\]
Good
\[12x^2-(9+4)x-3\]
\[12x^2-9x+4x-3\]
\[3x(4x-3)+1(4x-3)\]
Looks really great so far!!
Oh but wait ...
Yes?
\(12x^2-(9+4)x-3\) Start with this step. This step is correct. Then continue from here but remember to DISTRIBUTE the negative.
\[12x^2-9x-4x-3\]
Very good. Now you may continue.
\[(12x^2-9x)-(4x-3)\]
\[3x(4x-3)-1(4x-3)\]
Ah, I see what you did now. Should have caught that one. You keep forgetting to properly replace the five with an equivalent expression. 5 = 9 - 4 not 9 + 4
Because 9 + 4 = 13 which is clearly not 5 so you have to be careful when replacing
\(12x^2-(9-4)x-3\) This is correct
What?
9 - 4 = 9 + 4 = Answer that above then we can continue
5 13
Okay, do you understand now which expression is the correct expression to replace with 5?
9-4
Very good. So now, can you continue from here? \(12x^2-(9-4)x-3\)
\[12x^2-9x-4x-3\]
Well, then you'd have the same expression as before. When you distribute the negative the signs within parenthesis flip flop.
The positive number becomes negative. The negative number becomes positive.
\[12x^2-9x+4x-3\]
There we go. Now you may finish finally.
Ugh... I feel dumb.. I’m still getting \[3x(4x-3)1(4x-3)\]
There is a plus missing. Otherwise it is correct.
\(\color{#0cbb34}{\text{Originally Posted by}}\) @EndersWorld Ugh... I feel dumb.. I’m still getting \[3x(4x-3)+1(4x-3)\] \(\color{#0cbb34}{\text{End of Quote}}\)
Yes, very good. Now factor out the binomial common to each factorization.
\[(3x+1)(4x-3)\]
I need to go.
Okay, well keep practicing. You're making huge progress. Never give up!
Join our real-time social learning platform and learn together with your friends!