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# Factoring completely with a common factor

We can factor quadratics by first pulling out a common factor so the result looks like a(x+b)(x+c). Created by Sal Khan.

## Want to join the conversation?

• Does a have to be greater than b when a≠b? For example, 8(x+2)(x+4) versus 8(x+4)(x+2).
• I didn't really understand what you were asking, but 8(x+2)(x+4) and 8(x+4)(x+2) are exactly the same thing. Remember that the order in which you multiply two or more things together doesn't change the final product.
a*b*c = b*a*c = c*a*b

Hope this helps.
• To use this method does a have to be greater than 1?
• It doesn't need to be greater than 1, but if it wasn't greater than 1 then it would be pointless. If the only greatest common factor was 1 for the expression, then you can't really factor the expression therefore making the expression prime.
• in shouldn't be 4((x-3)*(x+1))? i'm thinking:
4(x-3)*(x+1)
(4x-12)*(x+1)
4x^2-8x-12

also, i found strange how in the alternative path of the second example "3(-x^2+7x-10)". it seem like there is no answer to: a+b=7; a*b=-10. did i miss something?
• At Sal says that you can factor the thing(what should I call it? A trinomial multiplied by a constant), he says you can factor with the "x squared" being negative. So how can you factor the trinomial with the x squared leading term being negative? Thank you.
-3x^2+21x-30 = (-3x+6)(x-5)
Find 2 number that gives 90 when multiplied and 21 when added.
15 and 6
-3x^2+15x+6x-30
-3x(x-5)+6(x-5)
(-3x+6)(x-5)
Is this still correct?
When checked it still gives me the first -3x^2+21x-30. But the final answer way different than the one shown in the video
• You are correct, but yours is not the completely factored form of this polynomial. Sal factors out 3 at the very beginning, and you can see this in yours: in -3x+6, 3 is common.

Hope this helps.
• How do we 'reverse' the process? eg at , how do we apply the distributive property to end up with the original trinomial 4x^2-8x-12? Does the 4 multiply just the first set of brackets or both of them ? i.e. 4(x-3) and 4(x+1) or 4(x-3) and (x+1) remains as it is before expanding?
• To ´reverse´ the process the 4 is multiplied only to the first set of brackets. Once you multiply the 4 to the first set of brackets multiply the remaining 2 brackets together to get back to the original trinomial. Think of 4 as just a factor. For example, if you had 2x3x4, you would multiply from left to right or use the commutative property to multiply whichever numbers seem more compatible. The trinomial in the video is the same except with variables, and variables are just unknown numbers. You could have 4(x-3) first then (4x-12)(x+1) to get 4x^2-8x-12, or 4(x+1) first then (4x+4)(x-3) to get 4x^2-8x-12. You could even multiply (x+1) and (x-3) together first then multiply (x^2-2x-3) by 4 to get the original trinomial 4x^2-8x-12. All of these different combinations of multiplication work due to the commutative property.
• How do I do this with numbers that have no common factors? My maths teacher says it is possible but I do not understand how.
• Let me be more clear about my question: when the question asks for thee product of two binomials the question is x^2+8x+15 answer 4 and 6 or(x+3)and(x+5) how is this entered into the answer box ?
• what is the answer to 6x^2 -18x-60.

I got 6(x-60)(x+57)

followed the steps in the video best I could
(1 vote)
• If by "answer," you mean finding the roots, no, you are not correct.

First, take out the 6 because everything is divisible by 6. Thus:

6(X^2-3X-10)

Then, you can solve for the roots from there.

=6(X-5)(X+2)

You did not take out the 6 from all the numbers, only the X. Additionally, you made a mistake with the roots. I would recommend rewatching the video.
Keep up the good work!