Asked by Anonymous
I really need help factoring these special Trinomials.
This question really puzzled me. I thought I had it correct, but when I checked the answer at the back of the book, I was wrong, and so I would like to know what was it that I had done wrong.
This was the question:
x^2-4x+4
This is how I solved it:
x^2-4x+4
x^2-2x-2x+4
x(x-2)-2(x+2)
(x-2)(x+2)
- I could not get any further than this. I thought that this would be correct, but the actual answer is
(x-2)^2
How did they get that?
If there is an easier way to solve this, then can you please show me how. Cause solving things like I have above is difficult.
This question really puzzled me. I thought I had it correct, but when I checked the answer at the back of the book, I was wrong, and so I would like to know what was it that I had done wrong.
This was the question:
x^2-4x+4
This is how I solved it:
x^2-4x+4
x^2-2x-2x+4
x(x-2)-2(x+2)
(x-2)(x+2)
- I could not get any further than this. I thought that this would be correct, but the actual answer is
(x-2)^2
How did they get that?
If there is an easier way to solve this, then can you please show me how. Cause solving things like I have above is difficult.
Answers
Answered by
Reiny
Your error is in this line of yours
x(x-2)-2(x+2)
notice when you multiply it back out you don't get your original, and you don't have a "common" factor.
should have been
x(x-2)-2(x-2) to get back your x^2 - 4x + 4
= (x-2)(x-2)
= (x-2)^2
x(x-2)-2(x+2)
notice when you multiply it back out you don't get your original, and you don't have a "common" factor.
should have been
x(x-2)-2(x-2) to get back your x^2 - 4x + 4
= (x-2)(x-2)
= (x-2)^2
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