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The entire question is: Has the polynomial below been factored correctly and completely? x^5 - 16x^3 + 8x^2 - 128 x^3(x^2 - 16) + 8(x^2 - 16) (x^2 -16)(x^3 + 8)
My answer is: "No, the polynomial hasn't been factor correctly and completely. It hasn't been factored correctly because there isn't a variable (x) for every constant in the given polynomial." and I say that because at the end of 128, there is no x! and if the person wanted to solve it that way, the lowest variable there is, is x^2, not x^3
Also, it can be factored further
The polynomial has been correctly factored so far, but it is not fully factored since: 1) x^2 - 16 is the difference of two squares which can be factored, and 2) x^3 + 8 is the sum of two cubes which is also factorable.
I don't understand how it's fully factored since there is no X at the end of 128 so there is nothing in common between all those numbers..... @mathstudent55 @ivettef365
Well i understand how it can be factored further because that turns into difference of 2 perfect squares which gives me: (x-4)(x+2)(x+4)(x^2-2x+4)
Did I write it was fully factored? It's factored correctly so far, meaning that in the process of factoring it fully, no mistake has been made, but the process of factoring fully is not complete.
So no mistake was there? So my final answer is: No, the following polynomial has not been factored completely. It can be further factored to (x-4)(x+2)(x+4)(x^2-2x+4) because x^2-16 is the difference of 2 perfect squares, and can be further factored to (x+4)(x-4), and (x^3+8) is the sum of 2 perfect cubes, and can be factored to (x+2) (x^2 - 2x +4), giving (x + 2)(x - 2)(x + 4)(x^2 - 2x +4) the complete answer"
The difference of two squares part, x^2 - 16, give (x + 4)(x - 4) The sum of two cubes part, x^3 + 8, give (x + 2)(x^2 - 2x + 4)
Everything you wrote is correct except for the very last listing of the factors.
The correct factorization is (x + 4)(x - 4)(x + 2)(x^2 - 2x + 4)
I got the answer, thanks.
You put in a (x - 2) factor where it should be (x - 4) in your last line.