2x^3-18x-2x^2+18
We have to factor this by grouping and I got :
(2x^3-18x)-(2x^2+18)
2x(x^2-9)-2(x^2+9)
What's next? I also have to solve for x in a different problem. But first I need to completely factor and I know that there are more steps.
Thanks!
We have to factor this by grouping and I got :
(2x^3-18x)-(2x^2+18)
2x(x^2-9)-2(x^2+9)
What's next? I also have to solve for x in a different problem. But first I need to completely factor and I know that there are more steps.
Thanks!
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You have factored incorrectly to this point. (2x^3-18x)-(2x^2+18) is 2x^3-18x-2x^2-18), not 2x^3-18x-2x^2+18. The correct polynomial grouping is (2x^3-18x)-(2x^2-18). Factoring down further results in 2x(x^2-9)-2(x^2-9). Undoing the distributive property gives us:
(2x - 2)(x^2 - 9)
However, 2x-2 can be made 2(x-1) by factoring out 2 and (x^2 - 9) can be factored down to (x+3)(x-3) by difference of two squares. Your final factorization is:
2(x-1)(x+3)(x-3)
(2x - 2)(x^2 - 9)
However, 2x-2 can be made 2(x-1) by factoring out 2 and (x^2 - 9) can be factored down to (x+3)(x-3) by difference of two squares. Your final factorization is:
2(x-1)(x+3)(x-3)
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First you made a mistake when putting parentheses around second group.
Since you put negative outside these parentheses, you need to change sign of both terms. Therefore:
−2x² becomes 2x² inside parentheses (that's ok)
18 becomes −18 inside parentheses (this is where you went wrong)
2x³ − 18x − 2x² + 18
= (2x³ − 18x) − (2x² − 18)
= 2x (x² − 9) − 2 (x² − 9)
Now you can factor out (x² − 9), giving:
= (x² − 9) (2x − 2)
= (x − 3) (x + 3) * 2 (x − 1)
= 2 (x − 1) (x − 3) (x + 3)
Since you put negative outside these parentheses, you need to change sign of both terms. Therefore:
−2x² becomes 2x² inside parentheses (that's ok)
18 becomes −18 inside parentheses (this is where you went wrong)
2x³ − 18x − 2x² + 18
= (2x³ − 18x) − (2x² − 18)
= 2x (x² − 9) − 2 (x² − 9)
Now you can factor out (x² − 9), giving:
= (x² − 9) (2x − 2)
= (x − 3) (x + 3) * 2 (x − 1)
= 2 (x − 1) (x − 3) (x + 3)
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Your approach is a correct but more complicated way of doing this factoring.
1. You can reorder the expression to be: 2x^3-2x^2-18x+18.
2. Therefore you can group factor like this: 2x^2*(x-1) -18(x-1)
3. Then you can factor it all together: (2x^2-18)*(x-1)
4. But notice that you can factor a 2 from the first part of the previous expression: 2(x^2-9)*(x-1)
5. Then you see that there is a difference of squares which you can factor out to get your final answer: 2*(x-3)*(x+3)(x-1).
6. The Answer: 2*(x-3)*(x+3)(x-1).
1. You can reorder the expression to be: 2x^3-2x^2-18x+18.
2. Therefore you can group factor like this: 2x^2*(x-1) -18(x-1)
3. Then you can factor it all together: (2x^2-18)*(x-1)
4. But notice that you can factor a 2 from the first part of the previous expression: 2(x^2-9)*(x-1)
5. Then you see that there is a difference of squares which you can factor out to get your final answer: 2*(x-3)*(x+3)(x-1).
6. The Answer: 2*(x-3)*(x+3)(x-1).
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2x^3 - 18x - 2x^2 + 18
= 2x^3 - 2x^2 - 18x + 18
= (2x^2)(x - 1) - 18(x - 1)
= (x - 1)(2x^2 - 18)
= 2(x - 1)(x^2 - 9)
= 2(x - 1)(x^2 + 3x - 3x - 9)
= 2(x - 1)(x(x + 3) - 3(x + 3))
= 2(x - 1)(x + 3)(x - 3)
= 2x^3 - 2x^2 - 18x + 18
= (2x^2)(x - 1) - 18(x - 1)
= (x - 1)(2x^2 - 18)
= 2(x - 1)(x^2 - 9)
= 2(x - 1)(x^2 + 3x - 3x - 9)
= 2(x - 1)(x(x + 3) - 3(x + 3))
= 2(x - 1)(x + 3)(x - 3)
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2x^3 - 18x - 2x^2 + 18
= 2x^3 - 2x^2 - 18x + 18
= 2x^2(x - 1) - 18(x - 1)
= 2(x - 1)(x + 3)(x - 3)
= 2x^3 - 2x^2 - 18x + 18
= 2x^2(x - 1) - 18(x - 1)
= 2(x - 1)(x + 3)(x - 3)
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2x(x^2 - 9) - 2(x^2 - 9) = (2x - 2)(x + 3)(x - 3) = 2(x - 1)(x + 3)(x - 3)