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anonymous
 4 years ago
2. Divide. (2x3 – x2 – 24x + 12) ÷ (2x – 1) (1 point)x2 – 1
x2 – 12
2x – 1
3. Simplify. (1 point)
–2
4. What is the undefined value of ? (1 point)4
0
–3
–4
5. Multiply. (1 point)
6. Add. (1 point)
7. Add
(1 point)
8. Subtract. (1 point)1
0
–1
9. Solve. (1 point)x =
x =
x =
x =
10. Solve. (1 point)x = –4
x = 8
x = 6
x = –6
please hlp
anonymous
 4 years ago
2. Divide. (2x3 – x2 – 24x + 12) ÷ (2x – 1) (1 point)x2 – 1 x2 – 12 2x – 1 3. Simplify. (1 point) –2 4. What is the undefined value of ? (1 point)4 0 –3 –4 5. Multiply. (1 point) 6. Add. (1 point) 7. Add (1 point) 8. Subtract. (1 point)1 0 –1 9. Solve. (1 point)x = x = x = x = 10. Solve. (1 point)x = –4 x = 8 x = 6 x = –6 please hlp

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Have you done long division with polynomials before ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, you don't need to longdivide to do it. Just factor the larger polynomial by grouping. So if you group the first two terms as \[(2x^{3}x^{2})+(24x+12)\] now you can factor out \[x^{2}\] from the first group, and factor out 12 in the second group. You get \[x^{2}(2x1)12(2x1)\] Now because \[2x1\] is in both terms, then you can factor this out of each term and get \[(2x1)(x^{2}12)\] Therefore \[(2x^{3}x^{2}24x+12)\div (2x1) = \frac{2x^{3}x^{2}24x+12}{2x1}=\frac{(2x1)(x^{2}12)}{2x1}\] From this you can see that you can cancel a factor of \[2x1\] Therefore this becomes just \[x^{2}12\] which is the correct answer. I have no idea what the rest of the questions are asking, but you should try to do them yourself now that you have this much anyway.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What @AddemF described is called factoring by grouping. This allows you to factor the numerator and cancel any terms from the denominator that are common with them. I thought from the question you asked to divide but this simplifies to the same thing. Whichever method works best for you @jroop!
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