Here's the question you clicked on:
jroop
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
Have you done long division with polynomials before ?
Yeah, you don't need to long-divide 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 \[x^{2}(2x-1)-12(2x-1)\] Now because \[2x-1\] is in both terms, then you can factor this out of each term and get \[(2x-1)(x^{2}-12)\] Therefore \[(2x^{3}-x^{2}-24x+12)\div (2x-1) = \frac{2x^{3}-x^{2}-24x+12}{2x-1}=\frac{(2x-1)(x^{2}-12)}{2x-1}\] From this you can see that you can cancel a factor of \[2x-1\] 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.
What @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!