Divide by long division
divide x^3+4x^2-3x-12 by x^2-3
I am a bit confused. Please explain!!
look at this youtube clip, it resembles your problem
http://www.youtube.com/watch?v=FTRDPB1wR5Y
explain how to do 2x(-7)-10 divide (-2)=10 divide 5+(-7))
Sure! To divide the polynomial x^3+4x^2-3x-12 by x^2-3 using long division, you need to follow these steps:
Step 1: Write the dividend (x^3+4x^2-3x-12) and the divisor (x^2-3) as shown:
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x^2 - 3 | x^3 + 4x^2 - 3x - 12
Step 2: Divide the first term of the dividend by the first term of the divisor to get the first term of the quotient. In this case, divide x^3 by x^2 to get x.
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x^2 - 3 | x^3 + 4x^2 - 3x - 12
x
Step 3: Multiply the divisor by the first term of the quotient (x) and write the result below the dividend, aligned with the appropriate terms. In this case, multiply (x^2 - 3) by x, which gives x^3 - 3x.
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x^2 - 3 | x^3 + 4x^2 - 3x - 12
x^3 - 3x
Step 4: Subtract the result from step 3 from the dividend. In this case, subtract (x^3 - 3x) from (x^3 + 4x^2 - 3x - 12) to get 7x^2.
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x^2 - 3 | x^3 + 4x^2 - 3x - 12
x^3 - 3x
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7x^2
Step 5: Bring down the next term from the dividend. In this case, bring down (-12).
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x^2 - 3 | x^3 + 4x^2 - 3x - 12
x^3 - 3x
-----
7x^2
- 12
Step 6: Repeat steps 2-5 with the new dividend (7x^2 - 12). Divide the first term of this new dividend (7x^2) by the first term of the divisor (x^2) to get 7.
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x^2 - 3 | x^3 + 4x^2 - 3x - 12
x^3 - 3x + 7
-----
7x^2
- 12
Step 7: Multiply the divisor by the new term of the quotient (7) and write the result below the new dividend.
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x^2 - 3 | x^3 + 4x^2 - 3x - 12
x^3 - 3x + 7
-----
7x^2 - 21
Step 8: Subtract the result from step 7 from the new dividend. In this case, subtract (7x^2 - 21) from (7x^2 - 12) to get 9.
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x^2 - 3 | x^3 + 4x^2 - 3x - 12
x^3 - 3x + 7
-----
7x^2 - 21
- 7x^2 + 21
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-9
Step 9: Since there are no more terms to bring down, the resulting value is the remainder, which is -9.
Therefore, the quotient is x + 7 with a remainder of -9: (x + 7 - 9 / (x^2 - 3)
I hope this explanation helps! Let me know if you have any further questions.