How do you divide #(3x^4 + 2x^3 - 11x^2 - 2x + 5)/ (x^2 - 2) # using polynomial long division?

Answer 1

#(3x^4+2x^3-11x^2-2x+5)/(x^2-2)=3x^2+2x-5+(2x-5)/(x^2-2)#

#3x^4+2x^3-11x^2-2x+5|x^2-2# you ask what is it #(3x^4)/x^2# and you get #3x^2#
(REMEMBER #3x^2#)
now yow duplicate #3x^2# to #x^2-2 # and get #3x^4-6x^2# and do: #3x^4+2x^3-11x^2-2x+5# #-# #3x^4+0x^3-6x^2+0x+0# #=# #0+2x^3-5x^2-2x+5# so we have now #2x^3-5x^2-2x+5# which is very exiting because we don't have the power of 4 anymore!

As we repeat this now, please attempt to comply with:

First step (write it down): #2x^3-5x^2-2x+5|x^2-2#
Second step (divided strongest power): #(2x^3)/(x^2)=2x#
(REMEMBER #2x#)
Third step (Second step times the divider): #2x*(x^2-2)=2x^3-4x#
Forth step (which is first step minus third step): #2x^3-5x^2-2x+5# #-# #2x^3+0x^2-4x+0# #=# #0-5x^2+2x+5#

--and once more

First step (write it down): #-5x^2+2x+5|x^2-2#
Second step (divided strongest power): #(-5x^2)/(x^2)=-5#
(REMEMBER #-5#)
Third step (Second step times the divider): #-5*(x^2-2)=-5x^2+10#
Forth step (which is first step minus third step): #-5x^2+2x+5# #-# #-5x^2+10# #=# #0+2x-5#
All REMEMBERs are the result: #3x^2+2x-5#
BUT in that case, #2x-5# could not be divided so it remains as #(2x-5)/(x^2-2)#

SoooooooooooooOOOOOOOOOOOOOOOOOO:)

#(3x^4+2x^3-11x^2-2x+5)/(x^2-2)=3x^2+2x-5+(2x-5)/(x^2-2)#

Let's make sure:

#(3x^2+2x-5+(2x-5)/(x^2-2))(x^2-2)=# #=3x^4-6x^2+2x^3-4x-5x^2+10+2x-5=# #=3x^4+2x^3-11x^2-2x+5#

Thus, everything is OK.

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Answer 2

To divide (3x^4 + 2x^3 - 11x^2 - 2x + 5) by (x^2 - 2) using polynomial long division, follow these steps:

  1. Arrange the dividend and divisor in descending order of exponents. Dividend: 3x^4 + 2x^3 - 11x^2 - 2x + 5 Divisor: x^2 - 2

  2. Divide the first term of the dividend by the first term of the divisor. (3x^4) / (x^2) = 3x^2

  3. Multiply the divisor by the quotient obtained in step 2. 3x^2 * (x^2 - 2) = 3x^4 - 6x^2

  4. Subtract the result obtained in step 3 from the dividend. (3x^4 + 2x^3 - 11x^2 - 2x + 5) - (3x^4 - 6x^2) = 2x^3 - 5x^2 - 2x + 5

  5. Bring down the next term from the dividend. 2x^3 - 5x^2 - 2x + 5

  6. Repeat steps 2-5 until all terms have been divided. (2x^3) / (x^2) = 2x 2x * (x^2 - 2) = 2x^3 - 4x

    (2x^3 - 5x^2 - 2x + 5) - (2x^3 - 4x) = -5x^2 + 2x + 5

    (-5x^2) / (x^2) = -5 -5 * (x^2 - 2) = -5x^2 + 10

    (-5x^2 + 2x + 5) - (-5x^2 + 10) = -8

  7. The quotient is the sum of the quotients obtained in each step. Quotient: 3x^2 + 2x - 5

  8. The remainder is the final result after all divisions. Remainder: -8

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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