Two polynomials P and D are given. Use either synthetic or long division to divide P(x) by D(x), and express P in the form P(x) = D(x) · Q(x) + R(x).

Question

P(x)= 6x⁴+4x³+5x², D(x) = 2x²+1

Answer 1

So, the answer, in P(x) = D(x)•Q(x) + R(x) form, is

Note that this adds nothing new to the original answer. I just wanted to see if it was possible to display the process the way it might look on a sheet of paper.

Answer 2

Hello Addison P.,

i'll run through the "long division" form. Synthetic division is essentially the same thing, but tabularized using just the coefficients. Once you understand the long form (which is a lot like long division of numbers), you'll be able to do the other.

So, write the division like a long division problem. Put the divisor (2x^2 + 1) on the left, then the division marker, then the starting polynomial (6x^4 + 4x^3 + 5x^2) under the division marker top bar.

Next, unlike in a numbers division problem, you will only look initially at the leading terms of the D and P, and divide them. (6x^4 / 2x^2) = 3x^2. Write that term (3x^2) as the first term in your quotient, because it is.

Nest, complete the first "take-away": multiply the 3x^2 by the entire expression of D, and write that, appropriately lined up by powers of x, under the P. You should have [(2x^2 + 1)*3x^2 =] 6x^4 + 3x^2. Did you remember to put the 3x^2 there under the 5x^2 term, NOT under the 4x^3!

Next, subtract the entire term (6x^4 + 3x^2) from P. So that's like adding -6x^4 - 3x^2. I suggest you draw your subtraction problem "line", and write the result on the next line underneath that. Neatness counts, because otherwise you may lose track of your terms!

You should have (0x^4) +4x^3 + 2x^2 at this point.

OK, now you generate the next term of the quotient: look at 2x^2 (leading term in D) and 4x^3 (leading, non-zero term in current "remainder"). Do the division, you should get 2x. That is the next term in the quotient. So add a "+2x" to the quotient line. Next, complete the second "take-away": multiply that 2x term by the entire D, and write below the current "remainer". You should write: [(2x^2 + 1)*2x =] 4x^3 + 2x. Subtract that entire line from the line above it, you should be calculating: [4x^3 + 2x^2 -4x^3 - 2x] = 2x^2 - 2x. Writ that expression down, in the appropriate place laterally, i.e. keep individual vertical columns for x^4 terms, x^3 terms, x^2 terms, and so on.

OK, now ready for [(2x^2 - 2x) / (2x^2 + 1)] step: you should have just a "+1" as the next term in your quotient. Subtract [(2x^2 + 1)] from your current "remainder"; gives you -2x - 1.

At this point you have NO terms in the remainder greater than or equal to the highest-power term of your divisor. You are done; your quotient = 3x^2+2x+1 and remainder of -2x-1.

Note here a curious byproduct (as it were) of polynomial vs. straight-number divisions: you can have a negative remainer. You just did! You don't have that in straight-number divisions.

-- Cheers, --Mr. d.