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## dividing polynomials

4x-1/3x2-11x-26

Should it be (3x^2 - 11x - 26)/(4x - 1)?

Jackeria,

Looks like you are working with dividing and simplifying polynomials. The problem you listed looks like (4x - 1) / (3x^2 - 11x - 26). Before trying division, we always want to check if we can simplify or factor either the numerator or the denominator. Can you? Completing the square doesn't help here, nor does the quadratic formula. Nothing nice. Looks like we're doing Polynomial long division!

You can divide one polynomial P(x) by another Q(x) if the degree of P(x), that is, the highest power of x in P(x) is larger than the highest power in Q(x).

In this case, P(x) = 4x - 1. Q(x) is the denominator, (3x^2 - 11x - 26) The degree of P(x) is 1, and the degree of Q(x) is 2. So we can't divide P(x) / Q(x). We could divide the reciprocal, Q(x) / P(x), then flip that to get our answer. Then we could use Polynomial Long Division.

36x/4 - 41/16     r  -375/16

________________

(4x -1) | (3x^2 - 11x - 26)

- (3x^2 - 3/4 x)

--------------------

-41/4 x - 26

-(-41/4 x + 41/16)

-------------------

-374/16

Polynomial long division works like the long division you learned in school, except you have to keep track of the x's. Start with the largest terms, and work in.

What factor, F, times 4x * F = 3x^2?  If we divide, we get 3x/4. That's our first term. Multiply through, then subtract. Just like regular long division. This eliminates the 3x^2 term completely, and we can move to the next term.

Continuing in this way, we get to -375/16. We want our answer to only have positive powers of x in it, so this is our remainder. Our answer for Q(x)/P(x) is 3x/4 - 41/16  - (375/16)/(4x -1)

The reason we write (375/16) over (4x-1) is because it did not divide with 'evenly'. Just like 15/4 = 3 remainder 3/4 , or 11/2 = 5 remainder 1/2.

We are looking for P/Q, so we can just write the reciprocal (flip) of what we just found.

So (4x - 1) / (3x^2 - 11x - 26) = 1/[ 3x/4 - 41/16 - (375/16)/(4x -1)]

Hope this helps!

Clint V. Wyzant Tutor, Norman, OK