help with dividing a polnomial expressions?

Question

can some one help me with this problem
i had to use the ^ because i cant do superscripts
divide:
x^4 + x^3 - 5x^2 + 13x - 6 by x^2 + 3x - 2

and can some one do this one also

x^4 - 3x^3 + 6x - 5 by x^2 - 2x + 2

Answer 1

It *is* awfully hard to write algebra expressions on Yahoo Answers, isn't it? Oh well, we do what we can.

Addressing your first question, let's write it out like a long division problem. You'll have to imagine the horizontal bar above the larger polynomial, because I can't format it properly here:

x^2 + 3x - 2 ) x^4 + x^3 - 5x^2 + 13x - 6

Our first step is to look at the first term in each polynomial, and divide the smaller into the larger. x^4 / x^2 = x^2, so the first term we write in our quotient (above the bar) is x^2. You can write it directly above the x^4, if you wish, or above the 5x^2 term, your preference.

Now, we're going to multiply that "partial quotient" by the entire divisor (x^2 + 3x - 2), which gives us x^4 + 3x^3 - 2x^2. Write that "partial product" below the first three terms of the original dividend (x^4 + x^3 - 5x^2). Our next step is to subtract the partial product: (x^4 + x^3 - 5x^2) - (x^4 + 3x^3 - 2x^2) = -2x^3 - 3x^2. We then drop down the next term from the dividend, the 13x term, and repeat:

-2x^3 / x^2 = -2x (add that term to our quotient)

-2x * (x^2 + 3x - 2) = -2x^3 - 6x^2 + 4x

(-2x^3 - 3x^2 + 13x) - (-2x^3 - 6x^2 + 4x) = 3x^2 + 9x

Finally, we drop down the last term from the dividend, -6, and repeat once more. I'm sorry that I can't write this out the way your math book can... I really hope Yahoo Answers creates some sort of way of inputting formulas, or at least monospacing text entry!

3x^2 / x^2 = 3 (so we add a +3 term to our quotient)

3 * (x^2 + 3x - 2) = 3x^2 + 9x - 6

(3x^2 + 9x - 6) - (3x^2 + 9x - 6) = 0... no remainder!

So our final quotient is x^2 - 2x + 3, with no remainder.
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Now, when doing the 2nd problem, you'll want to write the larger polynomial, the dividend, as x^4 - 3x^3 + 0x^2 + 6x - 5. We incorporate that extra 0x^2 term so that we never find ourselves trying to subtract unlike terms.

Sorry there's no better way to write this out on Yahoo Answers. Hope that helped anyway!

Answer 2

Using long division of a polynomial

(x^4 + x^3 - 5x^2 + 13x - 6) / (x^2 + 3x - 2) = x^2 - 2x + 3 with no remainder

(x^4 - 3x^3 + 0x^2 +6x - 5) / (x^2 - 2x + 2) = x^2 + 6x + 20 + (66x+35)/(x^2 - 2x + 2)

Answer 3

x^2 + 3x - 2 divide into x^4 + x^3 - 5x^2 + 13x - 6

You do it long division style. It's not easy to type here, the answer is x^2 -2x+3.

The second one doesn't come out clean.
The answer is x^2 - x -4with a remainder of 3 over x^2 - 2x + 2

Answer 4

(x^4 + x^3 - 5x^2 + 13x - 6)/(x^2 + 3x - 2)
((x^2 - 2x + 3)(x^2 + 3x - 2))/(x^2 + 3x - 2)
x^2 - 2x + 3

ANS : x^2 - 2x + 3

Couldn't get the next one to work.

I used www.quickmath.com to find out how to work it. By the way, lots of people use ^ for superscripts

Answer 5

I'll do the first.
the solution must be of the type: (x²+nx+3) to make the first and last terms come out.

So, we multiply your given (x²+3x-2) by (x²+3), add that to nx (x²+3x-2) and set all that equal to the polynomial.
the x^4 term and the -6 term now drop out. we're left with:
(3+n)x^3 + (3n+1)x² + (9-2n)x = x^3 - 5x² + 13x.
n=-2

So the solution is (x²-2x+3)

-edit-

I decided to do the second too and got conflicting answers for n. I don't think it divides evenly.

Answer 6

from the polynmial x^4+x^3-5x^2+13x-6
take out the polynomial x^2+3x-2 as common..........
then u'll get answer as x^2-2x+3

Answer 7

(2x-4)(x+5)^2 /(2x + 10) = (2x-4)(x+5)^2 /2(x + 5) = one million/2[ (2x-4)(x+5)] Did you get what what grew to alter into into finished as much as right here I basically factorized the denominator and then chanced on (x + 5) is person-friendly in backside and real So it cancelled one yet yet another. one million/2[ (2x-4)(x+5)] =one million/2[2(x -2)(x + 5) } = (x -2) (x + 5)

Answer 8

1) Q: x^2 - 2x + 3
R: -18x + 6

2) Q: x^2 - x
R: 8x - 5
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For how to solve.. refer you mathematics books or email me.