Question

Long division method
3x

3 x square ​

Answer 1

Divide 2x3 – 9x2 + 15 by 2x – 5

First off, I note that there is a gap in the degrees of the terms of the dividend: the polynomial 2x3 – 9x2 + 15 has no x term. My work might get complicated inside the division symbol, so it is important that I make sure to leave space for a x-term column, just in case. (This is like a zero in, say, the hundreds place of the dividend holding that column open for subtractions under the long-division symbol.) I can create this space by turning the dividend into 2x3 – 9x2 + 0x + 15.

(This is a legitimate mathematical step. I've only added zero, so I haven't actually changed the value of anything.)

Now that I have all the "room" I might need for my work, I'll do the division. I start, as usual, with the long-division set-up:

long-division set-up

Dividing 2x3 by 2x, I get x2, so I put that on top. Then I multiply the x2 by the 2x – 5 to get 2x3 – 5x2, which I put underneath. Then I change the signs, add down, and carry down the 0x + 15 from the original dividend. This gives me –4x2 + 0x + 15 as my new bottom line:

(2x^3)/(2x) = x^2; (x^2)(2x – 5) = 2x^3 – 5x^2; (2x^3 – 9x^2 + 0x + 15) – (2x^3 – 5x^2) = –4x^2 + 0x + 15

Dividing –4x2 by 2x, I get –2x, which I put on top. Multiplying this –2x by 2x – 5, I get –4x2 + 10x, which I put underneath. Then I change the signs, add down, and carry down the +15 from the previous dividend. This gives me –10x + 15 as my new bottom line:

(–4x^2)/(2x) = –2x; (–2x)(2x – 5) = –4x^2 + 10x; (–4x^2 + 0x + 15) – (–4x^2 – 10x) = –10x + 15

Dividing –10x by 2x, I get –5, which I put on top. Multiplying –5 by 2x – 5, I get 10x + 25, which I put underneath. Then I change the signs and add down, which leaves me with a remainder of –10:

(–10x)/(2x) = –5; (–5)(2x – 5) = –10x + 25; (–10x + 15) – (–10x + 25) = –10

I need to remember to add the remainder to the polynomial part of the answer:

\mathbf{\color{purple}{\mathit{x}^2 - 2\mathit{x} - 5 + \dfrac{-10}{2\mathit{x} - 5}}}x

2

−2x−5+

2x−5

−10

Divide 4x4 + 1 + 3x3 + 2x by x2 + x + 2

First, I'll rearrange the dividend, so the terms are written in the usual order:

4x4 + 3x3 + 2x + 1

I notice that there's no x2 term in the dividend, so I'll create one by adding a 0x2 term to the dividend (inside the division symbol) to make space for my work.

long-division set-up: 4x^4 + 3x^3 + 0x^2 + 2x + 1 divided by x^2 + x + 2

Then I'll do the division in the usual manner. Dividing the 4x4 by x2, I get 4x2, which I put on top. Then I multiply through, and so forth, leading to a new bottom line:

(4x^4)/(x^2) = 4x^2; (4x^2)(x^2 + x + 2) = 4x^4 + 4x^3 + 8x^2; (4x^4 + 3x^3 + 0x^2 + 2x + 1) – (4x^4 + 4x^3 + 8x^2) = –x^3 – 8x^2 + 2x + 1, which is my new bottom line

Dividing –x3 by x2, I get –x, which I put on top. Then I multiply through, etc, etc:

(–x^3)/(x^2) = –x; (–x)(x^2 + x + 2) = –x^3 – x^2 – 2x; (–x^3 – 8x^2 + 2x + 1) – (–x^3 – x^2 – 2x) = –7x^2 + 4x + 1, which is my new bottom line

Dividing –7x2 by x2, I get –7, which I put on top. Then I multiply through, etc, etc:

(–7x^2)/(x^2) = –7; (–7)(x^2 + x + 2) = –7x^2 – 7x – 14; (–7x^2 + 4x + 1) – (–7x^2 – 7x – 14) = 11x + 15, which is my new bottom line

And then I'm done dividing, because the remainder is linear (11x + 15) while the divisor is quadratic. The quadratic can't divide into the linear polynomial, so I've gone as far as I can.

Then my answer is:

\mathbf{\color{purple}{4\mathit{x}^2 - \mathit{x} - 7 + \dfrac{11\mathit{x} + 15}{\mathit{x}^2 + \mathit{x} + 2}}}4x

2

−x−7+

x

2

+x+2

11x+15

To succeed with polyomial long division, you need to write neatly, remember to change your signs when you're subtracting, and work carefully, keeping your columns lined up properly. If you do this, then these exercises should not be very hard; annoying, maybe, but not hard.

You can use the Mathway widget below to practice finding doing long polynomial division. Try the entered exercise, or type in your own exercise. Then click the button and select "Divide Using Long Polynomial Division" to compare your answer to Mathway's.

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Hope it helps