« Multiplying Polynomials

Zeros of Polynomials »

Example 1:

Suppose you are given two polynomials, and we want to divide one polynomial by another. One method is long division, a process similar to long division of two whole numbers. I will use an example as I explain each step along the way.

Suppose we want to divide x² + 3x + 5 by x + 1. Setup the long division as you would do with whole numbers, with the first polynomial (called the dividend) under the long division line, and the polynomial we are dividing by (called the divisor) on the left:

$x + 1\left){\vphantom{1{{x^2} + 3x + 5}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{{{x^2} + 3x + 5}}}$

Make sure you write the terms left to right from highest degree to lowest degree for both the dividend and the divisor.

The long division process goes as follows: Imagine taking just the highest degree term from the dividend (in our example, x2) and dividing it by the highest degree term of the divisor (in our example, x). The result is the first term of our "quotient". In our example, the result will be x. Usually, you should write the answer above the term of the same degree as the result:

$x + 1\mathop{\left){\vphantom{1{{x^2} + 3x + 5}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{{{x^2} + 3x + 5}}}} \limits^{\displaystyle x}$

Now, take the result and multiply it by the entire divisor:

$(x)(x + 1) = {x^2} + x$

Write this result below the dividend, making sure you line up each term of the result under the term in the dividend with the same degree:

$$x + 1\mathop{\left){\vphantom{1{{x^2} + 3x + 5}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{{{x^2} + 3x + 5}}}} \limits^{\displaystyle x}$$ $${x^2} + x$$

Now, we have to subtract our result x² + x from the dividend. One way to do this without losing track of signs is to reverse all the signs of the terms of our result and add like terms:

$$x + 1\mathop{\left){\vphantom{1{{x^2} + 3x + 5}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{{{x^2} + 3x + 5}}}} \limits^{\displaystyle x}$$ $$\frac{{ - {x^2} - x}}{{2x}}$$

Note that the first term will always cancel out (and possibly others will as well). After writing what is left over, bring down the next term in the dividend that we haven't used yet:

$$x + 1\mathop{\left){\vphantom{1{{x^2} + 3x + 5}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{{{x^2} + 3x + 5}}}} \limits^{\displaystyle x}$$ $$\underline { - {x^2} - x}$$ $$2x + 5$$

Now, we repeat the process of long division, taking the highest degree of our new polynomial (which is 2x) and dividing it by the highest degree term of the divisor (again, x); the result is 2. That is our second term of our quotient, and we write it as follows:

$$x + 1\mathop{\left){\vphantom{1{{x^2} + 3x + 5}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{{{x^2} + 3x + 5}}}} \limits^{\displaystyle x+2}$$ $$\underline { - {x^2} - x}$$ $$2x + 5$$

As before, multiply 2 by x + 1 and write the result below 2x + 5 (lining up like terms), switch the signs, then add:

$$x + 1\mathop{\left){\vphantom{1{{x^2} + 3x + 5}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{{{x^2} + 3x + 5}}}} \limits^{\displaystyle {x + 2}}$$ $$\underline { - {x^2} - x}$$ $$2x + 5$$ $$\underline { - 2x - 2}$$ $$3$$

We stop once we have no more terms to bring down. The result from the last step is the remainder. So the quotient is x + 2 and our remainder is 3.

It is typical to write the answer as follows:

$\frac{{{x^2} + 3x + 5}}{{x + 1}} = x + 2 + \frac{3}{{x + 1}}$

Example 2:

Divide the polynomials x⁴ + 3x² - 5 and x² + 4x .

We first write in long division form

$${x^2} + 4x\left){\vphantom{1{{x^4} + 3{x^2} - 5}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{{{x^4} + 3{x^2} - 5}}}$$

Next decide what we need to multiply x² by to get x⁴. Since x² * x² = x⁴ we can write

$${x^2} + 4x\mathop{\left){\vphantom{1{{x^4} + 3{x^2} - 5}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{{{x^4} + 3{x^2} - 5}}}} \limits^{\displaystyle {{x^2}}}$$

Next, we multiply x² + 7x and x².

$${x^2} + 4x\mathop{\left){\vphantom{1{{x^4} + 3{x^2} - 5}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{{{x^4} + 3{x^2} - 5}}}} \limits^{\displaystyle {{x^2}}}$$ $${x^4} + 4{x^3}$$

Now subtract to get and bring down the 3x² to get

$${x^2} + 4x\mathop{\left){\vphantom{1{{x^4} + 3{x^2} - 5}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{{{x^4} + 3{x^2} - 5}}}} \limits^{\displaystyle {{x^2}}}$$ $$\underline {{x^4} + 4{x^3}}$$ $$- 4{x^3} + 3{x^2}$$

We repeat this process until the degree of the remainder is less than the degree of the denominator.

$${x^2} + 4x\mathop{\left){\vphantom{1{{x^4} + 3{x^2} - 5}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{{{x^4} + 3{x^2} - 5}}}} \limits^{\displaystyle {{x^2} - 4x + 19}}$$ $$\underline {{x^4} + 4{x^3}}$$ $$- 4{x^3} + 3{x^2}$$ $$\underline { - 4{x^3} - 16{x^2}}$$ $$19{x^2}$$ $$\underline {19{x^2} + 76x - 5}$$ $$- 76x + 5$$

Example 3:

Divide the polynomials x⁵+ 5x⁴ + 9x³ + 11x² + 12x + 13 and x + 2

$$x + 2\mathop{\left){\vphantom{1{{x^5} + 5{x^4} + 9{x^3} + 11{x^2} + 12x + 13}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{{{x^5} + 5{x^4} + 9{x^3} + 11{x^2} + 12x + 13}}}} \limits^{\displaystyle {{x^4} + 3{x^3} + 3{x^2} + 5x + 2}}$$ $$\underline {{x^5} + 2{x^4}}$$ $$3{x^4} + 9{x^3}$$ $$\underline {3{x^4} + 6{x^3}}$$ $$3{x^3} + 11{x^2}$$ $$\underline {3{x^3} + 6{x^2}}$$ $$5{x^2} + 12x$$ $$\underline {5{x^2} + 10x}$$ $$2x + 13$$ $$\underline {2x + 4}$$ $$9$$