Polynomials can be expressed in two primary ways: factored form and standard form. Each form exposes different properties of the function and can be converted from one to the other.

Factored form of a polynomial, which is the product of linear terms, is expressed as f\left(x\right) = a\left(x - r_1\right)\left(x - r_2\right)...\left(x - r_n\right), where r_1, r_2, ..., r_n are the zeros of the function. A benefit of factored form is that the real zeros, found by setting each factor equal to zero and solving for x, correspond to the x-intercepts on the graph of the function. Converting a polynomial to factored form allows us to easily identify these real zeros. To convert a polynomial from factored form to standard form, we multiply out the factors and combine like terms. This is an application of the distributive property.

The standard form of a polynomial is a_n x^n + a_{n - 1} x^{n - 1} + \ldots + a_1 x + a_0, where n is a non-negative integer and each a_i is a coefficient.

A polynomial expression labeled with its parts written as: p of x is equal to a sub n times x to the nth power plus a sub n minus 1 times x to the power of n minus 1 plus ellipsis plus a sub 2 times x squared plus a sub 1 times x plus a sub 0. a sub n times x to the nth power is the leading term, a sub n is the leading coefficient and the power n is the degree. In the term a sub n minus 1 times x to the power of n inus 1, a sub n minus 1 is the coefficient. The term a sub 2 times x squared is the quadratic term, a sub 1 times x is the linear term, and a sub 0 is the constant term.

This form reveals the end behavior of the function, determined by the sign and degree of the leading term a_nx^n. It also reveals the y-intercept which is a_0.

To multiply polynomials, we apply the distributive property which will require us to use the product of powers property of exponents when multiplying variables:

General example	\left(a_{n}x^{n}+\ldots+a_{0}\right)\left(b_{m}x^{m}+\ldots +b_{0}\right)=\\ \left(a_{n}b_{m}\right)x^{n+m}+\ldots+\left(a_{n}b_0\right)x^n+\ldots+\left(a_{0}b_m\right)x^m+\ldots+\left(a_0b_0\right)
Numerical example	\left(5x^2+2x+1\right)\left(x^2-3x+5\right)= 5x^4-13x^3+20x^2+7x+5

Because n and m were non-negative integers, n+m will also be non-negative. The exponents will still be constants, so the result is another polynomial.

Examples

Example 1

Write a polynomial expression in standard form for the perimeter of the rectangle shown.

A rectangle with a length of 1.6 x plus 9 halves y and width of one half x plus 0.4 y.

Worked Solution

Create a strategy

The perimeter of a shape is the sum of its side lengths. In this case, the shape is a rectangle, so we can add the two labeled side lengths and then double the result.

Apply the idea

\displaystyle P	\displaystyle =	\displaystyle 2\left(l+w\right)	Formula for perimeter
	\displaystyle =	\displaystyle 2\left(1.6x + \dfrac{9}{2}y + \dfrac{1}{2}x + 0.4y\right)	Substitute expressions for the length and width
	\displaystyle =	\displaystyle 2\left(1.6x + 4.5y + 0.5x + 0.4y\right)	Rewrite the fractions as decimals
	\displaystyle =	\displaystyle 2\left(2.1x + 4.9y\right)	Combine like terms
	\displaystyle =	\displaystyle 4.2x + 9.8y	Distributive property

Reflect and check

We could alternatively have doubled each side length first, then added the two results:

\displaystyle \text{Perimeter}	\displaystyle =	\displaystyle 2\left(1.6x + 4.5y\right) + 2\left(0.5x +0.4y\right)
	\displaystyle =	\displaystyle 3.2x + 9y + x + 0.8y	Distributive property
	\displaystyle =	\displaystyle 4.2x + 9.8y	Combine like terms

Doing so gives the same result for the perimeter.

Example 2

Consider the polynomial function h\left(a\right)=\left(3 + 2a\right)^3 .

Rewrite h\left(a\right) in standard form.

Worked Solution

Create a strategy

To simplify this expression we can rewrite the exponent as repeated multiplication, then carefully multiply each term in one factor by each term in the next factor.

Apply the idea

\displaystyle h\left(a\right)	\displaystyle =	\displaystyle \left(3 + 2a\right)^3
	\displaystyle =	\displaystyle \left(3 + 2a\right)\left(3 + 2a\right)\left(3 + 2a\right)	Rewrite the exponent as multiplication
	\displaystyle =	\displaystyle \left(9 + 12a + 4a^2\right)\left(3 + 2a\right)	Multiply the first two factors as a PST
	\displaystyle =	\displaystyle 27 + 36a + 12a^2 + 18a + 24a^2 + 8a^3	Multiply the two remaining factors
	\displaystyle =	\displaystyle 8a^3 + 36a^2 + 54a + 27	Combine like terms

Reflect and check

Notice that the leading term (the term with the highest power of the variable) for this polynomial is 8a^3, even though this is not the first term that is written.

Identify the y-intercept of h\left(a\right).

Worked Solution

Create a strategy

The y-intercept is revealed by looking at the constant term of the function when written standard form.

Apply the idea

From part (a) we have the standard form h\left(a\right) = 8a^3 + 36a^2 + 54a + 27. The constant term is 27, so the y-intercept is \left(0,27\right).

Reflect and check

We can confirm this by substituting 0 for a and evaluating.

h\left(0\right) = 8\left(0\right)^3 + 36\left(0\right)^2 + 54\left(0\right) + 27 = 0 + 0 + 0 + 27 = 27.

Describe the end behavior of h\left(x\right).

Worked Solution

Create a strategy

Use the fact that the leading coefficient is positive and the function has an odd degree to determine the end behavior.

Apply the idea

\lim_{x \to \infty} h\left(x\right) = \infty \\ \lim_{x \to -\infty} h\left(x\right) = -\infty

Idea summary

We can use the distributive property to convert a polynomial in factored form to standard form.

Factored form of polynomials

Factoring a polynomial is a process of expressing a polynomial as a product of its factors. In other words, it is the inverse process of multiplying polynomials.

We have learned several different strategies, including a few identities, that can help us factor a polynomial.

Factoring using Greatest Common Factor (GCF):
A greatest common factor from each term of a polynomial is factored out ax+ay+\ldots=a(x+y+\ldots)
Factoring by grouping:
A method for factoring an expression containing at least four terms, by grouping the terms in pairs and taking out common factors ax + ay + bx + by = a\left(x + y\right) + b\left(x + y\right) = \left(x+y\right)\left(a+b\right)
Factoring quadratic trinomials:
A trinomial that can be expressed as the product of two binomials ax^{2} + bx + c= \left(mx + p\right)\left(nx + q\right) where mn=a,pq=c and np+mq=b
Perfect square trinomials:
A trinomial that is formed by multiplying a binomial by itself a^{2} + 2 a b + b^{2} = \left(a + b\right)^{2} \text{ or } a^{2} - 2 a b + b^{2} = \left(a - b\right)^{2}
Difference of two squares:
The result of a perfect square being subtracted from another perfect square a^{2} - b^{2} = \left(a+b\right)\left(a-b\right)
Sum of two cubes:
Two perfect cube expressions being added a^{3} + b^{3} = \left(a+b\right)\left(a^2-ab+b^2\right)
Difference of two cubes:
The result of a perfect cube being subtracted from another perfect cube a^{3} - b^{3} = \left(a-b\right)\left(a^2+ab+b^2\right)

Examples

Example 3

Considerg\left(x\right)=6 x^{4} - 10x^{3} - 24x^{2} in factored form.

Rewrite g\left(x\right) in factored form.

Worked Solution

Create a strategy

First, we look for the greatest common factor for all three terms and factor it out. Then, from the simpler trinomial, we find the values of r and s that multiply to ac and add to b. After finding those values, we write the trinomial in the form ax^{2} + rx + sx + c and factor it by grouping.

Apply the idea

The greatest common factor of the three terms is 2x^{2}. We factor out 2x^{2} and get g\left(x\right)=2x^{2}\left(3 x^{2} - 5x - 12\right)

For 3 x^{2} - 5x - 12, we find two numbers that have a product of ac = \left(3\right)\left(-12\right) = -36 and sum of b = -5.

The factors of -36 include \pm1, \pm2, \pm3,\pm4,\pm6,\pm9,\pm18 and \pm36. Among these factors, 4 and -9 are the only numbers that add to -5 and multiply to -36. We can let r=4 and s=-9. Next, we write 3 x^{2} - 5x - 12 in the form ax^{2} + rx + sx + c. Substituting the values for r and s, we get

g\left(x\right)=3x^{2} +4x -9 x - 12

Now, we factor this expression by grouping

\displaystyle 3x^{2} +4x -9x -12	\displaystyle =	\displaystyle (3x^{2} +4x) +(-9x -12)	Group based on common factors
	\displaystyle =	\displaystyle x\left(3x + 4\right) - 3\left(3x + 4\right)	Factor out each GCF (x and -3)
	\displaystyle =	\displaystyle \left(x-3\right)\left(3x+4\right)	Factor out the common binomial factor

So the factored form is, g\left(x\right)=2x^{2}\left(x-3\right)\left(3x+4\right).

Reflect and check

We can check the answer by multiplying the factored form 2x^{2}\left(x-3\right)\left(3x+4\right).

\displaystyle 2x^{2}\left(x-3\right)\left(3x+4\right)	\displaystyle =	\displaystyle 2x^{2}\left(x\left(3x+4\right)-3\left(3x+4\right)\right)	Distributive property
	\displaystyle =	\displaystyle 2x^{2}\left(3x^{2} +4x -9x - 12\right)	Distributive property
	\displaystyle =	\displaystyle 6 x^{4} + 8x^{3} -18x^{3} - 24x^{2}	Distributive property
	\displaystyle =	\displaystyle 6 x^{4} - 10x^{3} - 24x^{2}	Combine like terms

Identify the zeros of g\left(x\right) and their multiplicities.

Worked Solution

Create a strategy

Set each factor found in part (a) equal to 0 and solve. The exponents on each factor indicate the multiplicity.

Apply the idea

In part (a) we found the factored form: g\left(x\right)=2x^{2}\left(x-3\right)\left(3x+4\right). We can find the zeros by setting each of these factors equal to zero and solving.

2x^2=0 gives x=0. The exponent tells us the multiplicity is 2.

x-3=0 gives x=3. The lack of exponent tells us the multiplicity is 1.

3x+4=0 gives x=-\dfrac{4}{3}. The lack of exponent tells us the multiplicity is 1.

Idea summary

The factoring methods and identities we can use to fully factor polynomials are

Factoring using the GCF
Factoring by grouping
Factoring quadratic trinomials
Perfect square trinomial identity
Difference of two squares identity
Sum of two cubes identity
Difference of two cubes identity

Polynomial division

Long division:

In general, when dividing polynomials where the divisor is not a monomial, we can use the process of polynomial long division.

When we divide a dividend, p\left(x\right), by a divisor, b\left(x\right), we can write the expression as the sum of the quotient, q\left(x\right), and the remainder, r\left(x\right), divided by the divisor. The notation for this is {\dfrac{p\left(x\right)}{b\left(x\right)}=q\left(x\right)+\dfrac{r\left(x\right)}{b\left(x\right)}}

If our division was done correctly, then p\left(x\right) = q\left(x\right)b\left(x\right) + r\left(x\right). Polynomial long division works in a very similar way to long division with whole numbers where we:

Divide the first terms of the dividend and divisor
Multiply the answer by the divisor
Subtract the product from the first part of the dividend
Bring down the next part of the dividend
Repeat steps 1-4 until there are no terms left to bring down

Note: Before performing long division, the terms of the divisor and dividend should first be arranged in descending order of exponents. In cases where there is no term corresponding to an exponent in the dividend or divisor, we use a placeholder term with a coefficient of 0.

For example, the long division of (2x^3 - 5x + 7)\div(x - 1) is shown below.

A figure showing the polynomial long division for 2 x cubed plus 0 x squared minus 5 x plus 7 divided by x minus 1. The expression 2 x cubed plus 0 x squared minus 5 x plus 7 is labeled Dividend, x minus 1 labeled Divisor, 2 x squared plus 2 x minus 3 labeled Quotient, and 4 labeled Remainder. Speak to your teacher for more information.

Step 1: 2x^3 \div x=2x^2

Step 2: 2x^2\left(x-1\right)=2x^3-2x^2

Step 3: \left(2x^3+0x^2\right)-\left(2x^3-2x^2\right)=2x^2

Step 4: Bring down -5x

Steps 1-4 are repeated two more times to find the quotient and the remainder.

For the example above, the dividend is p\left(x\right)=2x^3-5x+7, the divisor is b\left(x\right) =x-1, and we found the quotient to be q(x) =2x^2+2x-3 and remainder r(x) = 4. So, our final answer is \dfrac{2x^3-5x+7}{x-1}=2x^2+2x-3+\dfrac{4}{x-1}

Synthetic division:

There is an efficient method of polynomial division known as synthetic division that can be used only when the divisor is a linear expression in the form x\pm a, where a is a constant.

Consider the long division and synthetic division of \dfrac{3x^3+17x^2+6x-26}{x+5}:

Long division

Synthetic division

Synthetic division is a short-hand notation version of long division where only the coefficients are used. To begin the synthetic division, we write the coefficients of the terms in the divinded inside an upside down division table. To find the number that goes outside the table, we set the linear divisor equal to zero and solve for x.\begin{aligned}x+5&=0\\x&=-5\end{aligned}

Once the synthetic division is set up, we always bring the first number down. Then, we follow these steps:

Multiply the first number below the table by the number outside the table
Write the result under the second coefficient from the dividend
Add the numbers in the second column
Repeat until the last column has been added

A figure showing the partial steps of synthetic division for 3 x cubed plus 17 x squared plus 6 x minus 26 divided by x plus 5. Speak to your teacher for more information.

Step 1: Multiply 3\times -5

Step 2: Write -15 under 17

Step 3: Add 17+-15

Step 1: Multiply 2\times -5

Step 2: Write -10 under 6

Step 3: Add 6+-10

A figure showing the steps of synthetic division for 3 x cubed plus 17 x squared plus 6 x minus 26 divided by x plus 5. Speak to your teacher for more information.

Step 1: Multiply -4\times -5

Step 2: Write 20 under -26

Step 3: Add -26+20

The answers beneath the table will be the coefficients of the quotient, and the last number is always the remainder. Because we are dividing by a linear term, the degree of the quotient will be one less than the degree of the divisor.

Therefore, the quotient for the example is q\left(x\right)=3x^2+2x+4 and the remainder is r\left(x\right)=-6, just like we found when we solved using long division.

Examples

Example 4

Write \dfrac{x^3 + 7 x^2 + 14 x + 3}{x + 2} in terms of the quotient and remainder by using long division.

Worked Solution

Create a strategy

First, we need to check that the terms are arranged in descending order of exponents. In this problem, the terms in the dividend and divisor are in the correct order.

Next, we need to check that there are no missing terms. To do that, we look at the exponents. No terms are missing because the exponents in the dividend are 3,2,1,0 and the exponents in the divisor are 1, 0.

Apply the idea

We set up the long division with the dividend under the table and the divisor outside the table.

A figure showing a polynomial long division set up: the dividend x cubed plus 7 x squared plus 14 x plus 3 under the table and the divisor x plus 2 outside the table.

To begin the division, we only need to consider the first terms of the dividend and divisor.

A figure showing partial steps of the polynomial long division for x cubed plus 7 x squared plus 14 x plus 3 divided by x plus 2. Speak to your teacher for more information.

Dividing these, we get x^3\div x=x^2, and we write this above the table.

Then, we multiply the result by the divisor, x^2\left(x+2\right)=x^3+2x^2, and write this below the dividend.

Next, we subtract and bring down the next term. When subtracting, remember to distribute the negative sign across all terms.

Now, we repeat the steps from above.

Dividing the first terms, we get 5x^2 \div x =5x.

Multiplying this result by the divisor, we have 5x\left(x+2\right)=5x^2+2x.

Next, we subtract and bring down the next term.

Since we brought down the last term in the dividend, this will be the last time we repeat the steps.

A figure showing the polynomial long division for x cubed plus 7 x squared plus 14 x plus 3 divided by x plus 2. Speak to your teacher for more information.

Dividing the first terms gives us 4x \div x=4.

Multiplying this by the divisor gives us 4\left(x+2\right)=4x+8

Subtracting that from 4x+3 leaves a remainder of -5.

The long division gives us a result of x^2 + 5x + 4 as the quotient and -5 as the remainder. So we can rewrite the expression as \frac{x^3 + 7 x^2 + 14 x + 3}{x + 2} = x^2 + 5x + 4 + \frac{-5}{x + 2}

Reflect and check

The parentheses included in each of the subtraction steps are very important because it reminds us to distribute the negative sign to all terms. Not distributing the negative and adding the terms instead is a common mistake, so remember to always include the parentheses.

Example 5

Rewrite \dfrac{2x^3 - 3x^2 + 4x - 1}{x + 1}as the sum of the quotient and a remainder fraction by using synthetic division.

Worked Solution

Create a strategy

For synthetic division, we still need to check that the terms are arranged in descending order of exponents and there are no missing terms. If there is a missing term, we write a 0 coefficient in its place.

This problem does not have missing terms because the exponents in the dividend are 3,2,1,0 and the exponents in the divisor are 1,0.

Apply the idea

We set up the synthetic division by writing the coefficients of the dividend inside an upside down division table. Then, we find the number that goes outside the table by setting the divisor equal to zero and solving for x:

\begin{aligned}x+1&=0\\x&=-1\end{aligned}

Now, we bring down the first number and follow these steps until the last column has been added:

Multiply the number below the table by the number outside the table
Write the result under the next coefficient from the dividend
Add the numbers in the next column

A figure showing a synthetic division with negative 1 in the middle left corner. Top row has digits 2, negative 3, 4, negative 1. Middle row has negative 2 under the negative 3, 5 under the 4, and negative 9 under negative 1. Bottom row has 2, negative 5, 9, and negative 10.

These numbers below the table are the coefficients of the quotient. Remember the quotient will be one degree less than the degree of the dividend. The degree of the dividend is 3, so the exponents of the variable will begin from 2 and decrease by one until we get to the remainder.

Writing in the variables, we get a result of 2x^2 - 5x + 9 as the quotient and -10 as the remainder. So we can rewrite the expression as \frac{2x^3 - 3x^2 + 4x - 1}{x + 1} = 2x^2 - 5x + 9 - \frac{10}{x + 1}

Reflect and check

We can check the answer by determining if the dividend is equal to the product of the quotient and the divisor plus the remainder.

\displaystyle \left( \text{Quotient} \times \text{Divisor} \right) + \text{Remainder}	\displaystyle =	\displaystyle \text{Dividend}	Formula
\displaystyle \left[\left(2x^{2} - 5x + 9\right) \left(x + 1\right) \right]- 10	\displaystyle =	\displaystyle \left(2x^3+2x^2-5x^2-5x+9x+9\right)-10	Expand
	\displaystyle =	\displaystyle 2x^3-3x^2+4x-1	Combine like terms

This polynomial is the same as the dividend, so we know we did the synthetic division correctly.

Example 6

Determine an appropriate method for dividing the following expressions. Explain your choice.

\dfrac{3x^4-6x^3+19x^2-8x+20}{3x^2+4}

Worked Solution

Create a strategy

The denominator of this expression is non-linear, and it cannot be factored. Therefore, long division is the most efficient method that can be used to evaluate this division.

Apply the idea

The first thing we need to do is make sure the terms are in descending order and no terms are missing. The dividend is not missing any terms, but the divisor is missing an x-term. We need to add a 0 term, so the divisor becomes 3x^2+0x+4.

Now, we can divide the polynomials as follows:

A figure showing the polynomial long division for 3 x raised to the fourth power minus 6 x cubed plus 19 x squared minus 8 x plus 20 divided by 3 x squared plus 0 x plus 4. Speak to your teacher for more information.

Therefore, \dfrac{3x^4-6x^3+19x^2-8x+20}{3x^2+4}=x^2-2x+5.

Reflect and check

If we multiply both sides of the answer by 3x^2+4, we can show that the dividend is equal to the product of the divisor and the quotient. 3x^4-6x^3+19x^2-8x+20=\left(3x^2+4\right)\left(x^2-2x+5\right)

This division could have been done with algebraic manipulation, but it would have been difficult to see.

\displaystyle \dfrac{3x^4-6x^3+19x^2-8x+20}{3x^2+4}	\displaystyle =	\displaystyle \dfrac{3x^4+19x^2+20-6x^3-8x}{3x^2+4}	Rearrange the terms
	\displaystyle =	\displaystyle \dfrac{3x^4+19x^2+20-2x\left(3x^2+4\right)}{3x^2+4}	Factor -2x from last two terms
	\displaystyle =	\displaystyle \dfrac{3x^4+4x^2+15x^2+20-2x\left(3x^2+4\right)}{3x^2+4}	Rewrite 19x^2 as {4x^2+15x^2}
	\displaystyle =	\displaystyle \dfrac{x^2\left(3x^2+4\right)+5\left(3x^2+4\right)-2x\left(3x^2+4\right)}{3x^2+4}	Factor by grouping
	\displaystyle =	\displaystyle \dfrac{x^2\left(3x^2+4\right)}{3x^2+4}+\dfrac{5\left(3x^2+4\right)}{3x^2+4}+\dfrac{-2x\left(3x^2+4\right)}{3x^2+4}	Separate the fractions
	\displaystyle =	\displaystyle x^2+5-2x	Reduce by 3x^2+4
	\displaystyle =	\displaystyle x^2-2x+5	Rearrange the terms

\left(-x^4+16x^2+2x\right)\div \left(x-4\right)

Worked Solution

Create a strategy

All terms have a GCF of -x, so we can try factoring this: \dfrac{-x\left(x^3-16x-2\right)}{x-4}

However, the resulting trinomial cannot be factored further, making the algebraic manipulation more difficult and therefore, less efficient. Because the divisor is linear, synthetic division is the most efficient method for dividing these polynomials.

Apply the idea

To set up the synthetic division, we need to determine which terms are missing in the dividend and add a 0 coefficient in their place. Notice we are missing the x^3-term and the constant, so the coefficients in the table will be -1,0,16,2,0.

Next, we set the divisor equal to zero and solve for x:\begin{aligned}x-4&=0\\x&=4\end{aligned}

Now, we can perform the synthetic division:

A figure showing a synthetic division with 4 in the top left corner. Top row has digits negative 1, 0, 16, 2, 0. Middle row has negative 4 under the leftmost 0, negative 16 under the 16, 0 under the 2, and 8 under the rightmost 0. Bottom row has negative 1, negative 4, 0, 2, and 8.

The degree of the quotient will be one less than the degree of the dividend. Since the degree of the dividend is 4, the exponents of the terms in the quotient will begin from 3 and decrease by one.

Therefore, the quotient is -x^3-4x^2+2 and the remainder is -8. We can also write this as \dfrac{-x^4+16x^2+2x}{x-4}=-x^3-4x^2+2+\dfrac{8}{x-4}

Reflect and check

As we discussed when creating a strategy for this problem, algebraic manipulation for this one is possible. The easiest way to begin is to factor -x^2 from the first two terms:

\displaystyle \dfrac{-x^4+16x^2+2x}{x-4}	\displaystyle =	\displaystyle \dfrac{-x^2\left(x^2-16\right)+2x}{x-4}	Factor -x^2 from first two terms
	\displaystyle =	\displaystyle \dfrac{-x^2\left(x-4\right)\left(x+4\right)+2x}{x-4}	Factor difference of squares
	\displaystyle =	\displaystyle \dfrac{-x^2\left(x-4\right)\left(x+4\right)}{x-4}+\dfrac{2x}{x-4}	Separate the fractions
	\displaystyle =	\displaystyle -x^2\left(x+4\right)+\dfrac{2x}{x-4}	Reduce by x-4

Looking at the remainder, we see the degrees of the numerator and denominator are the same. This tells us that the remainder can actually be divided further. Let's use synthetic division to further simplify the remainder.

A figure showing a synthetic division with 4 in the top left corner. Top row has digits 2 and 0. Middle row has 8 under the 0. Bottom row has 2 and 8.

In other words, \dfrac{2x}{x-4}=2+\dfrac{8}{x-4}. Continuing where we left off:

\displaystyle -x^2\left(x+4\right)+\dfrac{2x}{x-4}	\displaystyle =	\displaystyle -x^2\left(x+4\right)+2+\dfrac{8}{x-4}	Rewrite \dfrac{2x}{x-4}
	\displaystyle =	\displaystyle -x^3-4x^2+2+\dfrac{8}{x-4}	Distributive property

\dfrac{4x^2-24x+36}{2x-6}

Worked Solution

Create a strategy

The first method we should check for is factoring, since that is usually the quickest method. Always look for a GCF before attempting to factor using any other method.

The numerator has a GCF of 4 and a GCF of 2 is in the denominator. We will factor these out, then try to factor the numerator further.

Apply the idea

\displaystyle \dfrac{4x^2-24x+36}{2x-6}

\displaystyle =

\displaystyle \dfrac{4\left(x^2-6x+9\right)}{2\left(x-3\right)}

Factor out the GCFs

Once the GCF has been factored out of the numerator, it is easier to see that this quadratic can be factored further. This is a perfect square trinomial, so we can use that identity to factor:

\displaystyle \dfrac{4\left(x^2-6x+9\right)}{2\left(x-3\right)}	\displaystyle =	\displaystyle \dfrac{4\left(x-3\right)\left(x-3\right)}{2\left(x-3\right)}	Factor the perfect square trinomial
	\displaystyle =	\displaystyle 2\left(x-3\right)	Reduce by common factor of 2\left(x-3\right)
	\displaystyle =	\displaystyle 2x-6	Distributive property

When 4x^2-24x+36 is divided by 2x-6, the result is 2x-6.

Reflect and check

From this result, we know 4x^2-24x+36=\left(2x-6\right)^2 which we can use to check the answer. Using the perfect square trinomial identity \left(A+B\right)^2=A^2+2AB+B^2 with A=2x and B=-6:

\displaystyle \left(2x-6\right)^2

\displaystyle =

\displaystyle 4x^2-24x+36

Perfect square trinomial expansion

We find the product of the quotient and the divisor is, in fact, equal to the dividend.

\dfrac{4x^2+6x-8}{2x+3}

Worked Solution

Create a strategy

This one can be solved using any method, so let's use long division.

Apply the idea

We set up and solve this as follows:

A figure showing the partial steps of the polynomial long division for 4 x squared plus 6 x minus 8 divided by 2 x plus 3. Speak to your teacher for more information.

At this point, we are supposed to repeat the steps of long division and divide -8 by the divisor. However, -8 is a constant which has a degree of 0 and cannot be divided by 2x+3 which has a degree of 1. So, we write a zero in the constant place value in the quotient, and continue with the remaining steps.

A figure showing the polynomial long division for 4 x squared plus 6 x minus 8 divided by 2 x plus 3. Speak to your teacher for more information.

The result of the given division is 2x+\dfrac{-8}{2x+3}.

Reflect and check

Show that synthetic can be used, but is more difficult because of leading coefficient

\displaystyle \dfrac{4x^2+6x-8}{2x+3}\div \dfrac{2}{2}	\displaystyle =	\displaystyle \dfrac{\frac{4x^2+6x-8}{2}}{\frac{2x+3}{2}}	Divide by \frac{2}{2}
	\displaystyle =	\displaystyle \dfrac{2x^2+3x-4}{x+\frac{3}{2}}	Evaluate the division

Now, we can use synthetic division with x=-\frac{3}{2} outside the table.

A figure showing a synthetic division with negative 3 halves in the top left corner. Top row has digits 2, 3, negative 4. Middle row has negative 3 under the 3 and 0 under negative 4. Bottom row has 2, 0, and negative 4.

This gives us an answer of 2x+\dfrac{-4}{x+\frac{3}{2}}. Normally, we do not leave fractions in the denominator, so we will rewrite the remainder by multiplying the numerator and denominator by 2:

\displaystyle \dfrac{-4}{x+\frac{3}{2}}\times \dfrac{2}{2}	\displaystyle =	\displaystyle \dfrac{-4\cdot 2}{\left(x+\frac{3}{2}\right)\cdot 2}	Multiply the remainder by \frac{2}{2}
	\displaystyle =	\displaystyle \dfrac{-8}{2x+3}	Evaluate the multiplication

So our final answer is 2x+\dfrac{-8}{2x+3}, just like we found before.

Idea summary

We can divide polynomials using long or synthetic division.

For long division, we first check for any missing terms in the dividend and divisor. If there is a missing term, we add a 0 coefficient in its place. Then, we follow these steps for the division:

Divide the first terms of the dividend and divisor
Multiply the answer by the divisor
Subtract the product from the first part of the dividend
Bring down the next part of the dividend
Repeat steps 1-4 until there are no terms left to bring down

Synthetic division is a short-hand notation of division that can only be used when the divisor is a linear expression in the form x\pm a where a is a constant.

We set up the synthetic divison by writing the coefficients of the divident in an upside down division table, and adding 0 coefficients for any missing terms. Then, we set the divisor equal to zero, solve for x, and write the result outside the table. Next, we follow these steps until the last column has been added:

Multiply the number below the table by the number outside the table
Write the result under the next coefficient from the dividend
Add the numbers in the next column

The Remainder and Factor theorems

Exploration

Divide the following polynomials:

\left(x^3+2x^2-5x-6\right)\div \left(x-1\right)
\left(x^3-7x^2-7x+20\right)\div \left(x+4\right)

Evaluate the following:

f\left(1\right) for f\left(x\right)=\left(x^3+2x^2-5x-6\right)
f\left(-4\right) for f\left(x\right)=\left(x^3-7x^2-7x+20\right)

Explain the relationship between the remainder of the division and the value of the function at the point.

In previous sections, when dividing p\left(x\right) by a linear divisor, x\pm a where a is a constant, we wrote our answers in the form \dfrac{p\left(x\right)}{x-a}=q\left(x\right)+\dfrac{r\left(x\right)}{x-a}

To check the answers we found for the quotient and remainder, we multiplied the quotient by the divisor and added the remainder. That is, p\left(x\right)=q\left(x\right)\left(x-a\right)+r\left(x\right)

Suppose x=a and the remainder is a constant R. If we substitute this into the above equation, we find \begin{aligned}p\left(a\right)&=q\left(a\right)\left(a-a\right)+R\\&=q\left(a\right)\cdot 0+R\\&=R\end{aligned}

This result shows that we can find the remainder of p\left(x\right)\div \left(x-a\right) without performing the division. It is known as the remainder theorem.

Remainder theorem

For a polynomial p\left(x\right) and a number a, the remainder on division by {x - a} is p\left(a\right).

The remainder theorem can be extended to the factor theorem which can help us find factors of a polynomial without having to fully factor the expression.

Factor theorem

For a polynomial p\left(x\right) and a number a, \left(x - a\right) is a factor of p\left(x\right) if and only if p\left(a\right) = 0.

Examples

Example 7

Using the remainder theorem, determine the remainder when p\left(x\right)=2x^3-4x^2+3x-1 is divided by 2x+1.

Worked Solution

Create a strategy

To use the remainder theorem, we need to substitute the value of x that makes 2x+1 equal to 0 into the polynomial.

\begin{aligned}2x+1&=0\\2x&=-1\\x&=-\dfrac{1}{2}\end{aligned}

By the remainder theorem, the answer of p\left(-\dfrac{1}{2}\right) is the remainder when p\left(x\right) is divided by 2x+1.

Apply the idea

\displaystyle p\left(x\right)	\displaystyle =	\displaystyle 2\left(-\dfrac{1}{2}\right)^3-4\left(-\dfrac{1}{2}\right)^2+3\left(-\dfrac{1}{2}\right)-1	Substitute x=-\dfrac{1}{2}
	\displaystyle =	\displaystyle 2\left(-\dfrac{1}{8}\right)-4\left(\dfrac{1}{4}\right)+3\left(-\dfrac{1}{2}\right)-1	Evaluate the exponents
	\displaystyle =	\displaystyle -\dfrac{1}{4}-1-\dfrac{3}{2}-1	Evaluate the multiplication
	\displaystyle =	\displaystyle -\dfrac{1}{4}-\dfrac{4}{4}-\dfrac{6}{4}-\dfrac{4}{4}	Create common denominators
	\displaystyle =	\displaystyle -\dfrac{15}{4}	Evaluate the subtraction

When p\left(x\right) is divided by \left(2x+1\right), the remainder is -\dfrac{15}{4}.

Reflect and check

The second part of the remainder theorem tells us if p\left(a\right)=0, then x-a is a factor. From that, we can also determine that if p\left(a\right)\neq 0, then x-a is not a factor of p\left(x\right).

Applying that to this problem, we can determine that \left(2x+1\right) is not a factor of p\left(x\right). This concept will be explored more in the next lesson.

Example 8

When the polynomials P\left(x\right)=x^4+5x^3-mx+n and Q\left(x\right)=mx^2+nx-1 are each divided by D\left(x\right)=x-1, the remainders are 7 and -6 respectively. Find the values of m and n.

Worked Solution

Create a strategy

By the remainder theorem, P\left(1\right)=7 and Q\left(1\right)=-6. We can use this to create two equations, then solve them for m and n.

Apply the idea

\begin{aligned}P\left(1\right)&=7\\\left(1\right)^4+5\left(1\right)^3-m\left(1\right)+n&=7\\1+5-m+n&=7\\-m+n&=1\end{aligned}

\begin{aligned} Q\left(1\right)&=-6\\m\left(1\right)^2+n\left(1\right)-1&=-6\\m+n&=-5\end{aligned}

The two equations we found are -m+n=1 and m+n=-5. This created a system of equations that we can solve by adding the equations together:

\begin{aligned}-m+n&=1\\ + m+n&=-5 \\ \hline 2n&=-4\\n&=-2 \end{aligned}

We can now substitute n=-2 into one of the equations and solve for m:\begin{aligned}m+\left(-2\right)&=-5\\m&=-3\end{aligned}

Reflect and check

If we describe this problem and the answer in context, we can say:

When x^4+5x^3+3x-2 is divided by x-1, the remainder is 7. And when -3x^2-2x-1 is divided by x-1, the remainder is -6.

Example 9

Use the Factor theorem to determine if the divisor is a factor of the dividend in the expression: \dfrac{2x^3+x^{2}-10x}{x-2}

Worked Solution

Create a strategy

Let p\left(x\right)=2x^3+x^2-10x. By the Factor theorem, if \left(x-2\right) is a factor, then p\left(2\right)=0.

Apply the idea

\displaystyle p\left(2\right)	\displaystyle =	\displaystyle 2\left(2\right)^3+\left(2\right)^2-10\left(2\right)	Substitute x=2
	\displaystyle =	\displaystyle 2\left(8\right)+4-10\left(2\right)	Evaluate the exponents
	\displaystyle =	\displaystyle 16+4-20	Evaluate the multiplication
	\displaystyle =	\displaystyle 0	Evaluate the addition and subtraction

Since p\left(2\right)=0, \left(x-2\right) is a factor of 2x^3+x^2-10x.

Reflect and check

We can use this information to fully factor the expression.

Since the divisor is of the form x - a, we can use synthetic division to find the result of \dfrac{p\left(x\right)}{x-2}.

A figure showing a synthetic division with 2 in the middle left corner. Top row has digits 2, 1, negative 10, and 0. Middle row has 4 under the 1, 10 under the negative 10, and 0 under the 0. Bottom row has 2, 5, 0, and 0.

The result of the synthetic division is 2x^2+5x, so we can rewrite the dividend as p\left(x\right)=\left(x-2\right)\left(2x^2+5x\right) Next, we can factor out the common factor of x from the second set of parentheses. p\left(x\right)=x\left(x-2\right)\left(2x+5\right)

Example 10

Given \left(x+4\right) is a factor of P\left(x\right)=6x^3+31x^2+25x-12, find the remaining factors and rewrite P\left(x\right) as a product of linear factors.

Worked Solution

Create a strategy

Since we already know \left(x+4\right) is a factor of P\left(x\right), we can divide P\left(x\right) by \left(x+4\right) and factor the resulting quadratic expression.

Apply the idea

The divisor is linear, so we can use synthetic division to divide \dfrac{P\left(x\right)}{x+4}.

A figure showing a synthetic division with negative 4 in the upper left corner. Top row has digits 6, 31, 25, and negative 12. Middle row has negative 24 under the 31, negative 28 under the 25, and 12 under the negative 12. Bottom row has 6, 7, negative 3, and 0.

The result of the synthetic division is 6x^2+7x-3. Now, we can factor this trinomal by looking for two values that have a product of ac=\left(6\right)\left(-3\right)=-18 and a sum of b=7.

The factors of -18 are \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18. Among these factors, 9 and -2 are the only numbers that add to 7 and multiply to -18.

We can let r=-2 and s=9. Next, we write 6x^2+7x-3 in the form ax^2+rx+sx+c. Substituting the values for r and s, we get 6x^2-2x+9x-3

Now, we factor this expression by grouping.

\displaystyle 6x^2-2x+9x-3	\displaystyle =	\displaystyle \left(6x^2-2x\right)+\left(9x-3\right)	Group based on common factors
	\displaystyle =	\displaystyle 2x\left(3x-1\right)+3\left(3x-1\right)	Factor out each GCF (2x and 3)
	\displaystyle =	\displaystyle \left(3x-1\right)\left(2x+3\right)	Factor out the common binomial factor

These are the remaining factors of P\left(x\right). Therefore, P\left(x\right)=\left(x+4\right)\left(3x-1\right)\left(2x+3\right).

Idea summary

To find the remainder of \dfrac{p\left(x\right)}{x-a} without performing the division, we can evaluate p\left(a\right). The result will be the value of the remainder.

We can use the Factor theorem to determine if a linear factor \left(x-a\right) is a factor of a polynomial p\left(x\right) by determining if p\left(a\right)=0.

1.11A Equivalent representations of polynomial and rational expressions

Introduction

Ideas

Standard form of polynomials

Examples

Example 1

Example 2

Factored form of polynomials

Examples

Example 3

Polynomial division

Examples

Example 4

Example 5

Example 6

The Remainder and Factor theorems

Exploration

Examples

Example 7

Example 8

Example 9

Example 10

Outcomes

1.11.A

1.11.B

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