Rational functions can also be rewritten in various forms to expose different characteristics. This can include vertical asymptotes, horizontal asymptotes, or slant asymptotes. Polynomial long division can be used to rewrite a rational function, allowing for the identification of slant asymptotes. Because rational functions are the quotient of two polynomials, the same strategies we used to rewrite polynomials can be used to rewrite rationals.

Examples

Example 1

Find the slant asymptote of h\left(x\right) = \dfrac{3x^3 - x^2 + 2x - 5}{x^2 - 1}.

Worked Solution

Create a strategy

Perform long division to find the quotient. This will be the slant asymptote (ignoring the remainder).

Apply the idea

Perform long division:

Worked out long division for 3x cubed minus x squared plus 2x minus 5 divided by x squared minus 1

We can see that the quotient of what divides evenly is 3x-1. This means the equation of the slant asymptote is y=3x-1.

Idea summary

By rewriting a rational function in different forms, we can identify various characteristics such as vertical and slant asymptotes.

Pascal's triangle

In geometry, we found combinations by using the following formula:

\displaystyle \binom{n}{r}={}_{n}C_{r}=\dfrac{n!}{(n-r)!r!}

\bm{n}

total number of objects

\bm{r}

number of objects being chosen

This formula uses factorials.

Factorial

A product of the first n positive integers. n!=n \cdot \left(n-1\right)\cdot \left(n-2\right)\cdot \ldots \cdot 2 \cdot 1

Example:

4!=4\cdot 3\cdot 2\cdot 1=24

Exploration

Calculate the following values:

{}_4C_0
{}_4C_1
{}_4C_2
{}_4C_3
{}_4C_4

Compare your answers to the triangle below.

A triangle made up of 6 rows of numbers. Starting from the topmost row: first row, 1; second row, 1, 1; third row, 1, 2, 1; fourth row, 1, 3, 3, 1; fifth row, 1, 4, 6, 4, 1; sixth row, 1, 5, 10, 10, 5, 1.

What do you notice?
What do you wonder?

The triangle we will explore today was discovered by French mathematician Blaise Pascal in the 17th century. Although we call it Pascal's triangle, students from other areas of the world know it by a different name.

In India, it is called the Staircase to Mount Meru and attributed to Indian mathematician Halayuda from the 10th century
In Iran, it is called Khayyam's Triangle and attributed to Arabic mathematician Al-Karaji and Omar Khayyam around the year 1000
In China, it is called Yanghui's triangle and attributed to Chinese mathematicians Jia Xian and Yanghui in the 11th century

Back in those days, there was often no means of sharing something published in one country with other countries. Because of this, mathematicians all over the world often discovered concepts around the same time and independent of each other. There are other names for it in other countries, but we call it Pascal's triangle.

To create the triangle, we begin with 1. This is called row 0.

A triangle made up of 2 rows of numbers. Top row, 1; bottom row, 1, 1.

In the next row, we begin and end with 1. This is row 1.

A triangle made up of 3 rows of numbers. Top row, 1; middle row, 1, 1; bottom row, 1, 2, 1. Arrows from both 1s in the middle row are pointing to 2 on the bottom row.

For the next row, we begin and end with 1, but we find the middle number by adding the two numbers above it. This is row 2.

A triangle made up of 5 rows of numbers. Starting from the topmost row: first row, 1; second row, 1, 1; third row, 1, 2, 1; fourth row, 1, 3, 3, 1; fifth row, 1, 4, 6, 4, 1. Speak to your teacher for more details.

This pattern continues to create the triangle. We always begin with 1 and end with 1. To get the number in the middle of the row, add the two numbers above it.

The row numbering begins from 0. The row number is always the same as the second element in the row.

Although there are many interesting mathematical phenomenons found within the triangle, the one we will focus on today is it's relationship to combination. In general, we can find the values of {}_n C_r in row number n (starting at row 0) and the r value is the element in the row, (also starting at 0).

Two triangles with an arrow pointing from the left triangle to the right triangle. The left triangle is made up of 5 rows of numbers. Starting from the topmost row: first row, 1; second row, 1, 1; third row, 1, 2, 1; fourth row, 1, 3, 3, 1; fifth row, 1, 4, 6, 4, 1. The right triangle is made up of 5 rows of combinations. Starting from the topmost row: first row, combination of 0 taken 0; second row, combination of 1 taken 0, combination of 1 taken 1; third row, combination of 2 taken 0, combination of 2 taken 1, combination of 2 taken 2; fourth row, combination of 3 taken 0, combination of 3 taken 1, combination of 3 taken 2, combination of 3 taken 3; fifth row, combination of 4 taken 0, combination of 4 taken 1, combination of 4 taken 2, combination of 4 taken 3, combination of 4 taken 4.

So, the value of {}_4C_2 is found in row 4 (remember to begin counting from 0) and is in position 2 (count from left to right starting from 0). We see {}_4C_2=6 and can confirm this using the combination formula:

\displaystyle {}_4C_2	\displaystyle =	\displaystyle \dfrac{4!}{(4-2)!2!}
	\displaystyle =	\displaystyle \dfrac{4!}{2!2!}
	\displaystyle =	\displaystyle \dfrac{4\cdot3\cdot2\cdot1}{2\cdot1\cdot2\cdot1}
	\displaystyle =	\displaystyle 6

The notation \binom{n}{r} may also be used for combinations.

Examples

Example 2

Find row 10 of Pascal's triangle.

Worked Solution

Create a strategy

In the exploration, we were given the triangle up to row 5. Now, we can continue adding the elements in the previous rows and continue until we reach row 10.

Apply the idea

Row 5 of Pascal's triangle is

The next row will begin and end with 1, and we will add the elements from row 5 to find the middle values.

Two rows of numbers. Top row: 1, 5, 10, 10, 5, 1; bottom row: 1, 6, 15, 20, 15, 6, 1. Arrows are pointing from numbers in the top row to the numbers in the bottom row. Speak to your teacher for more details.

Continue beginning and ending rows with 1 and adding the elements from the previous rows.

Five rows of numbers. Starting from the topmost row: first row, 1, 6, 15, 20, 15, 6, 1; second row, 1, 7, 21, 35, 35, 21, 7, 1; third row, 1, 8, 28, 56, 70, 56, 28, 8, 1; fourth row, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1; fifth row, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1.

The 10th row of Pascal's triangle is 1,10,45,120,210,252,210,120,45,10,1.

Reflect and check

Keep this triangle close by so you can refer to it when calculating combinations when n\leq10.

Example 3

Evaluate each expression using the given approach.

Evaluate \binom{6}{3} using Pascal's triangle.

Worked Solution

Create a strategy

In Pascal's triangle, \binom{n}{r} is the entry in position r in row n. We start counting the rows and entries from 0.

Apply the idea

We have n=6 and r=3. The entries of Pascal's triangle up to row 7 are as follows:

Pascal's triangle made up of 8 rows of numbers. Starting from the topmost row: first row, 1; second row, 1, 1; third row, 1, 2, 1; fourth row, 1, 3, 3, 1; fifth row, 1, 4, 6, 4, 1; sixth row, 1, 5, 10, 10, 5, 1; seventh row, 1, 6, 15, 20, 15, 6, 1; eighth row, 1, 7, 21, 35, 35, 21, 7, 1.

Row 6 is 1,\,6,\, 15,\, 20,\, 15,\, 6,\,1. Since the count for r begins at 0, \binom{6}{3} represents the 4th entry in the 6th row. So, \binom{6}{3}=20.

Reflect and check

We can check our answer by using the combination formula with n=6 and r=3.

\displaystyle \binom{n}{r}	\displaystyle =	\displaystyle \dfrac{n!}{(n-r)!r!}	State the combination formula
\displaystyle \binom{6}{3}	\displaystyle =	\displaystyle \dfrac{6!}{(6-3)!3!}	Substitute n=6 and r=3
	\displaystyle =	\displaystyle \dfrac{6!}{3!3!}	Evaluate the difference
	\displaystyle =	\displaystyle 20	Evaluate the factorials

Evaluate {}_{7}C_{4} using the combination formula.

Worked Solution

Create a strategy

The formula is {}_{n}C_{r}=\dfrac{n!}{(n-r)!r!}.

Apply the idea

We have n=7 and r=4.

\displaystyle {}_{n}C_{r}	\displaystyle =	\displaystyle \dfrac{n!}{(n-r)!r!}	State the combination formula
\displaystyle {}_{7}C_{4}	\displaystyle =	\displaystyle \dfrac{7!}{(7-4)!4!}	Substitute n=7 and r=4
	\displaystyle =	\displaystyle \dfrac{7!}{3!4!}	Evaluate the difference
	\displaystyle =	\displaystyle \dfrac{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{(3\cdot2\cdot1)(4\cdot3\cdot2\cdot1)}	Write out the expansion of the factorials
	\displaystyle =	\displaystyle \dfrac{7\cdot5}{1}	Simplify the common factors
	\displaystyle =	\displaystyle 35	Evaluate the product and quotient

Reflect and check

Finding row 7 of Pascal's triangle would be time consuming. We can also check this using a calculator by using the button marked with {}_{n}C_{r} or \text{nCr} which may require the use of the "shift" button. The calculator should confirm that {}_{7}C_{4}=35.

Idea summary

We can find the value of {}_nC_r or \binom{n}{r} using the combination formula

\displaystyle \binom{n}{r}={}_{n}C_{r}=\dfrac{n!}{(n-r)!r!}

\bm{n}

total number of objects

\bm{r}

number of objects being chosen

or by using Pascal's triangle. In general, we can find the values of {}_n C_r in row number n (starting at row 0) and the r-value is the element in the row, (also starting at 0).

Pascal's triangle made up of 11 rows of numbers. Starting from the topmost row: first row, 1; second row, 1, 1; third row, 1, 2, 1; fourth row, 1, 3, 3, 1; fifth row, 1, 4, 6, 4, 1; sixth row, 1, 5, 10, 10, 5, 1; seventh row, 1, 6, 15, 20, 15, 6, 1; eighth row, 1, 7, 21, 35, 35, 21, 7, 1; ninth row, 1, 8, 28, 56, 70, 56, 28, 8, 1; tenth row, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1; eleventh row, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1.

The binomial theorem

Exploration

Evaluate the following binomials:

(a+b)^0
(a+b)^1
(a+b)^2
(a+b)^3
(a+b)^4
(a+b)^5

What do you notice about the coefficients of each expansion?
What patterns do you notice in the exponents of each expansion?

To expand binomials in the form (a+b)^n, we previously used the distributive property to multiply {(a+b)} to itself n times. However, there are patterns in the expansions, and these patterns can be summarized into the binomial theorem.

In general, the binomial theorem says:

\displaystyle \left(a+b\right)^n=\sum_{r=0}^{n}\binom{n}{r}\left(a\right)^{n-r}\left(b\right)^r

\bm{n}

exponent of the binomial

\bm{a}

first term of the binomial

\bm{b}

second term of the bimomial

\bm{r}

an integer value from 0 to n

The symbol \sum is captial sigma, and we call it a summation. It tells us to begin by substituting the lower limit (r=0 for the binomial theorem) into the expression on the right side. This corresponds to the first term in the expansion. Then, we substitute r=1 for the second term, r=2 for the third term, and continue this pattern until we substitute r=n, the upper limit, for the (n+1)th term. Because it is a sum, we add each of the terms together.

\begin{aligned}\left(a+b\right)^n&=\sum_{r=0}^{n}\binom{n}{r}\left(a\right)^{n-r}\left(b\right)^r\\&=\binom{n}{0}a^nb^0+\binom{n}{1}a^{n-1}b^1+\ldots +\binom{n}{n-1}a^1b^{n-1}+\binom{n}{n}a^0b^n\end{aligned}

For example: \begin{aligned}(a+b)^4&=\sum_{r=0}^4\binom{4}{r}\cdot a^{4-r}\cdot b^r\\&=\binom{4}{0}\cdot a^{4-0} \cdot b^0 +\binom{4}{1}\cdot a^{4-1} \cdot b^1+\binom{4}{2}\cdot a^{4-2} \cdot b^2+\binom{4}{3}\cdot a^{4-3} \cdot b^3 +\binom{4}{4}\cdot a^{4-4} \cdot b^4\\&=a^4+4a^3b+6a^2b^2+4ab^3+b^4\end{aligned}

Looking at the exponents on the variables in an expansion, the exponents of the first term are integers that descend beginning from n and ending at 0. The exponents of the second term are integers that ascend beginning from 0 and ending at n.

Because the coefficients in the binomial theorem are combinations, the elements of Pascal's triangle can be used to evaluate the coefficients as row n will given the coefficients of the terms of \left(a+b\right)^n in order.

Pascal's triangle with each row next to the corresponding left parenthesis a plus b right parenthesis raised to n. The first row is just 1 corresponding to the coefficients of left parenthesis a plus b right parenthesis raised to 0, the second row is 1 1 corresponding to the coefficients of left parenthesis a plus b right parenthesis raised to 1, the third row is 1 2 1 corresponding to the coefficients of left parenthesis a plus b right parenthesis squared, the fourth row is 1 3 3 1 corresponding to the coefficients of left parenthesis a plus b right parenthesis cubed, and so on. The final row is 1 7 21 35 35 21 7 1 which corresponds to the coefficients of left parenthesis a plus b right parenthesis raised to 7.

We can use the binomial theorem to find a specific term in the expansion without having to fully expand the binomial. Each term in the expansion of (a+b)^n is of the form: \binom{n}{r}\cdot a^{n-r}\cdot b^r where r corresponds to the (r+1)th term since we begin couting the terms from 0.

So if we need to find the 10th term, we let r=9.

When fully expanded, each expansion will have (n+1) terms because we begin counting the terms from 0. So if we need to find the second to last term and n=12, there will be 13 total terms and the second to last term will be the 12th term. In this case, r=11.

Examples

Example 4

Use the binomial theorem to expand the following expressions.

(5x-7y)^2

Worked Solution

Create a strategy

Because the exponent is small, we can write out the 3 terms of the general binomial expansion with n=2:

\left(a+b\right)^2=\binom{2}{0}a^{2-0}b^0+\binom{2}{1}a^{2-1}b^1+\binom{2}{2}a^{2-2}b^2

For this problem, a=5x and b=-7y.

Apply the idea

Substituting a=5x and b=-7y gives:

\displaystyle \left(5x-7y\right)^2

\displaystyle =

\displaystyle \binom{2}{0}\left(5x\right)^{2-0}\left(-7y\right)^0+\binom{2}{1}\left(5x\right)^{2-1}\left(-7y\right)^1+\binom{2}{2}\left(5x\right)^{2-2}\left(-7y\right)^2

Row 2 of Pascal's triangle is 1 \text{ }2\text{ }1. We can replace these values with the respective combination values.

=1\left(5x\right)^{2}\left(-7y\right)^0+2\left(5x\right)^{1}\left(-7y\right)^1+1\left(5x\right)^{0}\left(-7y\right)^2

Finally, we evaluate each of the terms. Be sure to apply the power property of exponents, (ab)^m=a^mb^m, to the terms with parenthesis.

\begin{aligned}&=1\left(25x^2\right)\left(1\right)+2\left(5x\right)\left(-7y\right)+1\left(1\right)\left(49y^2\right)\\&=25x^2-70xy+49y^2\end{aligned}

Reflect and check

Previously, we would have used the distributive property to expand this expression.

\begin{aligned}\left(5x-7y\right)^2&=25x^2-35x-35x+49y^2\\&=25x^2-70x+49y^2\end{aligned}

Notice that this is the same answer as we found above. We have also worked with perfect square trinomials to expand binomials and factor them as A^2+2AB+B^2=\left(A+B\right)^2, but this statement follows directly from the binomial theorem.

\displaystyle \left(A+B\right)^2	\displaystyle =	\displaystyle A^2+2AB+B^2	Perfect square trinomial
\displaystyle (A+B)^2	\displaystyle =	\displaystyle \sum_{r=0}^2 \binom{2}{r}A^{2-r}B^r	Binomial theorem
	\displaystyle =	\displaystyle \binom{2}{0}A^{2-0}B^0+\binom{2}{1}A^{2-1}B^1+\binom{2}{2}A^{2-2}B^2	Perform the expansion
	\displaystyle =	\displaystyle A^2+2AB+B^2	Evaluate

\left(p+4\right)^{5}

Worked Solution

Create a strategy

This is a simple binomial, so we can use the patterns described above:

The coefficients of each term are found in row n of Pascal's triangle
The exponents of the first term descend from n to 0 for each term in the expansion
The exponents of the second term ascend from 0 to n for each term in the expansion

For this problem, n=5, the first term is p, and the second term is 4.

Apply the idea

The coefficients will be the values found in row 5 in Pascal's triangle which are: 1, \, 5,\, 10,\, 10,\, 5,\, 1

The exponents for p will begin from 5 and descend to 0; the exponents of 4 will begin from 0 and ascend to 5:

\displaystyle \left(p+4\right)^{5}	\displaystyle =	\displaystyle 1(p^{5})(4^0)+5(p^{4})(4^1)+10(p^{3}) (4^2)+10(p^{2})(4^3) +5(p^{1})(4^4)+1(p^{0}) (4^5)
	\displaystyle =	\displaystyle p^{5} +20p^{4}+160p^{3}+640p^{2}+1280p+1024

Reflect and check

If we had used the full binomial theorem and the combination formula instead of Pascal's triangle for the coefficients, it would have taken much longer to evaluate.

\displaystyle \left(p+4\right)^{5}	\displaystyle =	\displaystyle {}_{5}C_{0}\cdot p^{5-0} \cdot 4^0 +{}_{5}C_{1}\cdot p^{5-1} \cdot 4^1+{}_{5}C_{2}\cdot p^{5-2} \cdot 4^2 \\ +{}_{5}C_{3}\cdot p^{5-3} \cdot 4^3 +{}_{5}C_{4}\cdot p^{5-4} \cdot 4^4+{}_{5}C_{5}\cdot p^{5-5} \cdot 4^5	Perform the expansion
	\displaystyle =	\displaystyle \dfrac{5!}{(5-0)!0!}\cdot p^{5-0} \cdot 4^0 +\dfrac{5!}{(5-1)!1!}\cdot p^{5-1} \cdot 4^1\\+\dfrac{5!}{(5-2)!2!}\cdot p^{5-2} \cdot 4^2 +\dfrac{5!}{(5-3)!3!}\cdot p^{5-3} \cdot 4^3 \\+\dfrac{5!}{(5-4)!4!}\cdot p^{5-4} \cdot 4^4+\dfrac{5!}{(5-5)!5!}\cdot p^{5-5} \cdot 4^5	Rewrite {}_nC_r in factorial form
	\displaystyle =	\displaystyle \dfrac{5!}{5!0!}\cdot p^{5} \cdot 4^0 +\dfrac{5!}{4!1!}\cdot p^{4} \cdot 4^1+\dfrac{5!}{3!2!}\cdot p^{3} \cdot 4^2 \\+\dfrac{5!}{2!3!}\cdot p^{2} \cdot 4^3 +\dfrac{5!}{1!4!}\cdot p^{1} \cdot 4^4+\dfrac{5!}{0!5!}\cdot p^{0} \cdot 4^5	Evaluate the difference
	\displaystyle =	\displaystyle 1\cdot p^{5} \cdot 4^0 +5\cdot p^{4} \cdot 4^1+10\cdot p^{3} \cdot 4^2 \\+10\cdot p^{2} \cdot 4^3 +5\cdot p^{1} \cdot 4^4+1\cdot p^{0} \cdot 4^5	Evaluate the factorials
	\displaystyle =	\displaystyle p^{5} +20p^{4}+160p^{3}+640p^{2}+1280p+1024	Evaluate the exponents and multiplication

Example 5

Find the middle term in the expansion \left( 2x^{3} - 3y^{2}\right)^{8}.

Worked Solution

Create a strategy

Each term in the expansion (a+b)^n is of the form {}_{n}C_{r}\cdot a^{n-r} \cdot b^r.

Since n=8, there will be 9 terms in the expansion because we start counting from 0. The middle term will be the 5th term. The term numbers are in the form (r+1), so we need to use r=4 to find the 5th term.

Apply the idea

The expression \left( 2x^{3} - 3y^{2}\right)^{8} can be rewritten as \left( 2x^{3} +\left(- 3y^{2}\right)\right)^{8} to highlight that the second term is negative.

For (a+b)^n={}_{n}C_{r}\cdot a^{n-r} \cdot b^r, we have:

a=2x^{3}
b=- 3y^{2}
n=8
r=4

\displaystyle {}_{n}C_{r}\cdot a^{n-r} \cdot b^r	\displaystyle =	\displaystyle {}_{8}C_{4}\cdot \left(2x^3\right)^{8-4} \cdot \left(-3y^2\right)^4	Substitution
	\displaystyle =	\displaystyle \dfrac{8!}{(8-4)!4!} \cdot \left(2x^3\right)^{8-4} \cdot \left(-3y^2\right)^4	Rewrite {}_{8}C_{4} in factorial form
	\displaystyle =	\displaystyle \dfrac{8!}{4!4!} \cdot \left(2x^3\right)^{4} \cdot \left(-3y^2\right)^4	Evaluate the difference
	\displaystyle =	\displaystyle 70 \cdot \left(2x^3\right)^{4} \cdot \left(-3y^2\right)^4	Evaluate the factorials
	\displaystyle =	\displaystyle 70 \cdot 16x^{12} \cdot 81y^8	Evaluate the powers
	\displaystyle =	\displaystyle 90\,720x^{12}y^{8}	Evaluate the product

Reflect and check

An alternative solution would be to use Pascal's Triangle to get the row for the power of 8: 1 \quad 8 \quad 28 \quad 56 \quad 70 \quad 56 \quad 28 \quad 8 \quad 1

The middle term will use the 70 as the coefficient, as it is in the middle.

Counting down from 8 for the exponent on the first term of the binomial and counting up from 0 for the exponent on the second term of the binomial, we get that they both have exponent 4, giving us:70 \cdot \left(2x^3\right)^4 \left(-3y^2\right)^4=90\,720x^{12}y^{8}

Example 6

Find the term in the expansion of \left(2a-\dfrac{b}{3}\right)^9 that contains a^6.

Worked Solution

Create a strategy

Each term in the expansion (a+b)^n is of the form \binom{n}{r}\cdot a^{n-r} \cdot b^r where a is the first term of the binomial, b is the second term of the binomial, and n is the power of the binomial. For this binomial, a=2a, b=-\dfrac{b}{3}, and n=9.

Apply the idea

If we subsitute the known values into the general form for each term in the expansion, we get

\binom{n}{r}\cdot a^{n-r} \cdot b^r=\binom{9}{r}(2a)^{9-r}\left(-\dfrac{b}{3}\right)^r

Since we are looking for the term with a^6, we can solve for the value of r. Currently, the exponent of the term with a is 9-r which we need to be equal to 6. That means, r must be 3.

\displaystyle \binom{9}{r}(2a)^{9-r}\left(-\dfrac{b}{3}\right)^r	\displaystyle =	\displaystyle \binom{9}{3}(2a)^{9-3}\left(-\dfrac{b}{3}\right)^3	Substitute r=3
	\displaystyle =	\displaystyle \dfrac{9!}{(9-3)!3!}(2a)^{9-3}\left(-\dfrac{b}{3}\right)^3	Rewrite \binom{9}{3} in factorial form
	\displaystyle =	\displaystyle \dfrac{9!}{6!3!}(2a)^{6}\left(-\dfrac{b}{3}\right)^3	Evalute the difference
	\displaystyle =	\displaystyle 84(2a)^{6}\left(-\dfrac{b}{3}\right)^3	Evaluate the factorials
	\displaystyle =	\displaystyle 84\left(64a^{6}\right)\left(-\dfrac{b^3}{27}\right)	Evalute the exponents
	\displaystyle =	\displaystyle -\dfrac{1792}{9}a^6b^3	Evaluate the coefficient

Reflect and check

This term is the (r+1)th term, which means it is the 4th term in the expansion. In general, we can use the form \binom{n}{r}\cdot a^{n-r} \cdot b^r to help us find a specific term in the expansion without having to fully expand the binomial.

Idea summary

The binomial theorem says:

\displaystyle \left(a+b\right)^n=\sum_{r=0}^{n}\binom{n}{r}\left(a\right)^{n-r}\left(b\right)^r

\bm{n}

exponent of the binomial

\bm{a}

first term of the binomial

\bm{b}

second term of the binomial

\bm{r}

an integer value from 0 to n

Each individual term is of the form\binom{n}{r}(a)^{n-r} (b)^rand we can use this expression to find the (r+1)th term of the expansion.

1.11B Equivalent representations of polynomial and rational expressions

Introduction

Ideas

Equivalent representations of rational functions

Examples

Example 1

Pascal's triangle

Exploration

Examples

Example 2

Example 3

The binomial theorem

Exploration

Examples

Example 4

Example 5

Example 6

Outcomes

1.11.A

1.11.C

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