Precalculus by Richard Wright
He heals the brokenhearted and binds up their wounds. Psalms 147:3 NIV
Summary: In this section, you will:
SDA NAD Content Standards (2018): PC.6.4, PC.7.2
A binomial expression is two terms such as x + y. You already know how to work with binomials including how to calculate exponents of binomials by multiplying the binomial by itself. However, calculating higher exponents of binomials can be long and tedious. The binomial theorem is a shortcut to expand exponents of binomials.
The first 6 powers of (x + y)n are given in the triangle below. The plus signs + between the terms have been removed to simplify the diagram. The coefficients make a triangle called Pascal's Triangle. Blaise Pascal wrote a treatise on the triangle in 1654. However, it was first known in Persia by Al-Karaji and China by Yang Hui around 1000 AD. It is created by adding the two upper numbers to get the one below. Such as the 3 + 3 in the n = 3 row give the 6 in the n = 4 row.
$$ \begin{array}{lcr} (x + y)^0 & 1 & n = 0 \\ (x + y)^1 & 1x \quad 1y & n = 1 \\ (x + y)^2 & 1x^2 \quad 2xy \quad 1y^2 & n = 2 \\ (x + y)^3 & 1x^3 \quad 3x^2y \quad 3xy^2 \quad 1y^3 & n = 3 \\ (x + y)^4 & 1x^4 \quad 4x^3y \quad 6x^2y^2 \quad 4xy^3 \quad 1y^4 & n = 4 \\ (x + y)^5 & 1x^5 \quad 5x^4y \quad 10 x^3y^2 \quad 10 x^2y^3 \quad 5xy^4 \quad 1y^5 & n = 5 \\ & \nearrow \qquad \nearrow \qquad \nearrow \quad \qquad \nearrow \qquad \nearrow \qquad \nearrow \quad & \\ & r=0 \quad r=1 \quad r=2 \quad r=3 \quad r=4 \quad r=5 \quad & \end{array} $$
Notice there are some patterns in the triangle.
The binomial coefficients, or combinations, are calculated by \(_nC_r = \frac{n!}{(n-r)!r!}\). This can also be represented \(\left(\begin{matrix} n \\ r \end{matrix}\right)\). Most calculators have a combination function built in.
$$ _nC_r = \frac{n!}{(n-r)!r!} $$
Or another notation.
$$ \left(\begin{matrix} n \\ r \end{matrix}\right) = \frac{n!}{(n-r)!r!} $$
Entered as
n nCr r
Entered as
\(\left(\begin{matrix}n \\ r\end{matrix}\right)\)
Evaluate \(_6C_3\).
You can either use the formula or the button on the calculator. First the formula. \(n = 6, r = 3\)
$$ _nC_r = \frac{n!}{(n-r)!r!} $$
$$ _6C_3 = \frac{6!}{(6-3)!3!} = 20 $$
You might rather use the calculator function.
On a TI-84, type
6 nCr 3 ENTER
On a NumWorks, input
$$ \left(\begin{matrix} 6 \\ 3 \end{matrix}\right) $$
The output is 20.
Evaluate (a) \(\left(\begin{matrix} 12 \\ 7 \end{matrix}\right)\) and (b) \(\left(\begin{matrix} 5 \\ 2 \end{matrix}\right)\)
Evaluate (a) \(_{13}C_{10}\) and (b) \(\left(\begin{matrix} 8 \\ 1 \end{matrix}\right)\)
286; 8
The binomial theorem is a shortcut for expanding powers of binomials.
$$ (a + b)^n = \sum_{r=0}^n \ _nC_r a^{n-r} b^r $$
Expand \((x + 3)^4\).
Compare \((x + 3)^4\) to \((a + b)^n\) to see that \(a = x, b = 3, n = 4\). Fill in the binomial theorem with r starting at 0.
$$ \sum_{r=0}^n \ _nC_r a^{n-r} b^r $$
$$ \color{blue}{\ _4C_0 x^{4-0} 3^0} + \color{purple}{\ _4C_1 x^{4-1} 3^1} + \color{red}{\ _4C_2 x^{4-2} 3^2} + \color{orange}{\ _4C_3 x^{4-3} 3^3} + \color{green}{\ _4C_4 x^{4-4} 3^4} $$
$$ \color{blue}{1·x^4·1} + \color{purple}{4·x^3·3} + \color{red}{6·x^2·9} + \color{orange}{4·x^1·27} + \color{green}{1·x^0·81} $$
$$ \color{blue}{x^4} + \color{purple}{12x^3} + \color{red}{54x^2} + \color{orange}{108x} + \color{green}{81} $$
Expand \((2 - x^3)^5\).
Compare \((2 - x^3)^5\) to \((a + b)^n\) to see that \(a = 2, b = -x^3, n = 5\). Fill in the binomial theorem with r starting at 0.
$$ \sum_{r=0}^n \ _nC_r a^{n-r} b^r $$
$$ \color{blue}{\ _5C_0 2^{5-0} (-x^3)^0} + \color{purple}{\ _5C_1 2^{5-1} (-x^3)^1} + \color{red}{\ _5C_2 2^{5-2} (-x^3)^2} + \color{orange}{\ _5C_3 2^{5-3} (-x^3)^3} + \color{green}{\ _5C_4 2^{5-4} (-x^3)^4} + \color{brown}{\ _5C_5 2^{5-5} (-x^3)^5} $$
$$ \color{blue}{1·32·1} + \color{purple}{5·16·(-x^3)} + \color{red}{10·8·(x^6)} + \color{orange}{10·4·(-x^9)} + \color{green}{5·2·(x^{12})} + \color{brown}{1·1·(-x^{15})} $$
$$ \color{blue}{32} \color{purple}{- 80x^3} + \color{red}{80x^6} \color{orange}{- 40x^9} + \color{green}{10x^{12}} \color{brown}{- x^{15}} $$
Expand \((2x + 3)^3\).
\(8x^3 + 36x^2 + 54x + 27\)
Find the coefficient of the term \(x^5y^3\) in the expansion of \((3x - 2y)^8\).
Use the binomial theorem to find the coefficients. \(a = 3x, b = -2y, n = 8\) Because only one coefficient is needed, only one value of r is required. The binomial theorem for each term is
$$ _nC_r a^{n-r}\color{blue}{b^r} $$
Compare this to the desired term \(x^5\color{blue}y^3\), the exponent of the y should be the same as the exponent of the b part of the formula. Thus \(r=3\) and we only have to work through the one term.
$$ _8C_3 (3x)^{8-3} (-2y)^3 $$
$$ 56 (243x^5) (-8y^3) $$
$$-108864x^5y^3 $$
The coefficient of the \(x^5y^3\) is –108864.
Find the coefficient of the term \(x^4y^6\) in the expansion of \((x - 5y)^{10}\).
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$$ _nC_r = \frac{n!}{(n-r)!r!} $$
Or another notation.
$$ \left(\begin{matrix} n \\ r \end{matrix}\right) = \frac{n!}{(n-r)!r!} $$
Entered as
n nCr r
Entered as
\(\left(\begin{matrix}n \\ r\end{matrix}\right)\)
$$ (a + b)^n = \sum_{r=0}^n \ _nC_r a^{n-r} b^r $$
Helpful videos about this lesson.
Evaluate the combination.
Expand the binomial.
Find the specific coefficient of the binomial expansion.
Expand and simplify the difference quotient \(\displaystyle \frac{f(x+h)-f(x)}{h}\).
Mixed Review