Question

A) What is the expansion formula for \[{(a - b)^2}\]
B) Expand \[{(x - 3)^3}\]
C) In rectangle ABCD, If \[l\left( {AB} \right) = 8cm\], \[l\left( {BC} \right) = 6cm\] then find \[l\left( {CD} \right)\& l\left( {AC} \right)\]
D) What are the factors of \[{x^2} - 16\]

Answer 1

Hint: The Binomial Theorem states that, where n is a positive integer,
\[{(x + y)^n} = \sum\limits_{k = 0}^n {n{C_k}{x^{n - k}}{y^k}} = {x^n} + n{C_1}{x^{n - 1}}y + n{C_2}{x^{n - 2}}{y^2} + ......\]
Formula used- $n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$

Complete step-by-step answer:
a) We have to find the expansion formula for \[{(a - b)^2}\]
\[{(a - b)^2} = (a - b)(a - b)\]
Multiply the two factors we have,
\[ = {a^2} - ab - ab + {b^2}\]
Simplifying the above equation,
\[ = {a^2} - 2ab + {b^2}\]
Hence, the expansion formula for \[{(a - b)^2}\]is \[{a^2} - 2ab + {b^2}\].

b) We have to Expand \[{(x - 3)^3}\]
From Binomial theorem we have,
\[{(x + y)^n} = \sum\limits_{k = 0}^n {n{C_k}{x^{n - k}}{y^k}} = {x^n} + n{C_1}{x^{n - 1}}y + n{C_2}{x^{n - 2}}{y^2} + ......\]
Putting y=-3 and apply binomial theorem to get the expansion of \[{(x - 3)^3}\] \[{\{ x + ( - 3)\} ^3} = \sum\limits_{k = 0}^3 {3{C_k}{x^{3 - k}}{{( - 3)}^k}} \]
                      \[ = {x^3} + 3{C_1}{x^{3 - 1}}( - 3) + 3{C_2}{x^{3 - 2}}{( - 3)^2} + 3{C_3}{x^{3 - 3}}{( - 3)^3}\]
Using the formula we mention in hint $n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$,
\[ = {x^3} - \dfrac{{3!}}{{1!(3 - 1)!}}3{x^2} + \dfrac{{3!}}{{2!(3 - 2)!}}9x - \dfrac{{3!}}{{3!}} \times 27\]
Simplifying that,
\[ = {x^3} - \dfrac{{3 \times 2 \times 1}}{{2 \times 1}}3{x^2} + \dfrac{{3 \times 2 \times 1}}{{2 \times 1}}9x - 27\]
Multiply and divide the terms, we get,
\[ = {x^3} - 9{x^2} + 27x - 27\]
Hence expanding \[{(x - 3)^3}\] we get,
\[{x^3} - 9{x^2} + 27x - 27\]

c) In rectangle ABCD, If \[l\left( {AB} \right) = 8cm\], \[l\left( {BC} \right) = 6cm\] then find \[l\left( {CD} \right)\& l\left( {AC} \right)\]
We have to find out, \[l\left( {CD} \right)\& l\left( {AC} \right)\]

For a rectangle the opposite sides are of the same length.
Therefore, \[AB{\text{ }} = {\text{ }}CD,{\text{ }}BC{\text{ }} = {\text{ }}DA\]
The measure of sides,
\[\begin{array}{*{20}{l}}
  {l\left( {AB} \right) = 8cm,{\text{ }}l\left( {BC} \right) = 6cm} \\
  {{\text{So}},{\text{ }}l\left( {AB} \right) = l\left( {CD} \right) = 8cm}
\end{array}\]
Thus, \[l\left( {CD} \right) = 8cm\].
The diagonal is AC, since ABCD is a rectangle so the angles are right angle.
Applying Pythagoras theorem,
\[A{C^2} = A{B^2} + B{C^2}\]
Substituting the values,
\[AC = \sqrt {{8^2} + {6^2}} \]
\[ = \sqrt {64 + 36} \]

   = \sqrt {100} \\
   = 10{\text{ cm}} \\
Then the length of AC is 10cm.
\[l\left( {AC} \right) = 10cm.\]

D) We have to find out the factor of \[{x^2} - 16\]
Applying the formula,
\[{a^2} - {b^2} = (a + b)(a - b)\]
Using the above formula,
\[{x^2} - 16\]\[ = {x^2} - {4^2}\]
              \[ = (x + 4)(x - 4)\]
So the factors are \[\left( {x + 4} \right){\text{ and }}\left( {x - 4} \right)\]

Note: A combination is a grouping or subset of items.
For a combination,
\[C(n,r){ = ^n}{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\]
Where, factorial n is denoted by \[n!\]and defined by
\[n! = n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right) \ldots \ldots .2.1\]