Question

Factorize the following:
I.\[{a^4} - {b^4}\]
II.\[{p^4} - 81\]
III.\[{x^4} - {\left( {y + z} \right)^4}\]
IV.\[{x^4} - {\left( {x - z} \right)^4}\]
V.\[{a^4} - 2{a^2}{b^2} + {b^4}\]

Answer 1

Hint: Use the difference of the square formula to factorize the expressions.
Factorization is the process of finding the factors of the expressions or the numbers. A number or quantity, when multiplied with another number, produces the given number or expression is known as the factor.
In this question, first, arrange the terms of the given expression with the term with the same variables and use the difference of the square formula to factorize the expressions. One of the arithmetic identity used here to solve the question is \[\left( {{x^2} - {y^2}} \right) = \left( {x + y} \right)\left( {x - y} \right)\]

Complete step-by-step answer:
I.\[{a^4} - {b^4}\]
We can write the expression as
\[{a^4} - {b^4} = {\left( {{a^2}} \right)^2} - {\left( {{b^2}} \right)^2}\]
Now use the difference of the square formula which is given as\[\left( {{x^2} - {y^2}} \right) = \left( {x + y} \right)\left( {x - y} \right)\], so the expression will become
\[
  {a^4} - {b^4} = {\left( {{a^2}} \right)^2} - {\left( {{b^2}} \right)^2} \\
   = \left( {{a^2} - {b^2}} \right)\left( {{a^2} + {b^2}} \right) \\
 \]
In the factor, \[\left( {{a^2} - {b^2}} \right)\]we can again use the difference of the square formula; hence the factor will be
\[
  {a^4} - {b^4} = {\left( {{a^2}} \right)^2} - {\left( {{b^2}} \right)^2} \\
   = \left( {{a^2} - {b^2}} \right)\left( {{a^2} + {b^2}} \right) \\
   = \left( {{a^2} + {b^2}} \right)\left( {a + b} \right)\left( {a - b} \right) \\
 \]

II.\[{p^4} - 81\]
We can write the expression as
\[{p^4} - 81 = {\left( {{p^2}} \right)^2} - {\left( {{3^2}} \right)^2}\]
Now use the difference of the square formula which is given as\[\left( {{x^2} - {y^2}} \right) = \left( {x + y} \right)\left( {x - y} \right)\], so the expression will become
\[
  {p^4} - 81 = {\left( {{p^2}} \right)^2} - {\left( {{3^2}} \right)^2} \\
   = \left( {{p^2} - {3^2}} \right)\left( {{p^2} + {3^2}} \right) \\
 \]
In the factor, \[\left( {{p^2} - {3^2}} \right)\]we can again use the difference of the square formula; hence the factor will be
\[
  {p^4} - 81 = {\left( {{p^2}} \right)^2} - {\left( {{3^2}} \right)^2} \\
   = \left( {{p^2} - {3^2}} \right)\left( {{p^2} + {3^2}} \right) \\
   = \left( {{p^2} + 9} \right)\left( {p + 3} \right)\left( {p - 3} \right) \\
 \]

III.\[{x^4} - {\left( {y + z} \right)^4}\]
We can write the expression as
\[{x^4} - {\left( {y + z} \right)^4} = {\left( {{x^2}} \right)^2} - {\left( {{{\left( {y + z} \right)}^2}} \right)^2}\]
Now use the difference of the square formula which is given as\[\left( {{x^2} - {y^2}} \right) = \left( {x + y} \right)\left( {x - y} \right)\], so the expression will become
\[
  {x^4} - {\left( {y + z} \right)^4} = {\left( {{x^2}} \right)^2} - {\left( {{{\left( {y + z} \right)}^2}} \right)^2} \\
   = \left( {{x^2} + {{\left( {y + z} \right)}^2}} \right)\left( {{x^2} - {{\left( {y + z} \right)}^2}} \right) \\
 \]
In the factor, \[\left( {{x^2} - {{\left( {y + z} \right)}^2}} \right)\] we can again use the difference of the square formula; hence the factor will be
\[
  {x^4} - {\left( {y + z} \right)^4} = {\left( {{x^2}} \right)^2} - {\left( {{{\left( {y + z} \right)}^2}} \right)^2} \\
   = \left( {{x^2} + {{\left( {y + z} \right)}^2}} \right)\left( {{x^2} - {{\left( {y + z} \right)}^2}} \right) \\
   = \left( {{x^2} + {{\left( {y + z} \right)}^2}} \right)\left( {x + \left( {y + z} \right)} \right)\left( {x - \left( {y + z} \right)} \right) \\
 \]

IV.\[{x^4} - {\left( {x - z} \right)^4}\]
We can write the expression as
\[{x^4} - {\left( {x - z} \right)^4} = {\left( {{x^2}} \right)^2} - {\left( {{{\left( {x - z} \right)}^2}} \right)^2}\]
Now use the difference of the square formula which is given as\[\left( {{x^2} - {y^2}} \right) = \left( {x + y} \right)\left( {x - y} \right)\], so the expression will become
\[
  {x^4} - {\left( {x - z} \right)^4} = {\left( {{x^2}} \right)^2} - {\left( {{{\left( {x - z} \right)}^2}} \right)^2} \\
   = \left( {{x^2} + {{\left( {x - z} \right)}^2}} \right)\left( {{x^2} - {{\left( {x - z} \right)}^2}} \right) \\
 \]
In the factor, \[\left( {{x^2} - {{\left( {x - z} \right)}^2}} \right)\] we can again use the difference of the square formula; hence the factor will be
\[
  {x^4} - {\left( {x - z} \right)^4} = {\left( {{x^2}} \right)^2} - {\left( {{{\left( {x - z} \right)}^2}} \right)^2} \\
   = \left( {{x^2} + {{\left( {x - z} \right)}^2}} \right)\left( {{x^2} - {{\left( {x - z} \right)}^2}} \right) \\
   = \left( {{x^2} + {{\left( {x - z} \right)}^2}} \right)\left( {x + \left( {x - z} \right)} \right)\left( {x - \left( {x - z} \right)} \right) \\
   = \left( {{x^2} + {x^2} + {z^2} - 2xz} \right)\left( {2x + z} \right)z \\
   = \left( {2{x^2} + {z^2} - 2xz} \right)\left( {2x + z} \right)z \\
 \]

V.\[{a^4} - 2{a^2}{b^2} + {b^4}\]
We can write the expression as
\[{a^4} - 2{a^2}{b^2} + {b^4} = {\left( {{a^2}} \right)^2} - 2{a^2}{b^2} + {\left( {{b^2}} \right)^2}\]
Now use the square formula\[{\left( {x - y} \right)^2} = {x^2} - 2xy + {y^2}\], where \[{x^2} = {a^2}\]and \[{y^2} = {b^2}\]
\[
  {a^4} - 2{a^2}{b^2} + {b^4} = {\left( {{a^2}} \right)^2} - 2{a^2}{b^2} + {\left( {{b^2}} \right)^2} \\
   = {\left( {{a^2} - {b^2}} \right)^2} \\
 \]
Now use the difference of the square formula which is given as\[\left( {{x^2} - {y^2}} \right) = \left( {x + y} \right)\left( {x - y} \right)\], so the expression will become
\[
  {a^4} - 2{a^2}{b^2} + {b^4} = {\left( {{a^2} - {b^2}} \right)^2} \\
   = \left( {{a^2} - {b^2}} \right)\left( {{a^2} + {b^2}} \right) \\
 \]
Now again, use the difference of the square formula which is given as\[\left( {{x^2} - {y^2}} \right) = \left( {x + y} \right)\left( {x - y} \right)\], so the expression becomes
\[
  {a^4} - 2{a^2}{b^2} + {b^4} = {\left( {{a^2} - {b^2}} \right)^2} \\
   = \left( {{a^2} - {b^2}} \right)\left( {{a^2} + {b^2}} \right) \\
   = \left( {{a^2} + {b^2}} \right)\left( {a + b} \right)\left( {a - b} \right) \\
 \]

Note: Students must be careful while using the arithmetic identity as \[\left( {{x^2} - {y^2}} \right) = \left( {x + y} \right)\left( {x - y} \right)\] as if the expression of x and y gets interchanged then, the factorization will yield a negative sign in front of them.