Question

Factorize the following:
1) ${{x}^{2}}+2x-63$
2) $6{{x}^{2}}-17x+12$
3) $25{{y}^{2}}+110y+121$
4) ${{x}^{2}}+16{{y}^{2}}+64{{z}^{2}}-8xy-64yz+16zx$
5) $25{{x}^{2}}+36{{y}^{2}}+49{{z}^{2}}+60xy-84yz-70zx$
6) $8+64{{x}^{3}}$
7) $8{{x}^{3}}+27{{y}^{3}}+36{{x}^{2}}y+54x{{y}^{2}}$
8) $343{{a}^{3}}-{{b}^{3}}-147{{a}^{2}}b+21a{{b}^{2}}$
9) $3\sqrt{3}{{a}^{3}}+27{{b}^{3}}-{{c}^{3}}+9\sqrt{3}abc$

Answer 1

Hint:
Factorization means breaking or splitting up a number into smaller numbers, which when multiplied together form the same original number. In addition to this, the smaller numbers are known as the factors.

Complete step by step solution:
Now, factoring the given numbers:
1) ${{x}^{2}}+2x-63$
$\begin{align}
  & \Rightarrow {{x}^{2}}+9x-7x-63 \\
 & \Rightarrow x\left( x+9 \right)-7\left( x+9 \right) \\
 & \therefore \left( x-7 \right)\left( x+9 \right) \\
\end{align}$
2) $6{{x}^{2}}-17x+12$
\[\begin{align}
  & \Rightarrow 6{{x}^{2}}-8x-9x+12 \\
 & \Rightarrow 2x\left( 3x-4 \right)-3\left( 3x-4 \right) \\
 & \therefore \left( 2x-3 \right)\left( 3x-4 \right) \\
\end{align}\]
3) $25{{y}^{2}}+110y+121$
$\begin{align}
  & \Rightarrow 25{{y}^{2}}+110y+121 \\
 & \Rightarrow 25{{y}^{2}}+55y+55y+121 \\
 & \Rightarrow 5y\left( 5y+11 \right)+11\left( 5y+11 \right) \\
 & \Rightarrow \left( 5y+11 \right)\left( 5y+11 \right) \\
 & \therefore {{\left( 5y+11 \right)}^{2}} \\
\end{align}$
4) ${{x}^{2}}+16{{y}^{2}}+64{{z}^{2}}-8xy-64yz+16zx$
\[\begin{align}
  & \Rightarrow {{\left( x \right)}^{2}}+{{\left( -4y \right)}^{2}}+{{\left( 8z \right)}^{2}}+2\left( x \right)\left( -4y \right)+2\left( -4y \right)\left( 8z \right)+2\left( 8z \right)\left( x \right) \\
 & \text{we know that, }{{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca \\
 & \Rightarrow {{\left( x-4y+8z \right)}^{2}} \\
\end{align}\]
5) $25{{x}^{2}}+36{{y}^{2}}+49{{z}^{2}}+60xy-84yz-70zx$
$\begin{align}
  & \Rightarrow {{\left( 5x \right)}^{2}}+{{\left( 6y \right)}^{2}}+{{\left( -7z \right)}^{2}}+2\left( 5x \right)\left( 6y \right)+2\left( 6y \right)\left( -7z \right)+2\left( -7z \right)\left( 5x \right) \\
 & \text{We know that, }{{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca \\
 & \Rightarrow {{\left( 5x+6y-7x \right)}^{2}} \\
\end{align}$
6) $8+64{{x}^{3}}$
$\begin{align}
  & \Rightarrow {{\left( 2 \right)}^{3}}+{{\left( 4x \right)}^{3}} \\
 & \text{We know that, }{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right) \\
 & \Rightarrow \left( 2+4x \right)\left( {{\left( 2 \right)}^{2}}-\left( 2 \right)\left( 4x \right)+{{\left( 4x \right)}^{2}} \right) \\
 & \Rightarrow \left( 2+4x \right)\left( 4-8x+16{{x}^{2}} \right) \\
\end{align}$
7) $8{{x}^{3}}+27{{y}^{3}}+36{{x}^{2}}y+54x{{y}^{2}}$
$\begin{align}
  & \Rightarrow {{\left( 2x \right)}^{3}}+{{\left( 3y \right)}^{3}}+3{{\left( 2x \right)}^{2}}\left( 3y \right)+3\left( 2x \right){{\left( 3y \right)}^{2}} \\
 & \text{We know that, }{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{2}}+3{{a}^{2}}b+3a{{b}^{2}} \\
\end{align}$
Thus, by using this identity
$\Rightarrow {{\left( 2x+3y \right)}^{3}}$
8) $343{{a}^{3}}-{{b}^{3}}-147{{a}^{2}}b+21a{{b}^{2}}$
$\begin{align}
  & \Rightarrow {{\left( 7a \right)}^{3}}-{{\left( b \right)}^{3}}-3{{\left( 7a \right)}^{2}}\left( b \right)+3\left( 7a \right){{\left( b \right)}^{2}} \\
 & \text{Using the identity, }{{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}} \\
 & \Rightarrow {{\left( 7a-b \right)}^{3}} \\
\end{align}$
9) $3\sqrt{3}{{a}^{3}}+27{{b}^{3}}-{{c}^{3}}+9\sqrt{3}abc$
$\begin{align}
  & \Rightarrow {{\left( \sqrt{3}a \right)}^{3}}+{{\left( 3b \right)}^{3}}+{{\left( -c \right)}^{3}}-3\left( \sqrt{3}a \right)\left( 3b \right)\left( -c \right) \\
 & \text{Now, using the identity } \\
 & \Rightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right) \\
 & \text{Applying the identity,} \\
 & \Rightarrow \left( \sqrt{3}a+3b-c \right)\left( {{\left( \sqrt{3}a \right)}^{2}}+{{\left( 3b \right)}^{2}}+{{\left( -c \right)}^{2}}-\left( \sqrt{3}a \right)\left( 3b \right)-\left( 3b \right)\left( -c \right)-\left( -c \right)\left( \sqrt{3}a \right) \right) \\
 & \Rightarrow \left( \sqrt{3}a+3b-c \right)\left( 3{{a}^{2}}+9{{b}^{2}}+{{c}^{2}}-3\sqrt{3}ab+3bc+\sqrt{3}ca \right) \\
\end{align}$

Note:
Various algebraic identities are also used, in factoring the given number. Thus, knowledge about various algebraic identities is also required while factoring a number. A student must also pay attention while putting the mathematical signs in the equations.