question_answer

           

Factorise:

(a) \[{{a}^{4}}-{{b}^{4}}\]

(b) \[{{p}^{4}}-81\]

(c) \[{{x}^{4}}-{{(y+2)}^{4}}\]

(d) \[{{x}^{4}}-{{(x-z)}^{4}}\]

(e) \[{{a}^{4}}-2{{a}^{2}}{{b}^{2}}+{{b}^{4}}\]

Answer:

(a) Using   \[{{a}^{2}}-{{b}^{2}}=(a-b)(a+b)\]

\[{{a}^{4}}-{{b}^{4}}={{({{a}^{2}})}^{2}}-{{({{b}^{2}})}^{2}}\]

\[=({{a}^{2}}+{{b}^{2}})({{a}^{2}}-{{b}^{2}})\]

\[=({{a}^{2}}+{{b}^{2}})(a+b)(a-b)\]

(b)   \[{{p}^{4}}-81={{\left( {{p}^{2}} \right)}^{2}}-{{\left( 9 \right)}^{2}}\]

\[=({{p}^{2}}+9)\left( {{p}^{2}}-9 \right)\] \[[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)]~\]

\[=\left( {{p}^{2}}+9 \right)\left( p-3 \right)\left( p+3 \right)~~\]

(c)   \[{{x}^{4}}-{{(y+2)}^{4}}={{({{x}^{2}})}^{2}}-{{[{{(y+2)}^{2}}]}^{2}}\]

\[=[({{x}^{2}})+{{(y+2)}^{2}}]\,\,[({{x}^{2}})-{{(y+2)}^{2}}]\]

\[=[{{(x)}^{2}}+{{(y+2)}^{2}}][(x-y-z)(x+y+2)]\]

(d)   \[{{x}^{4}}-{{(x-z)}^{4}}={{({{x}^{2}})}^{2}}-{{[{{(x-z)}^{2}}]}^{2}}\]

\[=[{{x}^{2}}-{{(x-z)}^{2}}]\,[{{x}^{2}}+{{(x-z)}^{2}}]\]

\[=\left[ \left( x-x+z \right)\left( x+x-\text{ }z \right) \right]\text{ }\left[ ({{x}^{2}}+{{(x-z)}^{2}} \right]~\]

\[=\text{ }z\left( 2x-z \right)\text{ }\left[ {{x}^{2}}+{{\left( x \right)}^{2}}+{{\left( z \right)}^{2}}-2xz \right]\]

\[=z(2x-z)[2{{x}^{2}}-2xz+{{z}^{2}}]\]

(e)    \[{{a}^{4}}-2{{a}^{2}}{{\text{b}}^{2}}+{{b}^{4}}={{\left( {{a}^{2}} \right)}^{2}}+{{\left( {{b}^{2}} \right)}^{2}}-2\left( {{a}^{2}} \right)\left( {{b}^{2}} \right)\]

\[={{({{a}^{2}}-{{b}^{2}})}^{2}}\]

\[=[({{a}^{2}}-{{b}^{2}})({{a}^{2}}+{{b}^{2}})]\]

\[=[(a-b)(a+b)({{a}^{2}}+{{b}^{2}})]\]