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